# Soluções Halliday 5ª edição volume 3

Instructor Solutions Manual for Physics by Halliday, Resnick, and Krane

The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or I have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach There are some traditional formula, such as x 0x

which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer I adopt a di erent approach for rounding of signi cant gures than previous authors; in partic- ular, I usually round intermediate answers. As such, some of my answers will di er from those in Exercises and Problems which are enclosed in a box also appear in the Student’s Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications of the answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Student’s Solution Manual for these exercises and problems. However, the material from the Student’s Solution Manual must not be copied.

E25-1 The charge transferred is Q = (2:5 104 C=s)(20 106 s) = 5:0 101 C:

E25-2 Use Eq. 25-4: s (8:99109Nm2=C2)(26:3106C)(47:1106C) r = = 1:40 m (5:66 N)

E25-3 Use Eq. 25-4: (8:99109Nm2=C2)(3:12106C)(1:48106C) F = = 2:74 N: (0:123 m)2

E25-4 (a) The forces are equal, so m1a1 = m2a2, or m2 = (6:31107kg)(7:22 m=s2)=(9:16 m=s2) = 4:97107kg: (b) Use Eq. 25-4: s (6:31107kg)(7:22 m=s2)(3:20103m)2 q = = 7:201011C (8:99109Nm2=C2) E25-5 (a) Use Eq. 25-4, 1 q1q2 1 (21:3 C)(21:3 C) F = = = 1:77 N 40 r2 4(8:851012 C2=N m2) (1:52 m)2 12

(b) In part (a) we found F12; to solve part (b) we need to rst nd F13. Since q3 = q2 and We must assess the direction of the force of q3 on q1; it will be directed along the line which connects the two charges, and will be directed away from q3. The diagram below shows the directions.

F 23 F 12 F 23 F net F 12 From this diagram we want to nd the magnitude of the net force on q1. The cosine law is appropriate here: net 12 13 12 13 = 9:40 N2;

E25-6 Originally F0 = CQ2 = 0:088 N, where C is a constant. When sphere 3 touches 1 the 0 charge on both becomes Q0=2. When sphere 3 the touches sphere 2 the charge on each becomes (Q0 + Q0=2)=2 = 3Q0=4. The force between sphere 1 and 2 is then F = C(Q0=2)(3Q0=4) = (3=8)CQ2 = (3=8)F0 = 0:033 N: 0

The forces on q3 are F~ 31 and F~ 32. These forces are given by the vector form of Coulomb’s Law, Eq. 25-5, E25-7

1 q3q1 1 q3q1 40 r2 40 (2d)2 31 1 q3q2 1 q3q2 F~ 32 = ^r32 = ^r32: 40 r2 40 (d)2 32

These two forces are the only forces which act on q3, so in order to have q3 in equilibrium the forces must be equal in magnitude, but opposite in direction. In short, 1 q3q1 1 q3q2 40 (2d)2 40 (d)2 q1 q2 ^r31 = ^r32: 41 Note that ^r31 and ^r32 both point in the same direction and are both of unit length. We then get

q1 = 4q2: E25-8 The horizontal and vertical contributions from the upper left charge and lower right charge are straightforward to nd. The contributions from the upper left charge require slightly more work. pp The diagonal distance is 2a; the components will be weighted by cos 45 = 2=2. The diagonal charge will contribute pp 1 (q)(2q) 2 2 q2 40 ( 2a)2 2 80 a2 pp 1 (q)(2q) 2 2 q2 Fy = p ^j = ^j: 40 ( 2a)2 2 80 a2 (a) The horizontal component of the net force is then p 1 (2q)(2q) 2 q2 4 a2 8 a2 0p 0 4 + 2=2 q2 40 a2 = (4:707)(8:99109N m2=C2)(1:13106C)2=(0:152 m)2^i = 2:34 N^i: (b) The vertical component of the net force is then p 1 (q)(2q) 2 q2 40 a2 80 a 2 p 2 + 2=2 q2 80 a2 = (1:293)(8:99109Nm2=C2)(1:13106C)2=(0:152m)2^j=0:642N^j:

E25-9 The magnitude of the force on the negative charge from each positive charge is F = (8:99109N m2=C2)(4:18106C)(6:36106C)=(0:13 m)2 = 14:1 N: The force from each positive charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30) = 1:73. The net force is then 24:5 N.

E25-10 Let the charge on one sphere be q, then the charge on the other sphere is Q = (52:6 106C) q. Then 1 qQ 40 r2 (8:99109Nm2=C2)q(52:6106C q) = (1:19 N)(1:94 m)2:

E25-11 This problem is similar to Ex. 25-7. There are some additional issues, however. It is easy enough to write expressions for the forces on the third charge

qq 40 r2 31 1 q3q2 F~ 32 = ^r32: 40 r2 32 Then 1 q3q1 1 q3q2 40 r2 40 r2 31 32 q1 q2 ^r31 = 2 ^r32: r2 r 31 32

The only way to satisfy the vector nature of the above expression is to have ^r31 = ^r32; this means that q3 must be collinear with q1 and q2. q3 could be between q1 and q2, or it could be on either side. Let’s resolve this issue now by putting the values for q1 and q2 into the expression:

(1:07 C) (3:28 C) r2 r2 31 32 r2 ^r31 = (3:07)r2 ^r32: 32 31

Since squared quantities are positive, we can only get this to work if ^r31 = ^r32, so q3 is not between q1 and q2. We are then left with 32 31 so that q3 is closer to q1 than it is to q2. Then r32 = r31 + r12 = r31 + 0:618 m, and if we take the square root of both sides of the above expression, p p (0:618 m) = (3:07)r31 r31;

E25-12 The magnitude of the magnetic force between any two charges is kq2=a2, where a = 0:153 m. The force between each charge is directed along the side of the triangle; but from symmetry only the component along the bisector is of interest. This means that we need to weight the above answer by a factor of 2 cos(30) = 1:73. The net force on any charge is then 1:73kq2=a2. The distance from any charge to the center of the equilateral triangle is x, given by x2 = (a=2)2 + (d x)2. Then x = a2=8d + d=2 = 0:644a: The angle between the strings and the plane of the charges is , given by sin = x=(1:17 m) = (0:644)(0:153 m)=(1:17 m) = 0:0842;

The force of gravity on each ball is directed vertically and the electric force is directed horizontally. The two must then be related by tan = F E=F G;

so or q = 1:29107C: On any corner charge there are seven forces; one from each of the other seven charges. The net force will be the sum. Since all eight charges are the same all of the forces will be repulsive. We need to sketch a diagram to show how the charges are labeled.

E25-13 2 6 8 4 7 5 The magnitude of the force of charge 2 on charge 1 is 1 q2 40 r2 12 1

3 where r12 = a, the length of a side. Since both charges are the same we wrote q2. By symmetry we expect that the magnitudes of F12, F13, and F14 will all be the same and they will all be at right angles to each other directed along the edges of the cube. Written in terms of vectors the forces

would be 1 q2 40 a2 1 q2 40 a2 1 q2 F~ 14 = k^: 40 a2 The force from charge 5 is 1 q2 15 4 r2 0 15 and is directed along the side diagonal away from charge 5. The distance r15 is also the side diagonal distance, and can be found from 15 then 1 q2 F15 = : 40 2a2 By symmetry we expect that the magnitudes of F15, F16, and F17 will all be the same and they will all be directed along the diagonals of the faces of the cube. In terms of components we would have 1 q2 p p 40 2a2 1 q2 p p 40 2a2 1 q2 p p F~ 17 = ^i= 2 + ^j= 2 : 40 2a2 The last force is the force from charge 8 on charge 1, and is given by 1 q2 40 r2 18 and is directed along the cube diagonal away from charge 8. The distance r18 is also the cube diagonal distance, and can be found from 18 then in term of components 1 q2 p p p F~ 18 = ^i= 3 + ^j= 3 + k^= 3 : 40 3a2 We can add the components together. By symmetry we expect the same answer for each com- ponents, so we’ll just do one. How about ^i. This component has contributions from charge 2, 6, 7, and 8: 1 q2 121 40 a2 1 2 2 3 3 or 1 q2 (1:90) 40 a2 p The three components add according to Pythagoras to pick up a nal factor of 3, so q2 F net = (0:262) : 0a2

E25-14 (a) Yes. Changing the sign of y will change the sign of Fy; since this is equivalent to putting the charge q0 on the \other” side, we would expect the force to also push in the \other” (b) The equation should look Eq. 25-15, except all y’s should be replaced by x’s. Then 1 q0 q Fx = p : 40 x x2 + L2=4 (c) Setting the particle a distance d away should give a force with the same magnitude as 1 q0 q F= p : 40 d d2 + L2=4 This force is directed along the 45 line, so Fx = F cos 45 and Fy = F sin 45. p (d) Let the distance be d = x2 + y2, and then use the fact that Fx=F = cos = x=d. Then x 1 x q0 q Fx = F = : d 40 (x2 + y2 + L2=4)3=2 and y 1 y q0 q Fy = F = : d 40 (x2 + y2 + L2=4)3=2 E25-15 (a) The equation is valid for both positive and negative z, so in vector form it would read 1 q0 q z F~ = Fzk^ = k^: 40 (z2 + R2)3=2 (b) The equation is not valid for both positive and negative z. Reversing the sign of z should p reverse the sign of Fz, and one way to x this is to write 1 = z= z2. Then

1 2q0 qz 1 1 F~ = Fzk^ = p p k^: 40 R2 z2 z2

E25-16 Divide the rod into small di erential lengths dr, each with charge dQ = (Q=L)dr. Each di erential length contributes a di erential force 1 q dQ 1 qQ dF = = dr: 40 r2 40 r2L Integrate: Z Z x+L 1 qQ x 40 r2L 1 qQ 1 1 = 40 L x x + L E25-17 You must solve Ex. 16 before solving this problem! q0 refers to the charge that had been called q in that problem. In either case the distance from q0 will be the same regardless of the sign of q; if q = Q then q will be on the right, while if q = Q then q will be on the left. Setting the forces equal to each other one gets

1 qQ 1 1 1 qQ 40 L x x + L 40 r2 or p r = x(x + L):

E25-18 You must solve Ex. 16 and Ex. 17 before solving this problem. If all charges are positive then moving q0 o axis will result in a net force away from the axis. If q = Q then both q and Q are on the same side of q0. Moving q0 closer to q will result in the attractive force growing faster than the repulsive force, so q0 will move away from equilibrium.

E25-19 We can start with the work that was done for us on Page 577, except since we are concerned with sin = z=r we would have 1 q0 dz z dFx = dF sin = : 2 2p 40 (y + z ) y2 + z2 We will need to take into consideration that changes sign for the two halves of the rod. Then ! q0 Z 0 z dz L=2 Z +z dz Fx = + ; 40 L=2 (y2 + z2)3=2 0 (y2 + z )3=2 2

Z L=2 q0 z dz 20 0 (y2 + z2)3=2 L=2 q0 1 20 y2 + z2 0 ! q0 1 1 = p : 20 y y2 + (L=2)2

E25-20 Use Eq. 25-15 to nd the magnitude of the force from any one rod, but write it as 1 qQ 4 r r2 + L2=4 0

where r2 = z2 + L2=4. The component of this along the z axis is Fz = F z=r. Since there are 4 rods, we have 1 qQz 1 qQz 0 r r + L =4 0 (z + L =4) z + L =2 2222222

Equating the electric force with the force of gravity and solving for Q, 0mg p qz putting in the numbers, (8:851012C2=Nm2)(3:46107kg)(9:8m=s2) p ((0:214m)2+(0:25m)2=4) (0:214m)2+(0:25m)2=2 (2:451012C)(0:214 m) so Q = 3:07106C:

In each case we conserve charge by making sure that the total number of protons is the same on both sides of the expression. We also need to conserve the number of neutrons. (a) Hydrogen has one proton, Beryllium has four, so X must have ve protons. Then X must be (b) Carbon has six protons, Hydrogen has one, so X must have seven. Then X is Nitrogen, N. (c) Nitrogen has seven protons, Hydrogen has one, but Helium has two, so X has 7 + 1 2 = 6 protons. This means X is Carbon, C.

E25-22 (a) Use Eq. 25-4: (8:99109Nm2=C2)(2)(90)(1:601019C)2 F = = 290 N: (121015m)2 (b) a = (290 N)=(4)(1:661027kg) = 4:41028m=s2.

E25-23 Use Eq. 25-4: (8:99109Nm2=C2)(1:601019C)2 F = = 2:89109N: (2821012m)2 E25-24 (a) Use Eq. 25-4: s (3:7109N)(5:01010m)2 q = = 3:201019C: (8:99109Nm2=C2) (b) N = (3:201019C)=(1:601019C) = 2.

E25-25 Use Eq. 25-4, 1 q1q2 ( 1 1:6 1019 C)( 1 1:6 1019 C) F = = 3 3 = 3:8 N: 40 r2 4(8:85 1012 C2=N m2)(2:6 1015 m)2 12

(b) The penny has enough electrons to make a total charge of 1:37105C. The fraction is then (1:15107C)=(1:37105C) = 8:401013:

E25-27 Equate the magnitudes of the forces: 1 q2 40 r2 so s (8:99109Nm2=C2)(1:601019C)2 r = = 5:07 m (9:111031kg)(9:81 m=s2)

E25-29 The mass of water is (250 cm3)(1:00 g/cm3) = 250 g. The number of moles of water is (250 g)=(18:0 g/mol) = 13:9 mol. The number of water molecules is (13:9 mol)(6:021023mol1) = 8:371024. Each molecule has ten protons, so the total positive charge is Q = (8:371024)(10)(1:601019C) = 1:34107C:

E25-30 The total positive charge in 0:250 kg of water is 1:34107C. Mary’s imbalance is then q1 = (52:0)(4)(1:34107C)(0:0001) = 2:79105C;

while John’s imbalance is The electrostatic force of attraction is then 1 q1q2 (2:79105)(4:86105) F = = (8:99109N m2=C2) = 1:61018N: 40 r2 (28:0 m) 2

(a) The gravitational force of attraction between the Moon and the Earth is GMEMM R2 where R is the distance between them. If both the Earth and the moon are provided a charge q, then the electrostatic repulsion would be E25-31

1 q2 FE = : 40 R2 Setting these two expression equal to each other, q2 40

which has solution p q = 5:71 1013 C: (b) We need (5:71 1013 C)=(1:60 1019 C) = 3:57 1032 protons on each body. The mass of protons needed is then (3:57 1032)(1:67 1027 kg) = 5:97 1065 kg: Ignoring the mass of the electron (why not?) we can assume that hydrogen is all protons, so we need that much hydrogen.

Assume that the spheres initially have charges q1 and q2. The force of attraction between them is 1 q1q2 40 r2 12 where r12 = 0:500 m. The net charge is q1 + q2, and after the conducting wire is connected each sphere will get half of the total. The spheres will have the same charge, and repel with a force of 1 (q + q ) 1 (q + q ) 1212212 F2 = = 0:0360 N: 40 r2 12

Since we know the separation of the spheres we can nd q1 + q2 quickly, P25-1

q q1 + q2 = 2 40r2 (0:0360 N) = 2:00 C 12 1 (2:00 C q2)q2 40 r2 12 22 0 = q2 + (2:00 C)q2 + (1:73 C) : 2 2

P25-2 The electrostatic force on Q from each q has magnitude qQ=40a2, where a is the length of thepside of the square. The magnitude of the vertical (horizontal) component of the force of Q on 0 (a) In order to have a zero net force on Q the magnitudes of the two contributions must balance, so p 2Q2 qQ 16 a2 4 a2 00 p (b) No.

P25-3 (a) The third charge, q3, will be between the rst two. The net force on the third charge will be zero if 1 q q3 1 4q q3 40 r2 40 r2 31 32 which will occur if 12 = r31 r32 The total distance is L, so r31 + r32 = L, or r31 = L=3 and r32 = 2L=3. Now that we have found the position of the third charge we need to nd the magnitude. The second and third charges both exert a force on the rst charge; we want this net force on the rst charge to be zero, so 1 q q3 1 q 4q 4 r2 4 r2 0 13 0 12 or q3 4q (L=3)2 L2 which has solution q3 = 4q=9. The negative sign is because the force between the rst and second charge must be in the opposite direction to the force between the rst and third charge. (b) Consider what happens to the net force on the middle charge if is is displaced a small distance z. If the charge 3 is moved toward charge 1 then the force of attraction with charge 1 will increase. But moving charge 3 closer to charge 1 means moving charge 3 away from charge 2, so the force of attraction between charge 3 and charge 2 will decrease. So charge 3 experiences more attraction to ward the charge that it moves toward, and less attraction to the charge it moves away from. Sounds unstable to me.

P25-4 (a) The electrostatic force on the charge on the right has magnitude q2 40×2 The weight of the ball is W = mg, and the two forces are related by F=W=tansin=x=2L: Combining, 2Lq2 = 40mgx3, so 2 1=3 qL x= : 20 (b) Rearrange and solve for q, s 2(8:851012C2=N m2)(0:0112 kg)(9:81 m=s2)(4:70102m)3 q = = 2:28108C: (1:22 m)

P25-5 (a) Originally the balls would not repel, so they would move together and touch; after touching the balls would \split” the charge ending up with q=2 each. They would then repel again. (b) The new equilibrium separation is (q=2)2 1=3 1=3 L1 x0 = = x = 2:96 cm: 20mg 4 P25-6 Take the time derivative of the expression in Problem 25-4. Then dx 2 x dq 2 (4:70102m) = = (1:20109C=s) = 1:65103m=s: dt 3 q dt 3 (2:28108C) P25-7 The force between the two charges is 1 (Q q)q F= : 40 r2 12

We want to maximize this force with respect to variation in q, this means nding dF =dq and setting it equal to 0. Then dF d 1 (Q q)q 1 Q 2q ==: dq dq 4 r2 4 r2 0 12 0 12 2

P25-8 Displace the charge q a distance y. The net restoring force on q will be approximately qQ 1 y qQ 16 F 2 = y: 40 (d=2)2 (d=2) 40 d3

Since F =y is e ectively a force constant, the period of oscillation is r 3 3 1=2 m 0m d T = 2 = : k qQ P25-9 Displace the charge q a distance x toward one of the positive charges Q. The net restoring force on q will be

qQ 1 1 40 (d=2 x)2 (d=2 + x)2 qQ 32 x: 40 d3

Since F =x is e ectively a force constant, the period of oscillation is r 3 3 1=2 m 0m d T = 2 = : k 2qQ

(b) Removing a positive Cesium ion is equivalent to adding a singly charged negative ion at that same location. The net force is then 0

where r is the distance between the Chloride ion and the newly placed negative ion, or p r = 3(0:20109m)2 The force is then (1:61019C)2 F = = 1:92109N: 4(8:851012C2=N m2)3(0:20109m)2

We can pretend that this problem is in a single plane containing all three charges. The magnitude of the force on the test charge q0 from the charge q on the left is 1 q q0 Fl = : 4 (a2 + R2) 0 P25-11

A force of identical magnitude exists from the charge on the right. we need to add these two forces as vectors. Only the components along R will survive, and each force will contribute an amount

R R2 + a2 so the net force on the test particle will be 2 q q0 R p: 40 (a2 + R2) R2 + a2

We want to nd the maximum value as a function of R. This means take the derivative, and set it equal to zero. The derivative is 2 2q q0 1 3R 40 (a2 + R2)3=2 (a2 + R2)5=2 which will vanish when a2 + R2 = 3R2;

E26-1 E = F=q = ma=q. Then E = (9:111031kg)(1:84109m=s2)=(1:601019C) = 1:05102N=C:

E26-2 The answers to (a) and (b) are the same! E26-3 F = W , or Eq = mg, so mg (6:64 1027 kg)(9:81 m=s2) E = = = 2:03 107 N=C: q 2(1:60 1019 C) The alpha particle has a positive charge, this means that it will experience an electric force which is in the same direction as the electric eld. Since the gravitational force is down, the electric force, and consequently the electric eld, must be directed up.

(b) F = Eq = (1:5103N=C)(1:601019C) = 2:41016N: (c) F = mg = (1:671027kg)(9:81 m=s2) = 1:61026N: (d) (2:41016N)=(1:61026N) = 1:51010:

E26-5 Rearrange E = q=40r2, q = 4(8:851012C2=N m2)(0:750 m)2(2:30 N=C) = 1:441010C:

E26-7 Use Eq. 26-12 for points along the perpendicular bisector. Then 1 p (3:56 1029 C m) E = = (8:99 109N m2=C2) = 1:95 104N=C: 4 x3 (25:4 109 m)3 0

E26-8 If the charges on the line x = a where +q and q instead of +2q and 2q then at the center of the square E = 0 by symmetry. This simpli es the problem into nding E for a charge +q at (a; 0) and q at (a; a). This is a dipole, and the eld is given by Eq. 26-11. For this exercise we have x = a=2 and d = a, so 1 qa 40 [2(a=2)2]3=2 or, putting in the numbers, E = 1:11105N=C.

E26-9 The charges at 1 and 7 are opposite and can be e ectively replaced with a single charge of 6q at 7. The same is true for 2 and 8, 3 and 9, on up to 6 and 12. By symmetry we expect the eld to point along a line so that three charges are above and three below. That would mean 9:30.

E26-10 If both charges are positive then Eq. 26-10 would read E = 2E+ sin , and Eq. 26-11 would look like 1qx 40 x2 + (d=2)2 x2 + (d=2)2 p

Treat the two charges on the left as one dipole and treat the two charges on the right as a second dipole. Point P is on the perpendicular bisector of both dipoles, so we can use Eq. 26-12 For the dipole on the left p = 2aq and the electric eld due to this dipole at P has magnitude 1 2aq El = 40 (x + a)3 E26-11

For the dipole on the right p = 2aq and the electric eld due to this dipole at P has magnitude 1 2aq Er = 40 (x a)3 The net electric eld at P is the sum of these two elds, but since the two component elds point in opposite directions we must actually subtract these values, E = Er El;

2aq 1 1 40 (x a)3 (x + a)3 aq 1 1 1 =: 20 x3 (1 a=x)3 (1 + a=x)3 We can use the binomial expansion on the terms containing 1 a=x, aq 1 20 x3 aq 1 20 x3 3(2qa2) =: 20×4

E26-12 Do a series expansion on the part in the parentheses 1 1 R2 2 R 1 1 1 = : p 22 1 + R2=z2 2 z 2z Substitute this in, R2 Q Ez = : 20 2z2 40z2

E26-13 At the surface z = 0 and Ez = =20. Half of this value occurs when z is given by 1z 2 z2 + R2 p which can be written as z2 + R2 = (2z)2. Solve this, and z = R= 3.

E26-14 Look at Eq. 26-18. The electric eld will be a maximum when z=(z2 + R2)3=2 is a maximum. Take the derivative of this with respect to z, and get 1 3 2z2 z2 + R2 3z2 =: (z2 + R2)3=2 2 (z2 + R2)5=2 (z2 + R2)5=2 p This will vanish when the numerator vanishes, or when z = R= 2.

E26-15 (a) The electric eld strength just above the center surface of a charged disk is given by Eq. 26-19, but with z = 0, E= 20 The surface charge density is = q=A = q=(R2). Combining, q = 2 R2E = 2(8:85 1012 C2=N m2)(2:5 102m)2(3 106 N=C) = 1:04 107C: 0

Notice we used an electric eld strength of E = 3 106 N=C, which is the eld at air breaks down (b) We want to nd out how many atoms are on the surface; if a is the cross sectional area of one atom, and N the number of atoms, then A = N a is the surface area of the disk. The number of atoms is A (0:0250 m)2 N = = = 1:31 1017 a (0:015 1018 m2) (c) The total charge on the disk is 1:04 107C, this corresponds to (1:04 107C)=(1:6 1019C) = 6:5 1011 electrons. (We are ignoring the sign of the charge here.) If each surface atom can have at most one excess electron, then the fraction of atoms which are charged is (6:5 1011)=(1:31 1017) = 4:96 106;

E26-16 Imagine switching the positive and negative charges. The electric eld would also need to switch directions. By symmetry, then, the electric eld can only point vertically down. Keeping only that component, Z =2 1 d 0 40 r2 2 =: 40 r2

E26-17 We want to t the data to Eq. 26-19, z Ez = 1 p : 20 z2 + R2 We can nd very easily if we assume that the measurements have no error because then at the surface (where z = 0), the expression for the electric eld simpli es to

E= : 20 Then = 2 E = 2(8:854 1012 C2=N m2)(2:043 107 N=C) = 3:618 104 C=m2. 0 Finding the radius will take a little more work. We can choose one point, and make that the reference point, and then solve for R. Starting with

and then rearranging, 20Ez z z2 + R2 20Ez 1 1 + (R=z)2 1 20Ez 1 + (R=z)2 1 (1 20Ez=)2 s R1 = 1: z (1 20Ez=)2

Using z = 0:03 m and Ez = 1:187 107 N=C, along with our value of = 3:618 104 C=m , we 2 nd s R1 = 1; z (1 2(8:8541012C2=Nm2)(1:187107N=C)=(3:618104C=m2))2 R = 2:167(0:03 m) = 0:065 m: (b) And now nd the charge from the charge density and the radius, q = R2 = (0:065 m)2(3:618 104 C=m2) = 4:80 C:

(b) Integrate: Z L+a 1 a 40 11 40 a L + a q1 40 a(L + a) (c) If a L then L can be replaced with 0 in the above expression.

E26-24 (a) The electric eld is zero nearer to the smaller charge; since the charges have opposite signs it must be to the right of the +2q charge. Equating the magnitudes of the two elds, 2q 5q 40×2 40(x + a)2 or p p 5x = 2(x + a);

which has solution p 2a x = p p = 2:72a: 5 2

E x d E26-26 (a) At point A, 1 q 2q 1 q 40 d2 (2d)2 40 2d2

At point B, 1 q 2q 1 6q 40 (d=2)2 (d=2)2 40 d2 At point C, 1 q 2q 1 7q 4 (2d)2 d2 4 4d2 00

(a) The electric eld does (negative) work on the electron. The magnitude of this work is W = F d, where F = Eq is the magnitude of the electric force on the electron and d is the distance through which the electron moves. Combining, E26-27

which gives the work done by the electric eld on the electron. The electron originally possessed a kinetic energy of K = 1 mv2, since we want to bring the electron to a rest, the work done must be 2 negative. The charge q of the electron is negative, so E~ and ~d are pointing in the same direction, By the work energy theorem, 1 W = K = 0 mv2: 2 We put all of this together and nd d, W mv2 (9:111031kg)(4:86 106 m=s)2 d = = = = 0:0653 m: qE 2qE 2(1:601019C)(1030 N=C) (b) Eq = ma gives the magnitude of the acceleration, and vf = vi + at gives the time. But vf = 0. Combining these expressions,

mv (9:111031kg)(4:86 106 m=s) t = i = = 2:69108 s: Eq (1030 N=C)(1:601019C) (c) We will apply the work energy theorem again, except now we don’t assume the nal kinetic energy is zero. Instead, W = K = Kf Ki;

and dividing through by the initial kinetic energy to get the fraction lost, W Kf Ki Ki Ki

But Ki = 1mv2, and W = qEd, so the fractional change is 2 W qEd (1:601019C)(1030 N=C)(7:88103m) = = = 12:1%: Ki 1mv2 1(9:111031kg)(4:86106m=s)2 22

E26-28 (a) a = Eq=m = (2:16104N=C)(1:601019C)=(1:671027kg) = 2:071012m=s2. pp (b) v = 2ax = 2(2:071012m=s2)(1:22102m) = 2:25105m=s:

E26-29 (a) E = 2q=40r2, or (1:88107C) E = = 5:85105N=C: 2(8:851012C2=N m2)(0:152 m=2)2

E26-30 (a) The average speed between the plates is (1:95102m)=(14:7109s) = 1:33106m=s. The speed with which the electron hits the plate is twice this, or 2:65106m=s. (b) The acceleration is a = (2:65106m=s)=(14:7109s) = 1:801014m=s2. The electric eld then has magnitude E = ma=q, or E = (9:111031kg)(1:801014m=s2)=(1:601019C) = 1:03103N=C: E26-31 The drop is balanced if the electric force is equal to the force of gravity, or Eq = mg. The mass of the drop is given in terms of the density by

4 m = V = r3: 3 Combining, mg 4r3g 4(851 kg=m3)(1:64106m)3(9:81 m=s2) q = = = = 8:111019C: E 3E 3(1:92105N=C) We want the charge in terms of e, so we divide, and get q (8:111019C) = = 5:07 5: e (1:601019C)

E26-32 (b) F = (8:99109N m2=C2)(2:16106C)(85:3109C)=(0:117m)2 = 0:121 N: (a) E2 = F=q = (0:121 N)=(2:16106C) = 5:60104N=C: 1 E1 = F=q2 = (0:121 N)=(85:3109C) = 1:42106N=C:

If each value of q measured by Millikan was a multiple of e, then the di erence between any two values of q must also be a multiple of q. The smallest di erence would be the smallest multiple, and this multiple might be unity. The di erences are 1.641, 1.63, 1.60, 1.63, 3.30, 3.35, 3.18, 3.24, all times 1019 C. This is a pretty clear indication that the fundamental charge is on the order of 1:6 1019 C. If so, the likely number of fundamental charges on each of the drops is shown below in a table arranged like the one in the book: E26-33

4 8 12 5 10 14 7 11 16 The total number of charges is 87, while the total charge is 142:69 1019 C, so the average charge per quanta is 1:64 1019 C.

E26-34 Because of the electric eld the acceleration toward the ground of a charged particle is not g, but g Eq=m, where the sign depends on the direction of the electric eld. (a)Ifthelowerplateispositivelychargedthena=gEq=m.Replaceginthependulumperiod expression by this, and then s L T = 2 : g Eq=m (b) If the lower plate is negatively charged then a = g + Eq=m. Replace g in the pendulum period expression by this, and then s L T = 2 : g + Eq=m

E26-35 The ink drop travels an additional time t0 = d=vx, where d is the additional horizontal distance between the plates and the paper. During this time it travels an additional vertical distance y0 = v t0, where v = at = 2y=t = 2yv =L. Combining, yyx

2yv t0 2yd 2(6:4104m)(6:8103m) y0 = x = = = 5:44104m; L L (1:6102m) so the total de ection is y + y0 = 1:18103m.

E26-36 (a) p = (1:48109C)(6:23106m) = 9:221015C m: Use = pE sin , where is the angle between ~p and E~ . For this dipole p = qd = 2ed or p = 2(1:6 1019 C)(0:78 109 m) = 2:5 1028 C m. For all three cases pE = (2:5 1028 C m)(3:4 106N=C) = 8:5 1022 N m: E26-37

(b) For the perpendicular case = 90, so sin = 1, and = 8:5 1022 N m:. (c) For the anti-parallel case = 180, so sin = 0, and = 0.

(d) Use Eq. 26-12 and F = Eq. Then 40x3F q 4(8:851012C2=N m2)(0:285m)3(5:221016N) (3:16106C) = 4:251022C m: E26-39 The point-like nucleus contributes an electric eld 1 Ze 40 r2 while the uniform sphere of negatively charged electron cloud of radius R contributes an electric eld given by Eq. 26-24, 1 Zer E = : 40 R3

The net electric eld is just the sum, Ze 1 r E= 40 r2 R3

E26-40 The shell theorem rst described for gravitation in chapter 14 is applicable here since both electric forces and gravitational forces fall o as 1=r2. The net positive charge inside the sphere of The net force on either electron will be zero when e2 eQ 4e2 d3 e2d d2 (d=2)2 d2 4R3 R3

P26-1 (a) Let the positive charge be located closer to the point in question, then the electric eld from the positive charge is 1q E+ = 40 (x d=2)2 The negative charge is located farther from the point in question, so 1q E = 40 (x + d=2)2 The net electric eld is the sum of these two elds, but since the two component elds point in opposite direction we must actually subtract these values, 1q1q 40 (z d=2)2 40 (z + d=2)2 1q 1 1 = 40 z2 (1 d=2z)2 (1 + d=2z)2 We can use the binomial expansion on the terms containing 1 d=2z, 1q 40 z2 1 qd = 20 z3

(b) The electric eld is directed away from the positive charge when you are closer to the positive charge; the electric eld is directed toward the negative charge when you are closer to the negative charge. In short, along the axis the electric eld is directed in the same direction as the dipole moment.

P26-2 The key to this problem will be the expansion of 1 1 3 zd 1 : (x2 + (z d=2)2)3=2 (x2 + z2)3=2 2 x2 + z2

p for d x2 + z2. Far from the charges the electric eld of the positive charge has magnitude 1q 4 x2 + (z d=2)2 0

the components of this are 1q x Ex;+ = ; x2 2 p 40 + z x2 + (z d=2)2 1 q (z d=2) Ez;+ = : x2 2 p 40 + z x2 + (z d=2)2 Expand both according to the rst sentence, then

1 xq 3 zd 40 (x2 + z2)3=2 2 x2 + z2 1 (z d=2)q 3 zd Ez;+ = 1 + : 40 (x2 + z2)3=2 2 x2 + z2 Similar expression exist for the negative charge, except we must replace q with q and the + in the parentheses with a , and z d=2 with z + d=2 in the Ez expression. All that is left is to add the expressions. Then

1 xq 3 zd 1 xq 3 zd Ex = 1 + + 1 ; 40 (x2 + z2)3=2 2 x2 + z2 40 (x2 + z2)3=2 2 x2 + z2 1 3xqzd 40 (x2 + z2)5=2 1 (z d=2)q 3 zd 1 (z + d=2)q 3 zd Ez = 1 + + 1 ; 40 (x2 + z2)3=2 2 x2 + z2 40 (x2 + z2)3=2 2 x2 + z2 1 3z2dq 1 dq 40 (x2 + z2)5=2 40 (x2 + z2)3=2 1 (2z2 x2)dq =: 40 (x2 + z2)5=2 p P26-3 (a) Each point on the ring is a distance z2 + R2 from the point on the axis in question. Since all points are equal distant and subtend the same angle from the axis then the top half of the ring contributes q1 z 4 (x2 + R2) z2 + R2 0

while the bottom half contributes a similar expression. Add, and q1 + q2 z q z Ez = = ; 4 (z2 + R2)3=2 4 (z2 + R2)3=2 00 (b) The perpendicular component would be zero if q1 = q2. It isn’t, so it must be the di erence q1 q2 which is of interest. Assume this charge di erence is evenly distributed on the top half of the ring. If it is a positive di erence, then E? must point down. We are only interested then in the vertical component as we integrate around the top half of the ring. Then Z q 1 (q1 2)= 4 z2 + R2 00 q1 q2 1 =: 22 z2 + R2 0

Add the contributions: 1 q 2q q 40 (z + d)2 z2 (z d)2 2 2 q 2d 3d 2d 3d 40z2 z z2 z z2 q 6d2 3Q 40z2 z2 40z4 where Q = 2qd2.

A monopole eld falls o as 1=r2. A dipole eld falls o as 1=r3, and consists of two oppositely charge monopoles close together. A quadrupole eld (see Exercise 11 above or read Problem 4) falls o as 1=r4 and (can) consist of two otherwise identical dipoles arranged with anti- parallel dipole moments. Just taking a leap of faith it seems as if we can construct a 1=r6 eld First we need an octopole which is constructed from a quadrupole. We want to keep things as simple as possible, so the construction steps are P26-5

2. The dipole is a charge +q at x = 0 and a charge q at x = a. We’ll call this a dipole at x = a=2 3. The quadrupole is the dipole at x = a=2, and a second dipole pointing the other way at x = a=2. The charges are then q at x = a, +2q at x = 0, and q at x = a.

4. The octopole will be two stacked, o set quadrupoles. There will be q at x = a, +3q at x = 0, 3q at x = a, and +q at x = 2a.

5. Finally, our distribution with a far eld behavior of 1=r6. There will be +q at x = 2a, 4q at x = a, +6q at x = 0, 4q at x = a, and +q at x = 2a.

P26-6 The vertical component of E~ is simply half of Eq. 26-17. The horizontal component is given by a variation of the work required to derive Eq. 26-16, 1 dz z 40 y2 + z2 y2 + z2 p

which integrates to zero if the limits are 1 to +1, but in this case, Z1 1 Ez = dEz = : 0 40 z

P26-7 (a) Swap all positive and negative charges in the problem and the electric eld must reverse direction. But this is the same as ipping the problem over; consequently, the electric eld must point parallel to the rod. This only holds true at point P , because point P doesn’t move when you ip the rod.

(b) We are only interested in the vertical component of the eld as contributed from each point on the rod. We can integrate only half of the rod and double the answer, so we want to evaluate

Ez (c) The previous expression expansion to P26-8 Evaluate 1 dz z 40 y2 + z2 p y2 + z2 0 p 2 (L=2)2 + y2 y =p: 40 y (L=2)2 + y 2

is exact. If y L, then the expression simpli es with a Taylor L2 40 y3 Z L=2

R 1 z dq Z 0 40 (z2 + r2)3=2 where r is the radius of the ring, z the distance to the plane of the ring, and dq the di erential charge on the ring. But r2 + z2 = R2, and dq = (2r dr), where = q=2R2. Then p Z R 2 r2 q R r dr 0 40 R5 q1 =: 40 3R2

The key statement is the second italicized paragraph on page 595; the number of eld lines through a unit cross-sectional area is proportional to the electric eld strength. If the exponent is n, then the electric eld strength a distance r from a point charge is P26-9

kq rn and the total cross sectional area at a distance r is the area of a spherical shell, 4r2. Then the number of eld lines through the shell is proportional to kq EA = 4r2 = 4kqr2n: rn Note that the number of eld lines varies with r if n 6= 2. This means that as we go farther from the point charge we need more and more eld lines (or fewer and fewer). But the eld lines can only start on charges, and we don’t have any except for the point charge. We have a problem; we really do need n = 2 if we want workable eld lines.

P26-10 The distance traveled by the electron will be d1 = a1t2=2; the distance traveled by the proton will be d2 = a2t2=2. a1 and a2 are related by m1a1 = m2a2, since the electric force is the same (same charge magnitude). Then d1 + d2 = (a1 + a2)t2=2 is the 5.00 cm distance. Divide by the proton distance, and then d1 + d2 a1 + a2 m2 = = + 1: d2 a2 m1

P26-11 This is merely a fancy projectile motion problem. vx = v0 cos while vy;0 = v0 sin . The x and y positions are x = vxt and 1 ax2 y = at2 + vy;0t = + x tan : 2 2v2 cos2 0

The acceleration of the electron is vertically down and has a magnitude of F Eq (1870 N=C)(1:61019C) a = = = = 3:2841014m=s2: m m (9:111031kg) We want to nd out how the vertical velocity of the electron at the location of the top plate. If we get an imaginary answer, then the electron doesn’t get as high as the top plate.

q p = 7:226105m=s: This is a real answer, so this means the electron either hits the top plate, or it misses both plates. The time taken to reach the height of the top plate is vy (7:226105m=s) (5:83106m=s) sin(39) t = = = 8:972109s: a (3:2841014m=s2)

In this time the electron has moved a horizontal distance of x = (5:83106m=s) cos(39)(8:972109s) = 4:065102m: This is clearly on the upper plate.

P26-12 Near the center of the ring z R, so a Taylor expansion yields z E= : 20 R2

The force on the electron is F = Ee, so the e ective \spring” constant is k = e=20R2. This means r r r k e eq != = = : m 20mR2 40mR3

P26-13 U = pE cos , so the work required to ip the dipole is W = pE [cos(0 + ) cos 0] = 2pE cos 0:

P26-14 If the torque on a system is given by j j = , where is a constant, then the frequency p of oscillation of the system is f = =I=2. In this case = pE sin pE, so p f = pE=I=2:

Use the a variation of the exact result from Problem 26-1. The two charge are positive, but since we will eventually focus on the area between the charges we must subtract the two eld contributions, since they point in opposite directions. Then P26-15

q11 Ez = 40 (z a=2)2 (z + a=2)2 and then take the derivative,

dEz q 1 1 = : dz 20 (z a=2)3 (z + a=2)3 Applying the binomial expansion for points z a,

dEz 8q 1 1 1 dz 20 a3 (2z=a 1)3 (2z=a + 1)3 8q 1 20 a3 8q 1 =: 0 a3 There were some fancy sign ips in the second line, so review those steps carefully! (b) The electrostatic force on a dipole is the di erence in the magnitudes of the electrostatic forces on the two charges that make up the dipole. Near the center of the above charge arrangement the electric eld behaves as dEz dz z=0

The net force on a dipole is dEz dEz F+ F = q(E+ E) = q Ez(0) + z+ Ez(0) z dz dz z=0 z=0

where the \+” and \-” subscripts refer to the locations of the positive and negative charges. This last line can be simpli ed to yield dEz dEz q (z+ z) = qd : dz dz z=0 z=0

(b) E = (2:0 m2)^j (2 N=C)^j = 4N m2=C: (c) E = (2:0 m2)^j [(3 N=C)^i + (4 N=C)k^] = 0: (d) In each case the eld is uniform so we can simply evaluate E = E~ A~ , where A~ has six parts, one for every face. The faces, however, have the same size but are organized in pairs with opposite directions. These will cancel, so the total ux is zero in all three cases.

E27-3 (a) The at base is easy enough, since according to Eq. 27-7, Z E = E~ dA~ :

There are two important facts to consider in order to integrate this expression. E~ is parallel to the axis of the hemisphere, E~ points inward while d~A points outward on the at base. E~ is uniform, so it is everywhere the same on the at base. Since E~ is anti-parallel to dA~ , E~ dA~ = E dA, then ZZ E = E~ dA~ = E dA:

Since E~ is uniform we can simplify this as ZZ E = E dA = E dA = EA = R2E:

The last steps are just substituting the area of a circle for the at side of the hemisphere. (b) We must rst sort out the dot product

E dA R We can simplify the vector part of the problem with E~ dA~ = cos E dA, so ZZ E = E~ dA~ = cos E dA

Once again, E~ is uniform, so we can take it out of the integral, ZZ E = cos E dA = E cos dA

We’ll integrate around the axis, from 0 to 2. We’ll integrate from the axis to the equator, from 0 to =2. Then Z Z 2 Z =2 E = E cos dA = E R2 cos sin d d : 00

Pulling out the constants, doing the integration, and then writing 2 cos sin as sin(2), Z =2 Z =2 E 00

Change variables and let = 2, then we have 1 Z E = R2E sin d = R2E: 02

E27-4 Through S1, E = q=0. Through S2, E = q=0. Through S3, E = q=0. Through S4, E = 0. Through S5, E = q=0.

E27-5 By Eq. 27-8, q (1:84 C) = = = 2:08105 N m2=C: E 0 (8:851012 C2=N m2) E27-6 The total ux through the sphere is E = (1 + 2 3 + 4 5 + 6)(103N m2=C) = 3103N m2=C: The charge inside the die is (8:851012C2=N m2)(3103N m2=C) = 2:66108C:

E27-7 The total ux through a cube would be q=0. Since the charge is in the center of the cube we expect that the ux through any side would be the same, or 1=6 of the total ux. Hence the ux through the square surface is q=60.

E27-8 If the electric eld is uniform then there are no free charges near (or inside) the net. The ux through the netting must be equal to, but opposite in sign, from the ux through the opening. The ux through the opening is Ea2, so the ux through the netting is Ea2.

E27-9 There is no ux through the sides of the cube. The ux through the top of the cube is (58 N=C)(100 m)2 = 5:8105N m2=C. The ux through the bottom of the cube is (110 N=C)(100 m)2 = 1:1106N m2=C: The total ux is the sum, so the charge contained in the cube is q = (8:851012C2=N m2)(5:2105N m2=C) = 4:60106C:

E27-10 (a) There is only a ux through the right and left faces. Through the right face R = (2:0 m )^j (3 N=C m)(1:4 m)^j = 8:4 N m2=C: 2

There are eight cubes which can be \wrapped” around the charge. Each cube has three external faces that are indistinguishable for a total of twenty-four faces, each with the same ux E. The total ux is q=0, so the ux through one face is E = q=240. Note that this is the ux through faces opposite the charge; for faces which touch the charge the electric eld is parallel to the surface, so the ux would be zero.

E27-11 E27-12 Use Eq. 27-11, = 20rE = 2(8:851012C2=N m2)(1:96 m)(4:52104N=C) = 4:93106C=m: E27-13 (a) q = A = (2:0106C=m2)(0:12 m)(0:42 m) = 3:17107C: (b) The charge density will be the same! q = A = (2:0 106C=m2)(0:08 m)(0:28 m) = 1:41107C:

E27-14 The electric eld from the sheet on the left is of magnitude El = =20, and points directly away from the sheet. The magnitude of the electric eld from the sheet on the right is the same, (a) To the left of the sheets the two elds add since they point in the same direction. This means (c) To the right of the sheets the two elds add since they point in the same direction. This means that the electric eld is E~ = (=0)^i.

E27-15 The electric eld from the plate on the left is of magnitude El = =20, and points directly toward the plate. The magnitude of the electric eld from the plate on the right is the same, but it (a) To the left of the plates the two elds cancel since they point in the opposite directions. This (b) Between the plates the two electric elds add since they point in the same direction. This (c) To the right of the plates the two elds cancel since they point in the opposite directions. This means that the electric eld is E~ = 0.

E27-16 The magnitude of the electric eld is E = mg=q. The surface charge density on the plates is = 0E = 0mg=q, or (8:851012C2=N m2)(9:111031kg)(9:81 m=s2) = = 4:941022C=m2: (1:601019C)

We don’t really need to write an integral, we just need the charge per unit length in the cylinder to be equal to zero. This means that the positive charge in cylinder must be +3:60nC=m. This positive charge is uniformly distributed in a circle of radius R = 1:50 cm, so E27-17

3:60nC=m 3:60nC=m = = = 5:09C=m3: R2 (0:0150 m)2

E27-18 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc

(a) For point P1 the charge enclosed is qenc = 1:26107C, so (1:26107C) E = = 3:38106N=C: 4(8:851012C2=N m2)(1:83102m)2 (b) Inside a conductor E = 0.

The proton orbits with a speed v, so the centripetal force on the proton is FC = mv2=r. This centripetal force is from the electrostatic attraction with the sphere; so long as the proton is outside the sphere the electric eld is equivalent to that of a point charge Q (Eq. 27-15), E27-19

1Q E= : 40 r2 If q is the charge on the proton we can write F = Eq, or mv2 1 Q =q r 40 r2

Solving for Q, 40mv2r q 4(8:851012 C2=N m2)(1:671027kg)(294103m=s)2(0:0113 m) (1:601019C) = 1:13109C:

E27-20 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc

(a) At r = 0:120 m qenc = 4:06108C. Then (4:06108C) E = = 2:54104N=C: 4(8:851012C2=N m2)(1:20101m)2 (b) At r = 0:220 m qenc = 5:99108C. Then

(5:99108C) E = = 1:11104N=C: 4(8:851012C2=N m2)(2:20101m)2 (c) At r = 0:0818 m qenc = 0 C. Then E = 0.

E27-21 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;

where L is the length of the cylinder. Note that = q=2rL represents a surface charge density. (a) r = 0:0410 m is between the two cylinders. Then (24:1106C=m2)(0:0322 m) E = = 2:14106N=C: (8:851012C2=N m2)(0:0410 m) (b) r = 0:0820 m is outside the two cylinders. Then (24:1106C=m2)(0:0322 m) + (18:0106C=m2)(0:0618 m) E = = 4:64105N=C: (8:851012C2=N m2)(0:0820 m) The negative sign is because it is pointing inward.

E27-22 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;

where L is the length of the cylinder. The charge enclosed is Z 2 2 qenc = dV = L r R

The electric eld is given by 2 2 2 2 L r R r R E= = : 20rL 20r At the surface, (2R)2 2 R 3R Es = = : 202R 40 Solve for r when E is half of this: 3R r2 R2 8 2r 0 = 4r2 3rR 4R2:

E27-23 The electric eld must do work on the electron to stop it. The electric eld is given by E = =20. The work done is W = F d = Eqd. d is the distance in question, so 20K 2(8:851012C2=N m2)(1:15105 eV) d = = = 0:979 m q (2:08106C=m2)e

E27-24 Let the spherical Gaussian surface have a radius of R and be centered on the origin. Choose the orientation of the axis so that the in nite line of charge is along the z axis. The electric eld is then directed radially outward from the z axis with magnitude E = =20, where is the perpendicular distance from the z axis. Now we want to evaluate I E = E~ dA~ ;

over the surface of the sphere. In spherical coordinates, dA = R2 sin d d , = R sin , and E~ dA~ = EA sin . Then I 2R E = sin R d d = : 20 0

(a) The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to E27-25

IZZ where L is the length of the cylinder. Now for the qenc part. If the (uniform) volume charge density is , then the charge enclosed in the Gaussian cylinder is ZZ qenc = dV = dV = V = r2L:

Combining, r2L=0 = E2rL or E = r=20: (b) Outside the charged cylinder the charge enclosed in the Gaussian surface is just the charge in the cylinder. Then ZZ qenc = dV = dV = V = R2L:

and and then nally R2 E= : 20r (b) E = q=0 = (1:52104C)=(8:851012C2=N m2) = 1:72107N m2=C: (c) E = =0 = (8:13106C=m2)=(8:851012C2=N m2) = 9:19105N=C

E27-27 (a) = (2:4106C)=4(0:65 m)2 = 4:52107C=m2: (b) E = = = (4:52107C=m2)=(8:851012C2=N m2) = 5:11104N=C: 0

E27-29 (a) The near eld is given by Eq. 27-12, E = =20, so (6:0106C)=(8:0102 m)2 E = 5:3107N=C: 2(8:851012 C2=N m2) (b) Very far from any object a point charge approximation is valid. Then 1 q 1 (6:0106C) E = = = 60N=C: 40 r2 4(8:851012 C2=N m2) (30 m)2

P27-1 For a spherically symmetric mass distribution choose a spherical Gaussian shell. Then III ~g dA~ = g dA = g dA = 4r2g:

Then g gr2 4G G or Gm g= : r2 P27-2 (a) The ux through all surfaces except the right and left faces will be zero. Through the left face, p l = EyA = b aa2: Through the right face, p r = EyA = b 2aa2: The net ux is then pp = ba5=2( 2 1) = (8830 N=C m1=2)(0:130 m)5=2( 2 1) = 22:3 N m2=C: (b) The charge enclosed is q = (8:851012C2=N m2)(22:3 N m2=C) = 1:971010C.

The net force on the small sphere is zero; this force is the vector sum of the force of gravity W , the electric force FE , and the tension T .

P27-3 T F E W These forces are related by Eq = mg tan : We also have E = =20, so 20mg tan q 2(8:851012 C2=N m2)(1:12106kg)(9:81 m=s2) tan(27:4) (19:7109C) = 5:11109C=m2:

P27-4 The materials are conducting, so all charge will reside on the surfaces. The electric eld inside any conductor is zero. The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc

(e) Since E = 0 inside the shell, qenc = 0, this requires that q reside on the inside surface. This is no charge on the outside surface.

P27-5 The problem has cylindrical symmetry, so use a Gaussian surface which is a cylindrical shell. The E~ eld will be perpendicular to the curved surface and parallel to the end surfaces, so Gauss’ law will simplify to IZZ qenc=0 = E~ dA~ = E dA = E dA = 2rLE;

(a) Outside the conducting shell qenc = +q 2q = q. Then E = q=20rL. The negative sign indicates that the eld is pointing inward toward the axis of the cylinder. (b) Since E = 0 inside the conducting shell, qenc = 0, which means a charge of q is on the inside surface of the shell. The remaining q must reside on the outside surface of the shell. (c) In the region between the cylinders qenc = +q. Then E = +q=20rL. The positive sign indicates that the eld is pointing outward from the axis of the cylinder.

P27-6 Subtract Eq. 26-19 from Eq. 26-20. Then z E= p : 20 z2 + R2

This problem is closely related to Ex. 27-25, except for the part concerning qenc. We’ll set up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on the axis of the physical cylinder. For Gaussian surfaces of radius r < R, there is no charge enclosed We've already worked out the integral P27-7

Z tube for the cylinder, and then from Gauss’ law, Z qenc = 0 E~ dA~ = 20rlE: tube

(a) When r < R there is no enclosed charge, so the left hand vanishes and consequently E = 0 (b) When r > R there is a charge l enclosed, so

P27-8 This problem is closely related to Ex. 27-25, except for the part concerning qenc. We’ll set up the problem the same way: the Gaussian surface will be a (imaginary) cylinder centered on the axis of the physical cylinders. For Gaussian surfaces of radius r < a, there is no charge enclosed We've already worked out the integral Z E~ dA~ = 2rlE;

tube for the cylinder, and then from Gauss’ law, Z qenc = 0 E~ dA~ = 20rlE: tube

(a) When r < a there is no enclosed charge, so the left hand vanishes and consequently E = 0 (b) When b > r > a there is a charge l enclosed, so

E= : 20r P27-9 Uniform circular orbits require a constant net force towards the center, so F = Eq = q=20r. The speed of the positron is given by F = mv2=r; the kinetic energy is K = mv2=2 = F r=2. Combining, q 40 (30109C=m)(1:61019C) 4((8:85 1012 C2=N m2) = 4:311017 J = 270 eV:

P27-10 = 20rE, so q = 2(8:851012C2=N m2)(0:014 m)(0:16 m)(2:9104N=C) = 3:6109C: P27-11 (a) Put a spherical Gaussian surface inside the shell centered on the point charge. Gauss’ law states I q E~ dA~ = enc : 0 Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, and it is everywhere the same on this surface. The dot product simpli es to E~ dA~ = E dA, while since E is a constant on the surface we can pull it out of the integral, and we end up with I q 0 H2 where q is the point charge in the center. Now dA = 4r , where r is the radius of the Gaussian surface, so q E= : 40r2 (b) Repeat the above steps, except put the Gaussian surface outside the conducting shell. Keep it centered on the charge. Two things are di erent from the above derivation: (1) r is bigger, and

(2) there is an uncharged spherical conducting shell inside the Gaussian surface. Neither change will a ect the surface integral or qenc, so the electric eld outside the shell is still q 40r2 (c) This is a subtle question. With all the symmetry here it appears as if the shell has no e ect; the eld just looks like a point charge eld. If, however, the charge were moved o center the eld inside the shell would become distorted, and we wouldn’t be able to use Gauss’ law to nd it. So Outside the shell, however, we can’t tell what is going on inside the shell. So the electric eld outside the shell looks like a point charge eld originating from the center of the shell regardless of where inside the shell the point charge is placed! (d) Yes, q induces surface charges on the shell. There will be a charge q on the inside surface (f) No, as the electric eld from the outside charge won’t make it through a conducting shell. (g) No, this is not a contradiction, because the outside charge never experienced any electrostatic attraction or repulsion from the inside charge. The force is between the shell and the outside charge.

P27-12 The repulsive electrostatic forces must exactly balance the attractive gravitational forces. Then 1 q2 m2 40 r2 r2 p or m = q= 40G: Numerically, (1:601019C) m = q = 1:86109kg: 4(8:851012C2=N m2)(6:671011N m2=kg2)

P27-13 The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III q =0 = E~ dA~ = E dA = E dA = 4r2E: enc

qenc = q + 4 r2dr, or R Zr qenc = q + 4 Ar dr = q + 2A(r2 a2): a

P27-14 (a) The problem has spherical symmetry, so use a Gaussian surface which is a spherical shell. The E~ eld will be perpendicular to the surface, so Gauss’ law will simplify to III qenc=0 = E~ dA~ = E dA = E dA = 4r2E:

(b) The electric eld in the hole is given by E~ h = E~ E~ b, where E~ is the eld from part (a) and E~ b is the eld that would be produced by the matter that would have been in the hole had the hole not been there. Then E~ b = ~b=30;

where ~b is a vector pointing from the center of the hole. Then ~r ~b E~ h = = (~r ~b): 30 30 30 But ~r ~b = ~a, so E~ h = ~a=30.

If a point is an equilibrium point then the electric eld at that point should be zero. If it is a stable point then moving the test charge (assumed positive) a small distance from the equilibrium point should result in a restoring force directed back toward the equilibrium point. In other words, there will be a point where the electric eld is zero, and around this point there will be an electric eld pointing inward. Applying Gauss’ law to a small surface surrounding our point P , we have a net inward ux, so there must be a negative charge inside the surface. But there should be nothing inside the surface except an empty point P , so we have a contradiction.

P27-15 P27-16 (a) Follow the example on Page 618. By symmetry E = 0 along the median plane. The charge enclosed between the median plane and a surface a distance x from the plane is q = Ax. Then E = Ax=0A = A=0: (b) Outside the slab the charge enclosed between the median plane and a surface a distance x from the plane is is q = Ad=2, regardless of x. The E = Ad=2=0A = d=20: P27-17 (a) The total charge is the volume integral over the whole sphere, Z Q = dV:

This is actually a three dimensional integral, and dV = A dr, where A = 4r2. Then Z Q = dV;

1 R4 = SR3: ZR r S2 0R 4S (b) Put a spherical Gaussian surface inside the sphere centered on the center. We can use Gauss’ law here because there is spherical symmetry in the entire problem, both inside and outside the Gaussian surface. Gauss’ law states I E~ dA~ = qenc : 0

Since there is spherical symmetry the electric eld is normal to the spherical Gaussian surface, and it is everywhere the same on this surface. The dot product simpli es to E~ dA~ = E dA, while since E is a constant on the surface we can pull it out of the integral, and we end up with I qenc 0 Now dA = 4r2, where r is the radius of the Gaussian surface, so H

qenc E= : 40r2 We aren’t done yet, because the charge enclosed depends on the radius of the Gaussian surface. We need to do part (a) again, except this time we don’t want to do the whole volume of the sphere, we only want to go out as far as the Gaussian surface. Then Z qenc = dV;

Zr r S2 0R 4S 1 4 R4 r4 = S : R Combine these last two results and S r4 40r2 R S r2 40 R Q r2 =: 40 R4 In the last line we used the results of part (a) to eliminate S from the expression.

P27-18 (a) Inside the conductor E = 0, so a Gaussian surface which is embedded in the conductor but containing the hole must have a net enclosed charge of zero. The cavity wall must then have a (b) The net charge on the conductor is +10:0 C; the charge on the outer surface must then be +13:0 C.

P27-19 (a) Inside the shell E = 0, so the net charge inside a Gaussian surface embedded in the (d) Yes.

Throughout this chapter we will use the convention that V (1) = 0 unless explicitly stated otherwise. Then the potential in the vicinity of a point charge will be given by Eq. 28-18, V = q=40r.

E28-1 (a) Let U12 be the potential energy of the interaction between the two \up” quarks. Then (2=3)2e(1:601019C) U12 = (8:99109N m2=C2) = 4:84105eV: (1:321015m) (b) Let U13 be the potential energy of the interaction between an \up” quark and a \down” quark. Then (1=3)(2=3)e(1:601019C) U13 = (8:99109N m2=C2) = 2:42105eV (1:321015m) Note that U13 = U23. The total electric potential energy is the sum of these three terms, or zero.

E28-2 There are six interaction terms, one for every charge pair. Number the charges clockwise from the upper left hand corner. Then p p U24 = q2=40( 2a): Add these terms and get 2 q2 q2

U = p 4 = 0:206 2 40a 0a The amount of work required is W = U .

(a) We build the electron one part at a time; each part has a charge q = e=3. Moving the rst part from in nity to the location where we want to construct the electron is easy and takes no work at all. Moving the second part in requires work to change the potential energy to 1 q1q2 40 r Bringing in the third part requires work against the force of repulsion between the third charge and both of the other two charges. Potential energy then exists in the form U13 and U23, where all three charges are the same, and all three separations are the same. Then U12 = U13 = U12, so the total potential energy of the system is 1 (e=3)2 3 (1:601019 C=3)2 U = 3 = = 2:721014 J 4 r 4(8:851012 C2=N m2) (2:821015 m) 0

(b) Dividing our answer by the speed of light squared to nd the mass, E28-3

2:72 1014 J m = = 3:02 1031 kg: (3:00 108 m=s)2

E28-4 There are three interaction terms, one for every charge pair. Number the charges from the left; let a = 0:146 m. Then (25:5109C)(17:2109C) 40a (25:5109C)(19:2109C) 40(a + x) (17:2109C)(19:2109C) U23 = : 40x Add these and set it equal to zero. Then (25:5)(17:2) (25:5)(19:2) (17:2)(19:2) a a+x x which has solution x = 1:405a = 0:205 m.

E28-5 The volume of the nuclear material is 4a3=3, where a = 8:01015m. Upon dividing in p half each part will have a radius r where 4r3=3 = 4a3=6. Consequently, r = a= 3 2 = 6:351015m. (a) The force of repulsion is (46)2(1:601019C)2 F = = 3000 N 4(8:851012C2=N m2)[2(6:351015m)]2

(b) The potential energy is (46)2e(1:601019C) U = = 2:4108eV 4(8:851012C2=N m2)2(6:351015m)

E28-6 This is a work/kinetic energy problem: 1 mv2 = qV . Then 20 s 2(1:601019C)(10:3103V) v0 = = 6:0107m=s: (9:111031kg)

(a) The energy released is equal to the charges times the potential through which the charge was moved. Then E28-7

U = qV = (30 C)(1:0 109 V) = 3:0 1010 J: (b) Although the problem mentions acceleration, we want to focus on energy. The energy will change the kinetic energy of the car from 0 to Kf = 3:0 1010 J. The speed of the car is then s r 2K 2(3:0 1010 J) v = = = 7100 m=s: m (1200 kg) (c) The energy required to melt ice is given by Q = mL, where L is the latent heat of fusion. Then Q (3:0 1010 J) m = = = 90; 100kg: L (3:33105J=kg)

E28-8 (a) U = (1:601019C)(1:23109V) = 1:971010J: E28-9 This is an energy conservation problem: 1 mv2 = qV ; V = q=40(1=r1 1=r2). Com- 2 bining, s q2 1 1

20m r1 r2 s (3:1106C)2 1 1

= ; 2(8:851012C2=N m2)(18106kg) (0:90103m) (2:5103m) = 2600 m=s:

E28-10 This is an energy conservation problem: 1 q2 1 m(2v)2 = mv2: 2 40r 2 Rearrange, q2 60mv2 (1:601019C)2 = = 1:42109m: 6(8:851012C2=N m2)(9:111031kg)(3:44105m=s)2)

E28-11 (a) V = (1:601019C)=4(8:851012C2=N m2)(5:291011m) = 27:2 V. (c) For uniform circular orbits F = mv2=r; the force is electrical, or F = e2=40r2. Kinetic energy is K = mv2=2 = F r=2, so e2 (1:601019C) K = = = 13:6 eV: 80r 8(8:851012C2=N m2)(5:291011m)

(d) The ionization energy is (K + U ), or Eion = [(13:6 eV) + (27:2 eV)] = 13:6 eV:

E28-12 (a) The electric potential at A is (5:0106 6 1 q1 q2 C) (2:010 C) V = + = (8:99109N m2=C) + = 6:0104V: A 40 r1 r2 (0:15 m) (0:05 m)

The electric potential at B is 1 q q (5:0106C) (2:0106C)

VB = 1 + 2 = (8:99109N m2=C) + = 7:8105V: 40 r2 r1 (0:05 m) (0:15 m)

E28-13 (a) The magnitude of the electric eld would be found from F (3:90 1015 N) E = = = 2:44 104 N=C: q (1:60 1019 C) (b) The potential di erence between the plates is found by evaluating Eq. 28-15,

Zb V = E~ d~s: a The electric eld between two parallel plates is uniform and perpendicular to the plates. Then E~ d~s = E ds along this path, and since E is uniform, Zb Zb Zb aaa

E28-14 V = Ex, so 20 2(8:851012C2=N m2) x = V = (48 V) = 7:1103m (0:12106C=m2)

E28-15 The electric eld around an in nitely long straight wire is given by E = =20r. The potential di erence between the inner wire and the outer cylinder is given by Zb V = (=20r) dr = (=20) ln(a=b): a

The electric eld near the surface of the wire is then given by V (855 V) E = = = = 1:32108V=m: 20a a ln(a=b) (6:70107m) ln(6:70107 m=1:05102 m) The electric eld near the surface of the cylinder is then given by V (855 V) E = = = = 8:43103V=m: 20a a ln(a=b) (1:05102m) ln(6:70107 m=1:05102 m) E28-16 V = Ex = (1:92105N=C)(1:50102m) = 2:88103V:

E28-17 (a) This is an energy conservation problem: 1 (2)(79)e2 (2)(79)e(1:601019C) K = = (8:99109N m2=C) = 3:2107 eV 40 r (7:01015m) (b) The alpha particles used by Rutherford never came close to hitting the gold nuclei.

E28-18 This is an energy conservation problem: mv2=2 = eq=40r, or s (1:601019C)(1:761015C) v = = 2:13104m=s 2(8:851012C2=N m2)(1:22102m)(9:111031kg)

E28-19 1 q 1 (1:16C) VA = = = 5060 V; 40 r 4(8:85 1012 C2=N m2) (2:06 m) and 1 q 1 (1:16C) VB = = = 8910 V; 40 r 4(8:85 1012 C2=N m2) (1:17 m) (b) The answer is the same, since when concerning ourselves with electric potential we only care about distances, and not directions.

E28-20 The number of \excess” electrons on each grain is 40rV 4(8:851012C2=N m)(1:0106m)(400 V) n = = = 2:8105 e (1:601019C) E28-21 The excess charge on the shuttle is q = 40rV = 4(8:851012C2=N m)(10 m)(1:0 V) = 1:1109C

E28-22 q = 1:37105C, so (1:37105C) V = (8:99109N m2=C2) = 1:93108V: (6:37106m) E28-23 The ratio of the electric potential to the electric eld strength is

V 1q 1q = = = r: E 40 r 40 r2

In this problem r is the radius of the Earth, so at the surface of the Earth the potential is V = Er = (100 V=m)(6:38106m) = 6:38108 V:

E28-24 Use Eq. 28-22: (1:47)(3:341030C m) V = (8:99109N m2=C2) = 1:63105V: (52:0109m)2 E28-25 (a) When nding VA we need to consider the contribution from both the positive and the negative charge, so 1 q VA = qa + 40 a + d There will be a similar expression for VB ,

1q VB = qa + : 40 a + d

1 q 1 q VA VB = qa + qa + ; 40 a + d 40 a + d q11 20 a a + d q a+d a 20 a(a + d) a(a + d) qd =: 20 a(a + d) (b) Does it do what we expect when d = 0? I expect it the di erence to go to zero as the two points A and B get closer together. The numerator will go to zero as d gets smaller. The denominator, however, stays nite, which is a good thing. So yes, Va VB ! 0 as d ! 0.

E28-26 (a) Since both charges are positive the electric potential from both charges will be positive. There will be no nite points where V = 0, since two positives can’t add to zero. (b) Between the charges the electric eld from each charge points toward the other, so E~ will pp vanish when q=x2 = 2q=(d x)2. This happens when d x = 2x, or x = d=(1 + 2).

p (a) V at C is 2(2:13106C) V = (8:99109N m2=C2) = 2:76106V (1:39102m) (c) Don’t forget about the potential energy of the original two charges! (2:13106C)2 U = (8:99109N m2=C2) = 2:08 J 0 (1:96102m) Add this to the answer from part (b) to get 7:35 J.

E28-28 The potential is given by Eq. 28-32; at the surface V s = R=20, half of this occurs when p 3R=4 = z:

We can nd the linear charge density by dividing the charge by the circumference,

Q 2R where Q refers to the charge on the ring. The work done to move a charge q from a point x to the origin will be given by E28-29

1Q1Q 4 R2 4 R2 + x2 00 qQ 1 1 = p : 4 R R2 + x2 0

Putting in the numbers, ! (5:931012C)(9:12109C) 1 1 = 1:861010J: 4(8:851012C2=N m2) 1:48m (1:48m)2 + (3:07m)2 p

E28-30 (a) The electric eld strength is greatest where the gradient of V is greatest. That is (b) The least absolute value occurs where the gradient is zero, which is between b and c and again between e and f .

E28-31 The potential on the positive plate is 2(5:52 V) = 11:0 V; the electric eld between the plates is E = (11:0 V)=(1:48102m) = 743 V=m.

E28-33 The radial potential gradient is just the magnitude of the radial component of the electric eld, @V Er = @r Then @V 1 q @r 40 r2 1 79(1:60 1019C) 4(8:85 1012 C2=N m2) (7:0 1015m)2 = 2:321021 V=m:

E28-34 Evaluate @V =@r, and Ze 1 r E= +2 : 40 r2 2R3

E28-35 Ex=@V=@x=2(1530V=m2)x.Atthepointinquestion,E=2(1530V=m2)(1:28 102m) = 39:2 V=m.

E28-36 Draw the wires so that they are perpendicular to the plane of the page; they will then \come out of” the page. The equipotential surfaces are then lines where they intersect the page, and they look like

E28-37 (a) jVB VAj = jW=qj = j(3:94 1019 J)=(1:60 1019 C)j = 2:46 V. The electric eld did work on the electron, so the electron was moving from a region of low potential to a region of (b) VC is at the same potential as VB (both points are on the same equipotential line), so (c) VC is at the same potential as VB (both points are on the same equipotential line), so VC VB = 0 V.

E28-38 (a) For point charges r = q=40V , so r = (8:99109N m2=C2)(1:5108C)=(30 V) = 4:5 m (b) No, since V / 1=r.

E28-39 The dotted lines are equipotential lines, the solid arrows are electric eld lines. Note that there are twice as many electric eld lines from the larger charge!

E28-41 This can easily be done with a spreadsheet. The following is a sketch; the electric eld is the bold curve, the potential is the thin curve.

sphere radius r

E28-42 Originally V = q=40r, where r is the radius of the smaller sphere. (a) Connecting the spheres will bring them to the same potential, or V1 = V2. (b) q1 + q2 = q; V1 = q1=40r and V2 = q2=402r; combining all of the above q2 = 2q1 and q1 = q=3 and q2 = 2q=3.

E28-43 (a) q = 4R2, so V = q=40R = R=0, or V = (1:601019C=m2)(6:37106m)=(8:851012C2=N m2) = 0:115 V (b) Pretend the Earth is a conductor, then E = =epsilon0, so

E = (1:601019C=m2)=(8:851012C2=N m2) = 1:81108V=m: E28-44 V = q=40R, so V = (8:99109N m2=C2)(15109C)=(0:16 m) = 850 V:

E28-45 (a) q = 40RV = 4(8:851012C2=N m2)(0:152 m)(215 V) = 3:63109C (b) = q=4R2 = (3:63109C)=4(0:152 m)2 = 1:25108C=m2.

(a) The total charge (Q = 57:2nC) will be divided up between the two spheres so that they are at the same potential. If q1 is the charge on one sphere, then q2 = Q q1 is the charge on the other. Consequently 1 q1 1 Q q1 40 r1 40 r2 Qr2 q1 = : r2 + r1 Putting in the numbers, we nd Qr1 (57:2 nC)(12:2 cm) q1 = = = 38:6 nC; r2 + r1 (5:88 cm) + (12:2 cm) (b) The potential on each sphere should be the same, so we only need to solve one. Then 1 q1 1 (38:6 nC) = = 2850 V: 4 r 4(8:85 1012 C2=N m2) (12:2 cm) 01 E28-47

(b) V = q=40r, so r = q=40V , and then r = (8:99109N m2=C2)(31:5109C)=(1:20103V) = 0:236 m: That is (0:236 m) (0:162 m) = 0:074 m above the surface.

E28-49 (a) Apply the point charge formula, but solve for the charge. Then 1q 40 r q = 4(8:85 1012 C2=N m2)(1 m)(106 V) = 0:11 mC:

Now that’s a fairly small charge. But if the radius were decreased by a factor of 100, so would the charge (1:10 C). Consequently, smaller metal balls can be raised to higher potentials with less (b) The electric eld near the surface of the ball is a function of the surface charge density, E = =0. But surface charge density depends on the area, and varies as r2. For a given potential, the electric eld near the surface would then be given by qV E= = = : 0 40r2 r Note that the electric eld grows as the ball gets smaller. This means that the break down eld is more likely to be exceeded with a low voltage small ball; you’ll get sparking.

E28-50 A \Volt” is a Joule per Coulomb. The power required by the drive belt is the product (3:41106V)(2:83103C=s) = 9650 W.

P28-1 (a) According to Newtonian mechanics we want K = 1 mv2 to be equal to W = qV 2 which means mv2 (0:511 MeV) V = = = 256 kV: 2q 2e (b) Let’s do some rearranging rst.

“# 1 1 2 K1 mc2 1 2 p K1 mc2 1 2 p 1p K +1 mc2 1 K 2 +1 mc2 and nally, s 1 = 1 K + 12 mc2 Putting in the numbers, v u1 t (256 keV) 2 +1 (511 keV) so v = 0:746c.

P28-2 (a) The potential of the hollow sphere is V = q=40r. The work required to increase the charge by an amount dq is dW = V =; dq. Integrating, e q e2 Z W = dq = : 0 40r 80r This corresponds to an electric potential energy of e(1:601019C) W = = 2:55105 eV = 4:081014J: 8(8:851012C2=N m2)(2:821015m) (b) This would be a mass of m = (4:081014J)=(3:00108m=s)2 = 4:531031kg.

P28-3 The negative charge is held in orbit by electrostatic attraction, or mv2 qQ =: r 40r2 The kinetic energy of the charge is 1 qQ K = mv2 = : 2 80r The electrostatic potential energy is qQ 40r so the total energy is qQ E= : 80r The work required to change orbit is then

qQ 1 1 W= : 80 r1 r2 R P28-4 (a) V = E dr, so r qr qr2 Z V = dr = : 4 R3 8 R3 00 0

(c) If instead of V = 0 at r = 0 as was done in part (a) we take V = 0 at r = 1, then V = q=40R on the surface of the sphere. The new expression for the potential inside the sphere will look like V = V 0 + V , where V 0 is the answer from part (a) and V is a constant so that the ss surface potential is correct. Then q qR2 3qR2 4 R 8 R3 8 R3 000

and then qr2 3qR2 q(3R2 r2) V= + = : 8 R3 8 R3 8 R3 000

P28-5 The total electric potential energy of the system is the sum of the three interaction pairs. One of these pairs does not change during the process, so it can be ignored when nding the change in potential energy. The change in electrical potential energy is then 2 2 2 qqq11 U = 2 2 = : 40rf 40ri 20 rf ri

In this case ri = 1:72 m, while rf = 0:86 m. The change in potential energy is then

11 U = 2(8:99109N m2=C2)(0:122 C)2 = 1:56108J (0:86 m) (1:72 m) The time required is t = (1:56108)=(831 W) = 1:87105s = 2:17 days:

P28-6 (a) Apply conservation of energy: qQ qQ 40d 40K (b) Apply conservation of energy: qQ 1 40(2d) 2 p so, combining with the results in part (a), v = K=m.

P28-7 (a) First apply Eq. 28-18, but solve for r. Then q (32:0 1012 C) r = = = 562 m: 40V 4(8:85 1012 C2=N m2)(512 V) (b) If two such drops join together the charge doubles, and the volume of water doubles, but the p radius of the new drop only increases by a factor of 3 2 = 1:26 because volume is proportional to The potential on the surface of the new drop will be 1 qnew 40 rnew 1 2qold 40 3 2 rold 1 qold = (2)2=3 = (2)2=3V old: 40 rold

P28-9 (a) The potential at any point will be the sum of the contribution from each charge, 1 q1 1 q2 40 r1 40 r2 where r1 is the distance the point in question from q1 and r2 is the distance the point in question from q2. Pick a point, call it (x; y). Since q1 is at the origin, p r1 = x2 + y2:

Since q2 is at (d; 0), where d = 9:60 nm, p r2 = (x d)2 + y2:

De ne the \Stanley Number” as S = 40V . Equipotential surfaces are also equi-Stanley surfaces. In particular, when V = 0, so does S. We can then write the potential expression in a sightly simpli ed form q1 q2 S= + : r1 r2 If S = 0 we can rearrange and square this expression.

q1 q2 r1 r2 r2 r2 q2 q2 12 x2 + y2 (x d)2 + y2 q2 q2 12

Let = q2=q1, then we can write x +y = ( 2 1)x2 + 2xd + ( 2 1)y2 = d2:

We complete the square for the ( 2 1)x2 + 2xd term by adding d2=( 2 1) to both sides of the equation. Then “# 2 d1 ( 2 1) x + + y2 = d2 1 + : 21 21 The center of the circle is at d (9:60 nm) = = 5:4 nm: 2 1 (10=6)2 1 (b) The radius of the circle is v u u 1+ 1 t 21 21 which can be simpli ed to j(10=6)j d = (9:6 nm) = 9:00 nm: 2 1 (10=6)2 1 (c) No.

P28-10 An annulus is composed of di erential rings of varying radii r and width dr; the charge on any ring is the product of the area of the ring, dA = 2r dr, and the surface charge density, or k 2k dq = dA = 2r dr = dr: r3 r2 The potential at the center can be found by adding up the contributions from each ring. Since we are at the center, the contributions will each be dV = dq=40r. Then b 1 k 2 a2 Z k dr k 1 b V= = := : 20 r3 40 a2 b2 40 b2a2 a

The total charge on the annulus is b 11 Z ba 2k Q = dr = 2k = 2k : a r2 a b ba

Combining, Q a+b V= : 80 ab q 1 1 1 40 r + d r r d q 1 1 40r 1 + d=r 1 d=r qdd 40r r r q 2d 1+ : 40r r P28-12 (a) Add the contributions from each di erential charge: dq = dy. Then y+L Z y+L V = dy = ln : y 40y 40 y

(b) Take the derivative: @V y L L Ey = = = : @y 40 y + L y2 40 y(y + L) (c) By symmetry it must be zero, since the system is invariant under rotations about the axis of the rod. Note that we can’t determine E? from derivatives because we don’t have the functional form of V for points o -axis! P28-13 (a) We follow the work done in Section 28-6 for a uniform line of charge, starting with Eq. 28-26, 1 dx 40 x2 + y2 ZL 1 kx dx 40 0 x2 + y2

kp L 40 0 kp = L2 + y2 y : 40 (b) The y component of the electric eld can be found from @V @y which (using a computer-aided math program) is ! ky Ey = 1 p : 40 L2 + y 2 (c) We could nd Ex if we knew the x variation of V . But we don’t; we only found the values of (d) We want to nd y such that the ratio

k k p L2 + y2 y = (L) 40 40 p is one-half. Simplifying, L2 + y2 y = L=2; which can be written as L2 + y2 = L2=4 + Ly + y2;

P28-14 The spheres are small compared to the separation distance. Assuming only one sphere at a potential of 1500 V, the charge would be q = 4 rV = 4(8:851012C2=N m)(0:150 m)(1500 V) = 2:50108C: 0

The potential from the sphere at a distance of 10.0 m would be (0:150 m) V = (1500 V) = 22:5 V: (10:0 m) This is small compared to 1500 V, so we will treat it as a perturbation. This means that we can assume that the spheres have charges of q = 4 rV = 4(8:851012C2=N m)(0:150 m)(1500 V + 22:5 V) = 2:54108C: 0

Calculating the fraction of excess electrons is the same as calculating the fraction of excess charge, so we’ll skip counting the electrons. This problem is e ectively the same as Exercise 28-47; we have a total charge that is divided between two unequal size spheres which are at the same potential on the surface. Using the result from that exercise we have Qr1 r2 + r1 where Q = 6:2 nC is the total charge available, and q1 is the charge left on the sphere. r1 is the radius of the small ball, r2 is the radius of Earth. Since the fraction of charge remaining is q1=Q, we can write q1 r1 r1 = = 2:0 108: Q r2 + r1 r2 P28-15

P28-16 The positive charge on the sphere would be q = 40rV = 4(8:851012C2=N m2)(1:08102m)(1000 V) = 1:20109C: The number of decays required to build up this charge is n = 2(1:20109C)=(1:601019C) = 1:501010:

The extra factor of two is because only half of the decays result in an increase in charge. The time required is t = (1:501010)=(3:70108s1) = 40:6 s:

P28-18 (a) Outside of an isolated charged spherical object E = q=40r2 and V = q=40r. Then E = V=r. Consequently, the sphere must have a radius larger than r = (9:15106V)=(100 (c) wv = (320106C=s), so (320106C=s) = = 2:00105C=m2: (0:485 m)(33:0 m=s)

(a) The charge which ows through a cross sectional surface area in a time t is given by q = it; where i is the current. For this exercise we have q = (4:82 A)(4:60 60 s) = 1330 C E29-1

E29-3 (a) j = nqv = (2:101014=m3)2(1:601019C)(1:40105m=s) = 9:41 A=m2: Since the ions have positive charge then the current density is in the same direction as the velocity. (b) We need an area to calculate the current.

(b) v = j=ne = (2:59105A=m2)=(8:491028=m3)(1:601019C) = 1:911015m=s: d

The current rating of a fuse of cross sectional area A would be imax = (440 A=cm2)A;

and if the fuse wire is cylindrical A = d2=4. Then E29-5 s 4 (0:552 A) d = = 4:00102 cm: (440 A=m2)

E29-6 Current density is current divided by cross section of wire, so the graph would look like:

4 I (A/mil^2 x10^-3) 3 2 1 50 100 150 200 d(mils)

E29-7 The current is in the direction of the motion of the positive charges. The magnitude of the current is i = (3:11018=s + 1:11018=s)(1:601019C) = 0:672 A:

E29-8 (a) The total current is i = (3:501015=s + 2:251015=s)(1:601019C) = 9:20104A: (b) The current density is j = (9:20104A)=(0:165103m)2 = 1:08104A=m2:

E29-9 (a) j = (8:70106=m3)(1:601019C)(470103m=s) = 6:54107A=m2: (b) i = (6:54107A=m2)(6:37106m)2 = 8:34107A:

E29-10 i = wv, so = (95:0106A)=(0:520 m)(28:0 m=s) = 6:52106C=m2:

The drift velocity is given by Eq. 29-6, j i (115 A) vd = = = = 2:71104m=s: ne Ane (31:2106m2)(8:491028=m3)(1:601019C) The time it takes for the electrons to get to the starter motor is x (0:855 m) t = = = 3:26103s: v (2:71104m=s) That’s about 54 minutes.

E29-11 E29-12 V = iR = (50103A)(1800 ) = 90 V: The resistance of an object with constant cross section is given by Eq. 29-13, L (11; 000 m) R = = (3:0 107 m) = 0:59 : A (0:0056 m2)

E29-14 The slope is approximately [(8:2 1:7)=1000] cm=C, so

1 = 6:5103 cm=C 4103=C 1:7 cm E29-15 (a) i = V =R = (23 V)=(15103 ) = 1500 A: (b) j = i=A = (1500 A)=(3:0103m)2 = 5:3107A=m2: (c) = RA=L = (15103 )(3:0103m)2=(4:0 m) = 1:1107 m. The material is possibly E29-13

E29-16 Use the equation from Exercise 29-17. R = 8 ; then T = (8 )=(50 )(4:3103=C) = 37 C: The nal temperature is then 57C.

E29-17 Start with Eq. 29-16, and multiply through by L=A, LL AA to get R R0 = R0 av(T T0): E29-18 The wire has a length L = (250)2(0:122 m) = 192 m. The diameter is 0.129 inches; the cross sectional area is then A = (0:129 0:0254 m)2=4 = 8:43106m2:

The resistance is R = L=A = (1:69108 m)(192 m)=(8:43106m2) = 0:385 : E29-19 If the length of each conductor is L and has resistivity , then L 4L RA = = D2=4 D2 and L 4L RB = = : (4D2=4 D2=4) 3D2 The ratio of the resistances is then RA = 3: RB

E29-20 R = R, so L =(d1=2)2 = L =(d =2)2. Simplifying, =d2 = =d2. Then 11 222 1122

d2 = (1:19103m)p(9:68108 m)=(1:69108 m) = 2:85103m: (b) V = iR = (6:00103A)(2:65106 ) = 1:59108V: (c) R = V=i = (1:59108V)=(750103A) = 2:12108 :

E29-22 Since V = iR, then if V and i are the same, then R must be the same. (a) Since R = R, 1L1=r2 = 2L2=r2, or =r2 = =r2. Then 1 21122 p riron=rcopper = (9:68108 m)(1:6910 m) = 2:39: 8

(b) Start with the de nition of current density: i V V j= = = : A RA L Since V and L is the same, but is di erent, then the current densities will be di erent.

Conductivity is given by Eq. 29-8, ~j = E~ . If the wire is long and thin, then the magnitude of the electric eld in the wire will be given by E V =L = (115 V)=(9:66 m) = 11:9 V=m: E29-23

We can now nd the conductivity, j (1:42104A=m2) = = = 1:19103( m)1: E (11:9 V=m)

E29-24 (a) vd = j=en = E=en: Then v = (2:701014= m)(120 V=m)=(1:601019C)(620106=m3 + 550106=m3) = 1:73102m=s: d

(b) j = E = (2:701014= m)(120 V=m) = 3:241014A=m2: E29-25 (a) R=L = =A, so j = i=A = (R=L)i=. For copper, j = (0:152103 =m)(62:3 A)=(1:69108 m) = 5:60105A=m2;

for aluminum, j = (0:152103 =m)(62:3 A)=(2:75108 m) = 3:44105A=m2: (b) A = L=R; if is density, then m = lA = l=(R=L). For copper, m = (1:0 m)(8960 kg=m3)(1:69108 m)=(0:152103 =m) = 0:996 kg;

for aluminum, m = (1:0 m)(2700 kg=m3)(2:75108 m)=(0:152103 =m) = 0:488 kg:

10 8 R (Kilo-ohms) 6 4 2 1 2 3 4 V(Volts)

E29-27 (a) The resistance is de ned as V (3:55 106 V=A2)i2 R = = = (3:55 106 V=A2)i: ii When i = 2:40 mA the resistance would be R = (3:55 106 V=A2)(2:40 103A) = 8:52 k : (b) Invert the above expression, and i = R=(3:55 106 V=A2) = (16:0 )=(3:55 106 V=A2) = 4:51 A:

E29-28 First, n = 3(6:021023)(2700 kg=m3)(27:0103kg) = 1:811029=m3. Then m (9:111031kg) = = = 7:151015s: ne2 (1:811029=m3)(1:601019C)2(2:75108 m)

E29-29 (a) E = E0=e = q=40eR2, so (1:00106C) E= = 4(8:851012C2=N m2)(4:7)(0:10 m)2

(b) E = E0 = q=40R2, so (1:00106C) E= = 4(8:851012C2=N m2)(0:10 m)2 (c) ind = 0(E0 E) = q(1 1=e)=4R2. Then (1:00106C) 1 ind = 1 = 6:23106C=m2: 4(0:10 m)2 (4:7)

E29-30 Midway between the charges E = q=0d, so q = (8:851012C2=N m2)(0:10 m)(3106V=m) = 8:3106C:

(a) At the surface of a conductor of radius R with charge Q the magnitude of the electric eld is given by 1 40 while the potential (assuming V = 0 at in nity) is given by

1 V = QR: 40 E29-31 The potential on the sphere that would result in \sparking” is V = ER = (3106N=C)R: (b) It is \easier” to get a spark o of a sphere with a smaller radius, because any potential on (c) The points of a lighting rod are like small hemispheres; the electric eld will be large near these points so that this will be the likely place for sparks to form and lightning bolts to strike.

If there is more current owing into the sphere than is owing out then there must be a change in the net charge on the sphere. The net current is the di erence, or 2 A. The potential on the surface of the sphere will be given by the point-charge expression, 1q 40 r and the charge will be related to the current by q = it. Combining, 1 it 40 r or 4 V r 4(8:85 1012 C2=N m2)(980 V)(0:13 m) 0 t = = = 7:1 ms: i (2 A) P29-1

P29-2 The net current density is in the direction of the positive charges, which is to the east. There are two electrons for every alpha particle, and each alpha particle has a charge equal in magnitude to two electrons. The current density is then = 1:0105C=m2:

P29-3 (a) The resistance of the segment of the wire is R = L=A = (1:69108 m)(4:0102m)=(2:6103m)2 = 3:18105 : The potential di erence across the segment is V = iR = (12 A)(3:18105 ) = 3:8104V: (c) The drift speed is v = j=en = i=Aen, so v = (12 A)=(2:6103m)2(1:61019C)(8:491028=m3) = 4:16105m=s: The electrons will move 1 cm in (1:0102m)=(4:16105m=s) = 240 s.

p (b) The speed of the particles in the beam is given by v = 2K=m, so p v = 2(22:4 MeV)=4(932 MeV=c2) = 0:110c: It takes (0:180 m)=(0:110)(3:00108m=s) = 5:45109s for the beam to travel 18.0 cm. The number of charges is then N = it=q = (250109A)(5:45109s)=(3:21019C) = 4260: (c) W = qV , so V = (22:4 MeV)=2e = 11:2 MV:

P29-5 (a) The time it takes to complete one turn is t = (250 m)=c. The total charge is q = it = (30:0 A)(950 m)=(3:00108m=s) = 9:50105C: (b) The number of charges is N = q=e, the total energy absorbed by the block is then U = (28:0109 eV)(9:50105C)=e = 2:66106J: This will raise the temperature of the block by T = U=mC = (2:66106J)=(43:5 kg)(385J=kgC) = 159 C: RR P29-6 (a) i = j dA = 2 jr dr;

Z i = 2 0Rj0(1 r=R)r dr = 2j0(R2=2 R3=3R) = j0R2=6: (b) Integrate, again: Z i = 2 0Rj0(r=R)r dr = 2j0(R3=3R) = j0R2=3:

P29-7 (a) Solve 20 = 0[1 + (T 20C)], or T = 20C + 1=(4:3103=C) = 250C: (b) Yes, ignoring changes in the physical dimensions of the resistor.

P29-8 The resistance when on is (2:90 V)=(0:310 A) = 9:35 . The temperature is given by T = 20C + (9:35 1:12 )=(1:12 )(4:5103=C) = 1650C:

Originally we have a resistance R1 made out of a wire of length l1 and cross sectional area A1. The volume of this wire is V1 = A1l1. When the wire is drawn out to the new length we have l2 = 3l1, but the volume of the wire should be constant so A2 = A1=3: The original resistance is l1 R1 = : A1 The new resistance is l2 3l1 A2 A1=3 or R2 = 54 .

P29-10 (a) i = (35:8 V)=(935 ) = 3:83102A: (d) E = (35:8 V)=(0:158 m) = 227 V=m: P29-9

P29-11 (a) = (1:09103 )(5:5103m)2=4(1:6 m) = 1:62108 m. This is possibly silver. (b) R = (1:62108 m)(1:35103m)4=(2:14102m)2 = 6:08108 :

P29-12 (a) L=L = 1:7105 for a temperature change of 1:0 C. Area changes are twice this, Take the di erential of RA = L: R dA+A dR = dL+L d, or dR = dL=A+L d=AR dA=A. For nite changes this can be written as R L A =+: RLA = = 4:3103. Since this term is so much larger than the other two it is the only signi cant e ect.

P29-13 We will use the results of Exercise 29-17, R R0 = R0 av(T T0):

To save on subscripts we will drop the \av” notation, and just specify whether it is carbon \c” or The disks will be e ectively in series, so we will add the resistances to get the total. Looking only at one disk pair, we have = R0;c + R0;i + (R0;c c + R0;i i) (T T0):

This last equation will only be constant if the coecient for the term (T T0) vanishes. Then

but R = L=A, and the disks have the same cross sectional area, so Lcc c + Lii i = 0;

or Lc i i (9:68108 m)(6:5103=C) = = = 0:036: Li c c (3500108 m)(0:50103=C)

P29-14 The current entering the cone is i. The current density as a function of distance x from the left end is then i j= : [a + x(b a)=L]2 The electric eld is given by E = j. The potential di erence between the ends is then ZL ZL i iL V = E dx = dx = 0 0 [a + x(b a)=L]2 ab

P29-15 The current is found from Eq. 29-5, Z where the region of integration is over a spherical shell concentric with the two conducting shells but between them. The current density is given by Eq. 29-10, ~j = E~ =;

and we will have an electric eld which is perpendicular to the spherical shell. Consequently, ZZ 11 i = E~ dA~ = E dA

By symmetry we expect the electric eld to have the same magnitude anywhere on a spherical shell which is concentric with the two conducting shells, so we can bring it out of the integral sign, and then Z2 1 4r E i = E dA = ;

where E is the magnitude of the electric eld on the shell, which has radius r such that b > r > a. The above expression can be inverted to give the electric eld as a function of radial distance, since the current is a constant in the above expression. Then E = i=4r2 The potential is given by Za V = E~ d~s;

b we will integrate along a radial line, which is parallel to the electric eld, so Za V = E dr;

b a i Z b 4r2 i a dr Z 4 b r i 1 1 = : 4 a b We divide this expression by the current to get the resistance. Then

11 R= 4 a b P29-16 Since =p=vd, p / vd. For an ideal gas the kinetic energy is proportional to the temperature, so / K / T .

We apply Eq. 30-1, E30-2 (a)C=V=q=(73:01012C)=(19:2V)=3:801012F: (b) The capacitance doesn’t change! (c) V = q=C = (2101012C)=(3:801012F) = 55:3 V: E30-3 q = CV = (26:0106F)(125 V) = 3:25103C: E30-4 (a) C = 0A=d = (8:851012F=m)(8:22102m)2=(1:3110 m) = 1:4310 F: 3 10 (b) q = CV = (1:431010F)(116 V) = 1:66108C.

Eq. 30-11 gives the capacitance of a cylinder, L (0:0238 m) C = 20 = 2(8:851012 F=m) = 5:4610 F: 13 ln(b=a) ln((9:15mm)=(0:81mm))

E30-6 (a) A = Cd=0 = (9:701012F)(1:20103m)=(8:851012F=m) = 1:32103m2: 0 (c) V = q0=C = [V ]0C0=C = [V ]0d=d0. Using this formula, the new potential di erence would be [V ]0 = (13:0 V)(1:10103m)=(1:20103m) = 11:9 V: The potential energy has changed by (11:9 V) (30:0 V) = 1:1 V.

E30-7 (a) From Eq. 30-8, (0:040 m)(0:038 m) C = 4(8:851012F=m) = 8:451011F: (0:040 m) (0:038 m) (b) A = Cd= = (8:451011F)(2:00103m)=(8:851012F=m) = 1:91102m2: 0

E30-8 Let a = b + d, where d is the small separation between the shells. Then ab (b + d)b ab d b2 40 = 0A=d: d

E30-1 E30-5 E30-9 The charge on each of the capacitors is then q = CV = (1:00 106 F)(110 V) = 1:10 104 C: If there are N capacitors, then the total charge will be N q, and we want this total charge to be 1:00 C. Then (1:00 C) (1:00 C) N = = = 9090: q (1:10 104 C)

E30-10 First nd the equivalent capacitance of the parallel part: C = C + C = (10:3106F) + (4:80106F) = 15:1106F: eq 1 2 Then nd the equivalent capacitance of the series part: 111 = + = 3:23105F1: C (15:1106F) (3:90106F) eq Then the equivalent capacitance of the entire arrangement is 3:10106F.

E30-11 First nd the equivalent capacitance of the series part: 111 = + = 3:05105F1: C (10:3106F) (4:80106F) eq The equivalent capacitance is 3:28106F. Then nd the equivalent capacitance of the parallel part: Ceq = C1 + C2 = (3:28106F) + (3:90106F) = 7:18106F: This is the equivalent capacitance for the entire arrangement.

E30-12 For one capacitor q = CV = (25:0106F)(4200 V) = 0:105 C. There are three capaci- tors, so the total charge to pass through the ammeter is 0:315 C.

E30-13 (a) The equivalent capacitance is given by Eq. 30-21, 111115 =+= + = Ceq C1 C2 (4:0F) (6:0F) (12:0F) (b) The charge on the equivalent capacitor is q = CV = (2:40F)(200 V) = 0:480 mC. For series capacitors, the charge on the equivalent capacitor is the same as the charge on each of the capacitors. This statement is wrong in the Student Solutions! (c) The potential di erence across the equivalent capacitor is not the same as the potential di erence across each of the individual capacitors. We need to apply q = CV to each capacitor using the charge from part (b). Then for the 4:0F capacitor, q (0:480 mC) C (4:0F) and for the 6:0F capacitor, q (0:480 mC) V = = = 80 V: C (6:0F) Note that the sum of the potential di erences across each of the capacitors is equal to the potential di erence across the equivalent capacitor.

E30-14 (a) The equivalent capacitance is Ceq = C1 + C2 = (4:0F) + (6:0F) = (10:0F): (c) For parallel capacitors, the potential di erence across the equivalent capacitor is the same as (b) For the 4:0F capacitor, q = CV = (4:0F)(200 V) = 8:0104 C;

and for the 6:0F capacitor, q = CV = (6:0F)(200 V) = 12:0104 C:

0A 0A d deq = = = : Ceq 3C 3

0A 0A deq = = = 3d: Ceq C=3 E30-16 (a) The maximum potential across any individual capacitor is 200 V; so there must be at least (1000 V)=(200 V) = 5 series capacitors in any parallel branch. This branch would have an equivalent capacitance of C = C=5 = (2:0106F)=5 = 0:40106F: eq (b) For parallel branches we add, which means we need (1:2106F)=(0:40106F) = 3 parallel branches of the combination found in part (a).

E30-17 Look back at the solution to Ex. 30-10. If C3 breaks down electrically then the circuit is (a) q = (10:3106F)(115 V) = 1:18103C: 1

E30-18 The 108F capacitor originally has a charge of q = (108106F)(52:4 V) = 5:66103C. After it is connected to the second capacitor the 108F capacitor has a charge of q = (108 106F)(35:8 V) = 3:87103C. The di erence in charge must reside on the second capacitor, so the capacitance is C = (1:79103C)=(35:8 V) = 5:00105F:

Consider any junction other than A or B. Call this junction point 0; label the four nearest junctions to this as points 1, 2, 3, and 4. The charge on the capacitor that links point 0 to point 1 is q1 = CV01; where V01 is the potential di erence across the capacitor, so V01 = V0 V1; where V0 is the potential at the junction 0, and V1 is the potential at the junction 1. Similar expressions For the junction 0 the net charge must be zero; there is no way for charge to cross the plates of the capacitors. Then q1 + q2 + q3 + q4 = 0; and this means E30-19

CV01 + CV02 + CV03 + CV04 = 0 or V01 + V02 + V03 + V04 = 0: Let V0i = V0 Vi, and then rearrange,

or 1 V0 = (V1 + V2 + V3 + V4) : 4 E30-20 U = uV = 0E2V=2, where V is the volume. Then

E30-21 The total capacitance is (2100)(5:0106F) = 1:05102F: The total energy stored is 11 U = C(V )2 = (1:05102F)(55103V)2 = 1:59107 J: 22 The cost is $0:03 (1:59107J) = $0:133: 3600103J

22 (b) (3:05106J)=(3600103J=kW h) = 0:847kW h: E30-23 (a) The capacitance of an air lled parallel-plate capacitor is given by Eq. 30-5, 0A (8:851012F=m)(42:0 104m2) C = = = 2:861011 F: d (1:30 103m) (b) The magnitude of the charge on each plate is given by q = CV = (2:861011 F)(625 V) = 1:79108 C: (c) The stored energy in a capacitor is given by Eq. 30-25, regardless of the type or shape of the capacitor, so 11 U = C(V )2 = (2:861011 F)(625 V)2 = 5:59 J: 22 (d) Assuming a parallel plate arrangement with no fringing e ects, the magnitude of the electric eld between the plates is given by Ed = V , where d is the separation between the plates. Then E=V=d=(625V)=(0:00130m)=4:81105V=m: (e) The energy density is Eq. 30-28, 11 u = 0E2 = ((8:851012F=m))(4:81105 V=m)2 = 1:02 J=m3: 22 E30-24 The equivalent capacitance is given by 1=Ceq = 1=(2:12106F) + 1=(3:88106F) = 1=(1:37106F):

The energy stored is U = 1 (1:37106F)(328 V)2 = 7:37102J: 2 E30-25 V=r = q=40r2 = E, so that if V is the potential of the sphere then E = V=r is the electric eld on the surface. Then the energy density of the electric eld near the surface is 12 2 1 (8:8510 F=m) (8150 V) u = 0E2 = = 7:41102J=m3: 2 2 (0:063 m)

E30-26 The charge on C3 can be found from considering the equivalent capacitance. q3 = (3:10 106F)(112 V) = 3:4710 C: The potential across C3 is given by [V ]3 = (3:4710 C)=(3:90 4 4 The potential across the parallel segment is then (112 V) (89:0 V) = 23:0 V. So [V ]1 = Then q1 = (10:3106F)(23:0 V) = 2:37104C and q2 = (4:80106F)(23:0 V) = 1:10104C:.

There is enough work on this problem without deriving once again the electric eld between charged cylinders. I will instead refer you back to Section 26-4, and state 1q 20 Lr E30-27

where q is the magnitude of the charge on a cylinder and L is the length of the cylinders. The energy density as a function of radial distance is found from Eq. 30-28, 1 1 q2 u = 0E2 = 2 82 L2r2 0

The total energy stored in the electric eld is given by Eq. 30-24, 1 q2 q2 ln(b=a) 2 C 2 20L where we substituted into the last part Eq. 30-11, the capacitance of a cylindrical capacitor. p We want to show that integrating a volume integral from r = a to r = ab over the energy density function will yield U=2. Since we want to do this problem the hard way, we will pretend we don’t know the answer, and integrate from r = a to r = c, and then nd out what c is. Then Z 1 2 c 2 L 1 q2 ZZZ 82 L2r2 a00 0 q2 c 2 L dr ZZZ 82 L2 r 0a00 2 Zc q dr 40L a r q2 c = ln : 40L a Now we equate this to the value for U that we found above, and we solve for c.

1 q2 ln(b=a) q2 c 2 2 20L 40L a p ab = c:

E30-28 (a) d = 0A=C, or d = (8:851012F=m)(0:350 m2)=(51:31012F) = 6:04103m: (b) C = (5:60)(51:31012 F) = 2:871010F.

E30-29 Originally, C1 = 0A=d1. After the changes, C2 = 0A=d2. Dividing C2 by C1 yields C2=C1 = d1=d2, so = d C2=d C1 = (2)(2:571012F)=(1:321012F) = 3:89: 21

E30-30 TherequiredcapacitanceisfoundfromU=1C(V)2,or 2 C = 2(6:61106J)=(630 V)2 = 3:331011F: The dielectric constant required is = (3:331011F)=(7:401012F) = 4:50. Try transformer oil.

Capacitance with dielectric media is given by Eq. 30-31, e0A C= : d The various sheets have di erent dielectric constants and di erent thicknesses, and we want to maximize C, which means maximizing e=d. For mica this ratio is 54 mm1, for glass this ratio is 35 mm1, and for paran this ratio is 0.20 mm1. Mica wins.

E30-31 E30-32 The minimum plate separation is given by d = (4:13103V)=(18:2106V=m) = 2:27104m: The minimum plate area is then dC (2:27104m)(68:4109F) A = = = 0:627 m2: 0 (2:80)(8:851012F=m)

E30-33 The capacitance of a cylindrical capacitor is given by Eq. 30-11, 1:0103m C = 2(8:851012F=m)(2:6) = 8:63108F: ln(0:588=0:11)

E30-34 (a) U = C0(V )2=2, C0 = e0A=d, and V=d is less than or equal to the dielectric strength (which we will call S). Then V = Sd and 1 2 so the volume is given by V = 2U=e0S2:

This quantity is a minimum for mica, so V = 2(250103J)=(5:4)(8:851012F=m)(160106V=m)2 = 0:41 m3: (b) e = 2U=V 0S2, so e = 2(250103J)=(0:087m3)(8:851012F=m)(160106V=m)2 = 25:

(a) The capacitance of a cylindrical capacitor is given by Eq. 30-11,

L C = 20e : ln(b=a) The factor of e is introduced because there is now a dielectric (the Pyrex drinking glass) between the plates. We can look back to Table 29-2 to get the dielectric properties of Pyrex. The capacitance of our \glass” is then E30-35

(0:15 m) C = 2(8:851012F=m)(4:7) = 7:31010 F: ln((3:8 cm)=(3:6 cm) (b) The breakdown potential is (14 kV/mm)(2 mm) = 28 kV.

e (a) Insert the slab so that it is a distance a above the lower plate. Then the distance between the slab and the upper plate is d a b. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top The bottom capacitor has a capacitance of C1 = 0A=a; while the top capacitor has a capacitance of C2 = 0A=(d a b): Adding these in series, E30-37

111 Ceq C1 C2 a dab 0A 0A db =: 0A So the capacitance of the system after putting the copper slab in is C = 0A=(d b): (b) The energy stored in the system before the slab is inserted is q2 q2 d Ui = = 2Ci 2 0A while the energy stored after the slab is inserted is q2 q2 d b Uf = = 2Cf 2 0A The ratio is U i=U f = d=(d b): (c) Since there was more energy before the slab was inserted, then the slab must have gone in willingly, it was pulled in!. To get the slab back out we will need to do work on the slab equal to q2 d q2 d b q2 b Ui Uf = = : 2 0A 2 0A 2 0A

E30-38 (a) Insert the slab so that it is a distance a above the lower plate. Then the distance between the slab and the upper plate is d a b. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is a, while that of the top The bottom capacitor has a capacitance of C1 = 0A=a; while the top capacitor has a capacitance of C2 = 0A=(d a b): Adding these in series,

111 Ceq C1 C2 a dab 0A 0A db =: 0A So the capacitance of the system after putting the copper slab in is C = 0A=(d b):

(b) The energy stored in the system before the slab is inserted is C (V )2 (V )2 A i0 Ui = = 2 2d while the energy stored after the slab is inserted is C (V )2 (V )2 A f0 Uf = = 2 2 db The ratio is U i=U f = (d b)=d: (c) Since there was more energy after the slab was inserted, then the slab must not have gone in willingly, it was being repelled!. To get the slab in we will need to do work on the slab equal to the energy di erence.

(V )2 A (V )2 A (V )2 Ab 000 Uf Ui = = : 2 d b 2 d 2 d(d b)

(a) E = V=d = CV=e0A, or (1121012F)(55:0 V) E = = 13400 V=m: (5:4)(8:851012F=m)(96:5104m2) (c) Q0 = Q(1 1=e) = (6:16109C)(1 1=(5:4)) = 5:02109C.

E30-40 (a) E = q=e0A, so (890109C) e = = 6:53 (1:40106V=m)(8:851012F=m)(110104m2) (b) q0 = q(1 1= ) = (890109C)(1 1=(6:53)) = 7:54107C: e

P30-1 The capacitance of the cylindrical capacitor is from Eq. 30-11, 20L C= : ln(b=a) If the cylinders are very close together we can write b = a + d, where d, the separation between the cylinders, is a small number, so 20L 20L C= = : ln ((a + d)=a) ln (1 + d=a) Expanding according to the hint, 20L 2a0L C = d=a d Now 2a is the circumference of the cylinder, and L is the length, so 2aL is the area of a cylindrical plate. Hence, for small separation between the cylinders we have 0A d which is the expression for the parallel plates.

P30-2 (a) C = 0A=x; take the derivative and dC 0 dA 0A dx dT x dT x2 dT 1 dA 1 dx =C : A dT x dT (b) Since (1=A)dA=dT = 2 a and (1=x)dx=dT = s, we need

= 2 = 2(23106=C) = 46106=C: sa Insert the slab so that it is a distance d above the lower plate. Then the distance between the slab and the upper plate is abd. Inserting the slab has the same e ect as having two capacitors wired in series; the separation of the bottom capacitor is d, while that of the top capacitor is abd. The bottom capacitor has a capacitance of C1 = 0A=d; while the top capacitor has a capacitance of C2 = 0A=(a b d): Adding these in series, 111 Ceq C1 C2 d abd 0A 0A ab =: 0A So the capacitance of the system after putting the slab in is C = 0A=(a b): P30-3

P30-4 The potential di erence between any two adjacent plates is V . Each interior plate has a charge q on each surface; the exterior plate (one pink, one gray) has a charge of q on the interior The capacitance of one pink/gray plate pair is C = 0A=d. There are n plates, but only n 1 plate pairs, so the total charge is (n 1)q. This means the total capacitance is C = 0(n 1)A=d.

P30-5 As far as point e is concerned point a looks like it is originally positively charged, and point d is originally negatively charged. It is then convenient to de ne the charges on the capacitors in terms of the charges on the top sides, so the original charge on C1 is q1;i = C1V0 while the original charge on C2 is q2;i = C2V0. Note the negative sign re ecting the opposite polarity of C2. (a) Conservation of charge requires q1;i + q2;i = q1;f + q2;f ;

but since q = CV and the two capacitors will be at the same potential after the switches are closed we can write C1 C2 V0 = V: C1 + C2

With numbers, (1:16 F) (3:22 F) V = (96:6 V) = 45:4 V: (1:16 F) + (3:22 F)

The negative sign means that the top sides of both capacitor will be negatively charged after the (b) The charge on C1 is C1V = (1:16 F)(45:4 V) = 52:7C: (c) The charge on C2 is C2V = (3:22 F)(45:4 V) = 146C:

P30-6 C2 and C3 form an e ective capacitor with equivalent capacitance Ca = C2C3=(C2 + C3). The charge on C1 is originally q0 = C1V0. After throwing the switch the potential across C1 is given by q1 = C1V1. The same potential is across Ca; q2 = q3, so q2 = CaV1. Charge is conserved, so q1 + q2 = q0. Combining some of the above, q0 C1 C1 + Ca C1 + Ca and then C2 C2(C2 + C3) q1 = 1 V0 = 1 V0: C1 + Ca C1C2 + C1C3 + C2C3 Similarly, 1 CaC1 1 1 1 q2 = V0 = + + V0: C1 + Ca C1 C2 C3 q3 = q2 because they are in series.

(a) If terminal a is more positive than terminal b then current can ow that will charge the capacitor on the left, the current can ow through the diode on the top, and the current can charge the capacitor on the right. Current will not ow through the diode on the left. The capacitors are Since the capacitors are identical and series capacitors have the same charge, we expect the capacitors to have the same potential di erence across them. But the total potential di erence across both capacitors is equal to 100 V, so the potential di erence across either capacitor is 50 V. The output pins are connected to the capacitor on the right, so the potential di erence across (b) If terminal b is more positive than terminal a the current can ow through the diode on the left. If we assume the diode is resistanceless in this con guration then the potential di erence across it will be zero. The net result is that the potential di erence across the output pins is 0 V. In real life the potential di erence across the diode would not be zero, even if forward biased. It will be somewhere around 0.5 Volts.

P30-7 P30-8 Divide the strip of width a into N segments, each of width x = a=N . The capacitance of each strip is C = 0ax=y. If is small then 111d = 1 x=d): y d + x sin d + x ( Since parallel capacitances add, 0a a a2 Z X 0 a C = C = (1 x=d)dx = 1 : d 0 d 2d

P30-9 (a) When S2 is open the circuit acts as two parallel capacitors. The branch on the left has an e ective capacitance given by 1111 Cl (1:010 F) (3:0106F) 7:510 F 6 7

while the branch on the right has an e ective capacitance given by 1111 =+=: Cl (2:0106F) (4:0106F) 1:3310 F 6

The charge on either capacitor in the branch on the left is while the charge on either capacitor in the branch on the right is q = (1:33106F)(12 V) = 1:6105C: (b) After closing S2 the circuit is e ectively two capacitors in series. The top part has an e ective capacitance of 6

while the e ective capacitance of the bottom part is C = (3:0106F) + (4:0106F) = (7:0106F): b

The e ective capacitance of the series combination is given by 1111 =+=: C (3:0106F) (7:0106F) 2:1106F eq

The charge on each part is q = (2:1106F)(12 V) = 2:52105C. The potential di erence across the top part is t

and then the charge on the top two capacitors is q1 = (1:0 106F)(8:4 V) = 8:4 106C and q = (2:0106F)(8:4 V) = 1:68105C. The potential di erence across the bottom part is 2

t and then the charge on the top two capacitors is q1 = (3:0 106F)(3:6 V) = 1:08 105C and 2

P30-10 Let V = Vxy. By symmetry V2 = 0 and V1 = V4 = V5 = V3 = V=2. Suddenly the problem is very easy. The charges on each capacitor is q1, except for q2 = 0. Then the equivalent capacitance of the circuit is q q1 + q4 Ceq = = = C1 = 4:0106F: V 2V1

P30-11 (a) The charge on the capacitor with stored energy U0 = 4:0 J is q0, where

q2 U0 = 0 : 2C When this capacitor is connected to an identical uncharged capacitor the charge is shared equally, so that the charge on either capacitor is now q = q0=2. The stored energy in one capacitor is then

q2 q2=4 1 U = = 0 = U0: 2C 2C 4 But there are two capacitors, so the total energy stored is 2U = U0=2 = 2:0 J. (b) Good question. Current had to ow through the connecting wires to get the charge from one capacitor to the other. Originally the second capacitor was uncharged, so the potential di erence across that capacitor would have been zero, which means the potential di erence across the con- necting wires would have been equal to that of the rst capacitor, and there would then have been energy dissipation in the wires according to P = i2R: That’s where the missing energy went.

P30-12 R = L=A and C = 0A=L. Combining, R = 0=C, or R = (9:40 m)(8:851012F=m)=(1101012F) = 0:756 :

2 R (b) U = u dV where dV Then Z 1 e2 e2 1 U = 4pi r2 dr = : R 3220r4 80 R

(c) R = e2=80mc2, or (1:601019C)2 R = = 1:401015m: 8(8:851012F=m)(9:111031kg)(3:00108m=s)2

2 P30-15 According to Problem 14, the force on a plate of a parallel plate capacitor is q2 F= : 20A The force per unit area is then F q2 2 A 20A2 20 where = q=A is the surface charge density. But we know that the electric eld near the surface of a conductor is given by E = =0, so F1 = 0E2: A2

P30-16 A small surface area element dA carries a charge dq = q dA=4R2. There are three forces on the elements which balance, so p(V0=V )dA + q dq=40R2 = p dA;

or pR3 + q2=162 R = pR3: 00 This can be rearranged as q2 = 162 pR(R3 R3): 00

P30-17 The magnitude of the electric eld in the cylindrical region is given by E = =20r; where is the linear charge density on the anode. The potential di erence is given by V = (=20) ln(b=a); where a is the radius of the anode b the radius of the cathode. Combining, E = V =r ln(b=a), this will be a maximum when r = a, so V = (0:180103m) ln[(11:0103m)=(0:180103m)](2:20106V=m) = 1630 V:

P30-18 This is e ectively two capacitors in parallel, each with an area of A=2. Then

0A=2 0A=2 0A e1 + e2 Ceq = e1 + e2 = : ddd2

We will treat the system as two capacitors in series by pretending there is an in nitesi- mally thin conductor between them. The slabs are (I assume) the same thickness. The capacitance of one of the slabs is then given by Eq. 30-31, e10A d=2 P30-19

where d=2 is the thickness of the slab. There would be a similar expression for the other slab. The equivalent series capacitance would be given by Eq. 30-21, 111 Ceq C1 C2 d=2 d=2 e10A e20A d e2 + e1 20A e1e2 20A e1e2 Ceq = : d e2 + e1 P30-20 Treat this as three capacitors. Find the equivalent capacitance of the series combination on the right, and then add on the parallel part on the left. The right hand side is

1 d d 2d e2 + e3 =+=: Ceq e20A=2 e30A=2 0A e2e3 Add this to the left hand side, and

e10A=2 0A e2e3 2d 2d e2 + e3 0A e1 e2e3 =+: 2d 2 e2 + e3

00 (c) W = U 0 U = 2U U = U = 0A(V )2=2d: (a) In the air gap we have 0E2V (8:851012F=m)(6:9103V=m)2(1:15102m2)(4:6103m) Ua = = = 1:11108J: 0 22 (b) The remaining 40% is in the slab.

P30-23 (a) C = 0A=d = (8:851012F=m)(0:118 m2)=(1:2210 m) = 8:5610 F. 2 11 (b) Use the results of Problem 30-24.

(4:8)(8:851012F=m)(0:118 m2) C0 = = 1:191010F (4:8)(1:22102m) (4:3103m)(4:8 1) (c) q = CV = (8:561011F)(120 V) = 1:03108C; since the battery is disconnected q0 = q. (d) E = q=0A = (1:03108C)=(8:851012F=m)(0:118 m2) = 9860 V=m in the space between (g) W = U0 U = q2(1=C 1=C0)=2, or (1:03108C)2 W = [1=(8:561011F) 1=(1:191010F)] = 1:73107J: 2

P30-24 The result is e ectively three capacitors in series. Two are air lled with thicknesses of x and d b x, the third is dielectric lled with thickness b. All have an area A. The e ective capacitance is given by 1 x dbx b C 0A 0A e0A 1b 0A e

E31-1 (5:12 A)(6:00 V)(5:75 min)(60 s/min) = 1:06104J: E31-3 If the energy is delivered at a rate of 110 W, then the current through the battery is P (110 W) i = = = 9:17 A: V (12 V) Current is the ow of charge in some period of time, so q (125 A h) i (9:2 A) which is the same as 13 hours and 36 minutes.

(a) (800 W h)=(2:0 W h) = 400 batteries, at a cost of (400)($0:80) = $320. (b) (800 W h)($0:12103 W h) = $0:096:

Go all of the way around the circuit. It is a simple one loop circuit, and although it does not matter which way we go around, we will follow the direction of the larger emf. Then E31-5

where i is positive if it is counterclockwise. Rearranging, Assuming the potential at P is VP = 100 V, then the potential at Q will be given by

VQ = VP (50 V) i(3:0 ) = (100 V) (50 V) (20 A)(3:0 ) = 10 V:

E31-6 (a) Req = (10 ) + (140 ) = 150 . i = (12:0 V)=(150 ) = 0:080 A. (c) Req = (10 ) + (20 ) = 30 . i = (12:0 V)=(30 ) = 0:400 A.

E31-7 (a) Req = (3:0 V 2:0 V)=(0:050 A) = 20 . Then R = (20 ) (3:0 ) (3:0 ) = 14 . (b) P = iV = i2R = (0:050 A)2(14 ) = 3:5102W.

(b) (1:20 A)(19:0 ) = 22:8 V is the potential di erence across R. Then an additional potential (c) The left side is positive; it is a reverse emf.

p E31-10 (a) The current in the resistor is (9:88 W)=(0:108 ) = 9:56 A. The total resistance of the circuit is (1:50 V)=(9:56 A) = 0:157 . The internal resistance of the battery is then (0:157 ) (b) (9:88 W)=(9:56 A) = 1:03 V.

E31-11 1 2 a b

3 4 Since the ammeter has no resistance the potential at a is the same as the potential at b. Con- sequently the potential di erence (V b) across both of the bottom resistors is the same, and the potential di erence (V t) across the two top resistors is also the same (but di erent from the bottom). We then have the following relationships: Vj = ij Rj ;

where the j subscript in the last line refers to resistor 1, 2, 3, or 4. For the top resistors, while for the bottom resistors, V3 = V4 implies i3 = i4: Then the junction rule requires i4 = 3i1=2, and the loop rule requires (i1)(2R) + (3i1=2)(R) = E or i1 = 2E=(7R): The current that ows through the ammeter is the di erence between i2 and i4, or 4E=(7R) 3E =(7R) = E =(7R).

E31-12 (a) De ne the current i1 as moving to the left through r1 and the current i2 as moving to the left through r2. i3 = i1 + i2 is moving to the right through R. Then there are two loop equations: E2 = (i3 i1)r2 + i3R: Multiply the top equation by r2 and the bottom by r1 and then add: r2E1 + r1E2 = i3r1r2 + i3R(r1 + r2);

which can be rearranged as r2E1 + r1E2 i3 = : r1r2 + Rr1 + Rr2 (b) There is only one current, so E1 + E2 = i(r1 + r2 + R);

(a) Assume that the current ows through each source of emf in the same direction as the emf. The the loop rule will give us three equations E31-13

E1 i1R1 + i3R1 E3 + i3R1 i1R1 = 0: The junction rule (looks at point a) gives us i1 + i2 + i3 = 0. Use this to eliminate i2 from the second loop equation, E2 + i1R2 + i3R2 + 2i3R1 E3 = 0;

and then combine this with the the third equation to eliminate i3, E1R2 E3R2 + 2i3R1R2 + 2E2R1 + 2i3R1R2 + 4i3R12 2E3R1 = 0;

or 2E3R1 + E3R2 E1R2 2E2R1 i3 = = 0:582 A: 4R1R2 + 4R2 1

Then we can nd i1 from E3 E2 i3R2 2i3R1 R2 Finally, we can nd i2 = (i1 + i3) = 0:0854 A: (b) Start at a and go to b ( nal minus initial!), +i2R2 E2 = 3:60 V:

E31-14 (a) The current through the circuit is i = E =(r + R). The power delivered to R is then P = iV = i2R = E2R=(r + R)2: Evaluate dP=dR and set it equal to zero to nd the maximum. Then dP r R dR (r + R)3 (b) When r = R the power is 1 E2 P = E2R = : (R + R)2 4r E31-15 (a) We rst use P = F v to nd the power output by the electric motor. Then P = The potential di erence across the motor is V m = E ir: The power output from the motor is the rate of energy dissipation, so P m = V mi: Combining these two expressions, 0 = (0:50 )i2 (2:0 V)i + (1:0 W):

Rearrange and solve for i, p (2:0 V) (2:0 V)2 4(0:50 )(1:0 W) i= ; 2(0:50 )

(b) The potential di erence across the terminals of the motor is V m = E ir which if i = 3:4 A yields V m = 0:3 V, but if i = 0:59 A yields V m = 1:7 V. The battery provides an emf of 2.0 V; it isn’t possible for the potential di erence across the motor to be larger than this, but both solutions seem to satisfy this constraint, so we will move to the next part and see what happens. (c) So what is the signi cance of the two possible solutions? It is a consequence of the fact that power is related to the current squared, and with any quadratics we expect two solutions. Both are possible, but it might be that only one is stable, or even that neither is stable, and a small perturbation to the friction involved in turning the motor will cause the system to break down. We will learn in a later chapter that the e ective resistance of an electric motor depends on the speed at which it is spinning, and although that won’t a ect the problem here as worded, it will a ect the physical problem that provided the numbers in this problem!

In parallel connections of two resistors the e ective resistance is less than the smaller resistance but larger than half the smaller resistance. In series connections of two resistors the e ective resistance is greater than the larger resistance but less than twice the larger resistance. Since the e ective resistance of the parallel combination is less than either single resistance and the e ective resistance of the series combinations is larger than either single resistance we can conclude that 3:0 must have been the parallel combination and 16 must have been the series The resistors are then 4:0 and 12 resistors.

E31-17 E31-18 Points B and C are e ectively the same point! (c) 0, since there is no resistance between B and C.

E31-19 Focus on the loop through the battery, the 3:0 , and the 5:0 resistors. The loop rule yields (12:0 V) = i[(3:0 ) + (5:0 )] = i(8:0 ): The potential di erence across the 5:0 resistor is then V = i(5:0 ) = (5:0 )(12:0 V)=(8:0 ) = 7:5 V:

E31-20 Each lamp draws a current of (500 W)=(120 V) = 4:17 A. Furthermore, the fuse can support (15 A)=(4:17 A) = 3:60 lamps. That is a maximum of 3.

E31-21 The current in the series combination is is = E=(R1 + R2). The power dissipated is Ps=iE=E2=(R1+R2): In a parallel arrangement R1 dissipates P1 = i1E = E2=R1. A similar expression exists for R2, sothetotalpowerdissipatedisPp=E2(1=R1+1=R2): The ratio is 5, so 5 = P p=P s = (1=R1 + 1=R2)(R1 + R2), or 5R1R2 = (R1 + R2)2. Solving for R2 yields 2:618R1 or 0:382R1. Then R2 = 262 or R2 = 38:2 .

E31-22 Combining n identical resistors in series results in an equivalent resistance of req = nR. Combining n identical resistors in parallel results in an equivalent resistance of req = R=n. If the resistors are arranged in a square array consisting of n parallel branches of n series resistors, then the e ective resistance is R. Each will dissipate a power P , together they will dissipate n2P . So we want nine resistors, since four would be too small.

E31-23 (a) Work through the circuit one step at a time. We rst \add” R2, R3, and R4 in parallel: 11111 =++= Re 42:0 61:6 75:0 18:7 We then \add” this resistance in series with R1, Re = (112 ) + (18:7 ) = 131 : (b) The current through the battery is i = E =R = (6:22 V)=(131 ) = 47:5 mA. This is also the current through R1, since all the current through the battery must also go through R1. The potential di erence across R1 is V1 = (47:5 mA)(112 ) = 5:32 V. The potential di erence across each of the three remaining resistors is 6:22 V 5:32 V = 0:90 V. The current through each resistor is then i4 = (0:90 V)=(75:0 ) = 12:0 mA:

E31-24 The equivalent resistance of the parallel part is r0 = R2R=(R2 + R): The equivalent resistance for the circuit is r = R1 + R2R=(R2 + R): The current through the circuit is i0 = E=r. The potential di erence across R is V = E i0R1, or V = E(1 R1=r);

R2 + R R1R2 + R1R + RR2 RR2 =E : R1R2 + R1R + RR2

Since P = iV = (V )2=R, RR2 P = E2 2 : (R1R2 + R1R + RR2)2 Set dP=dR = 0, the solution is R = R1R2=(R1 + R2).

(a) First \add” the left two resistors in series; the e ective resistance of that branch is 2R. Then \add” the right two resistors in series; the e ective resistance of that branch is also 2R. Now we combine the three parallel branches and nd the e ective resistance to be 11114 Re 2R R 2R 2R (b) First we \add” the right two resistors in series; the e ective resistance of that branch is 2R. We then combine this branch with the resistor which connects points F and H . This is a parallel connection, so the e ective resistance is 1113 Re 2R R 2R E31-25

This value is e ectively in series with the resistor which connects G and H , so the \total” is Finally, we can combine this value in parallel with the resistor that directly connects F and G according to 1138 Re R 5R 5R or Re = 5R=8.

E31-26 The resistance of the second resistor is r2 = (2:4 V)=(0:001 A) = 2400 . The potential di erence across the rst resistor is (12 V) (2:4 V) = 9:6 V. The resistance of the rst resistor is (9:6 V)=(0:001 A) = 9600 .

E31-27 See Exercise 31-26. The resistance ratio is r1 (0:95 0:1 V) r1 + r2 (1:50 V) or r2 (1:50 V) = 1: r1 (0:95 0:1 V) We can choose any standard resistors we want, and we could use any tolerance, but then we will need to check our results. 22 and 39 would work; as would 27 and 47 . There are other choices.

Consider any junction other than A or B. Call this junction point 0; label the four nearest junctions to this as points 1, 2, 3, and 4. The current through the resistor that links point 0 to point 1 is i1 = V01=R; where V01 is the potential di erence across the resistor, so V01 = V0 V1; where V0 is the potential at the junction 0, and V1 is the potential at the junction For the junction 0 the net current must be zero; there is no way for charge to accumulate on the junction. Then i1 + i2 + i3 + i4 = 0; and this means E31-28

V01=R + V02=R + V03=R + V04=R = 0 or V01 + V02 + V03 + V04 = 0: Let V0i = V0 Vi, and then rearrange,

or 1 V0 = (V1 + V2 + V3 + V4) : 4 The current through the radio is i = P=V = (7:5 W)=(9:0 V) = 0:83 A. The radio was left one for 6 hours, or 2:16104 s. The total charge to ow through the radio in that time is (0:83 A)(2:16104 s) = 1:8104 C.

E31-29 E31-30 The power dissipated by the headlights is (9:7 A)(12:0 V) = 116 W. The power required by the engine is (116 W)=(0:82) = 142 W, which is equivalent to 0:190 hp.

E31-31 (a) P = (120 V)(120 V)=(14:0 ) = 1030 W: E31-32 E31-33 We want to apply either Eq. 31-21, PR = i2R;

or Eq. 31-22, depending on whether we are in series (the current is the same through each bulb), or in parallel (the potential di erence across each bulb is the same. The brightness of a bulb will be measured by P , even though P is not necessarily a measure of the rate radiant energy is emitted from the bulb. (b) If the bulbs are in parallel then PR = (VR)2=R is how we want to compare the brightness. The potential di erence across each bulb is the same, so the bulb with the smaller resistance is (b) If the bulbs are in series then PR = i2R is how we want to compare the brightness. Both bulbs have the same current, so the larger value of R results in the brighter bulb. One direct consequence of this can be tried at home. Wire up a 60 W, 120 V bulb and a 100 W, 120 V bulb in series. Which is brighter? You should observe that the 60 W bulb will be brighter.

E31-35 (a) The bulb is on for 744 hours. The energy consumed is (100 W)(744 h) = 74:4 kW h, (c) i = P=V = (100 W)=(120 V) = 0:83 A.

E31-36 P = (V )2=r and r = r0(1 + T ). Then P0 (500 W) P = = = 660 W 1 + T 1 + (4:0104=C)(600C)

E31-37 (a) n = q=e = it=e, so n = (485103A)(95109s)=(1:61019C) = 2:881011:

(c) P p = ipV = (485103A)(47:7106V) = 2:3106W; while P = i V = (2:4105A)(47:7 aa 106V) = 1:14103W.

pp (a) i = P=r = (1:55 W)=(8:33103 ) = 13:6 A, so j = i=A = (13:6 A)=(5:12103m)2 = 1:66105A=m2: pp (b) V = P r = (1:55 W)(8:33103 ) = 0:114 V:

E31-39 (a) The current through the wire is The resistance of the wire is R = V =i = (75 V)=(64 A) = 1:17 : The length of the wire is then found from RA (1:17 )(2:6106 m2) L = = = 6:1 m: (5:0107 m) One could easily wind this much nichrome to make a toaster oven. Of course allowing 64 Amps to (b) We want to combine the above calculations into one formula, so RA AV =i A(V )2 P then (2:6106 m2)(110 V)2 L = = 13 m: (4800 W)(5:0107 m) Hmm. We need more wire if the potential di erence is increased? Does this make sense? Yes, it does. We need more wire because we need more resistance to decrease the current so that the same power output occurs.

E31-40 (a) The energy required to bring the water to boiling is Q = mCT. The time required is Q (2:1 kg)(4200 J=kg)(100C 18:5C) t = = = 2:22103s 0:77P 0:77(420 W) (b) The additional time required to boil half of the water away is mL=2 (2:1 kg)(2:26106J=kg)=2 t = = = 7340 s: 0:77P 0:77(420 W) E31-41 (a) Integrate both sides of Eq. 31-26;

Z q Z t dt dq 0 q EC 0 RC t t 0 RC 0 q EC t EC RC q EC EC

(b) Rearrange Eq. 31-26 in order to get q terms on the left and t terms on the right, then integrate;

q t dt ZZ dq q 0 RC q0 t t q0 RC 0 qt q0 RC q q0 q = q0et=RC :

6 (c) t = C ln(1 q=q0), so t = (2:56 s) ln(1 15:5106C=1:98105C) = 3:91 s:

E31-44 (a) V = E (1 et= C ), so C = (1:28106s)= ln(1 5:00 V=13:0 V) = 2:64106s (b) C = =R = (2:64106s)=(15:2103 ) = 1:731010F C

E31-45 (a) V = E et= C , so C = (10:0 s)= ln(1:06 V=100 V) = 2:20 s (b) V = (100 V)e17 s=2:20 s = 4:4102V.

E31-46 V = Eet= C and C = RC, so ttt R= = = : Cln(V=V0) (220109F)ln(0:8V=5V) 4:03107F If t is between 10:0 s and 6:0 ms, then R is between R = (10106s)=(4:03107F) = 24:8 ;

and R = (6103s)=(4:03107F) = 14:9103 : The charge on the capacitor needs to build up to a point where the potential across the capacitor is VL = 72 V, and this needs to happen within 0.5 seconds. This means that we want to solve E31-47

CVL = CE 1 eT=RC for R knowing that T = 0:5 s. This expression can be written as T (0:5 s) R = = = 2:35106 : Cln(1VL=E) (0:15C)ln(1(72V)=(95V))

pp (c) VC = V0et= , so C (1103C) 6 6 V = et=(1:010 )(1:010 F) = (1000 V)et=(1:0 s) C (1:0106F) (d) PR = (VR)2=R, so PR = (1000 V)2e2t=(1:0 s)=(1106!) = (1 W)e2t=(1:0 s):

E31-49 (a) i = dq=dt = E et= C =R, so (4:0 V) 6 6 i = e(1:0 s)=(3:010 )(1:010 F) = 9:55107A: (3:0106 ) (b) PC = iV = (E =R)et= (1 e ), so 2 C t= C

(4:0 V)2 6 6 6 6 PC = e(1:0 s)=(3:010 )(1:010 F) 1 e(1:0 s)=(3:010 )(1:010 F) = 1:08106W: (3:0106 ) (c) PR = i2R = (E2=R)e2t= C, so (4:0 V)2 6 6 PR = e2(1:0 s)=(3:010 )(1:010 F) = 2:74106W: (3:0106 )

(d) P = PR + PC, or P = 2:74106W + 1:08106W = 3:82106W E31-50 The rate of energy dissipation in the resistor is P = i2R = (E2=R)e2t= C: R

Evaluating Z 1 E2 Z 1 E2 0 R0 2 but that is the original energy stored in the capacitor.

P31-1 The terminal voltage of the battery is given by V = E ir; so the internal resistance is E V (12:0 V) (11:4 V) i (50 A) The resistance of the wire is given by V (3:0 V) i (50 A) What about the motor? Trying it, V (11:4 V) (3:0 V) i (50 A) so it appears to be within spec.

P31-2 Traversing the circuit we have so i = 2E =(r1 + r2 + R): The potential di erence across the rst battery is then

+ r2 + r2 + R This quantity will only vanish if r2 r1 + R = 0, or r1 = R + r2. Since r1 > r2 this is actually possible; R = r1 r2.

2r1 r2 r1 + R V1 = E ir1 = E 1 = E r1 + R r1

P31-3 V = E iri and i = E=(ri + R), so R ri + R There are then two simultaneous equations: (0:10 V)(500 ) + (0:10 V)ri = E (500 )

and with solution (a) ri = 1:5103 and (c) The cell receives energy from the sun at a rate (2:0 mW/cm2)(5:0 cm2) = 0:010 W. The cell converts energy at a rate of V 2=R = (0:16 V)2=(1000 ) = 0:26 %

P31-4 (a) The emf of the battery can be found from E = iri + V l = (10 A)(0:05 ) + (12 V) = 12:5 V (b) Assume that resistance is not a function of temperature. The resistance of the headlights is then rl = (12:0 V)=(10:0 A) = 1:2 : The potential di erence across the lights when the starter motor is on is V l = (8:0 A)(1:2 ) = 9:6 V;

and this is also the potential di erence across the terminals of the battery. The current through the battery is then E V (12:5 V) (9:6 V) ri (0:05 ) so the current through the motor is 50 Amps.

P31-5 (a) The resistivities are AA and B=rBA=L=(35:0106 )(91:0104m2)=(42:6m)=7:48109 m: (b)Thecurrentisi=V=(rA+rB)=(630V)=(111:2)=5:67106A.Thecurrentdensity is then j = (5:67106A)=(91:0104m2) = 6:23108A=m2: (c)EA=Aj=(1:63108 m)(6:23108A=m2)=10:2V=mandEB=Bj=(7:48109 (d)VA=EAL=(10:2V=m)(42:6m)=435VandVB=EBL=(4:66V=m)(42:6m)=198V.

P31-6 Set up the problem with the traditional presentation of the Wheatstone bridge problem. Then the symmetry of the problem ( ip it over on the line between x and y) implies that there is no current through r. As such, the problem is equivalent to two identical parallel branches each with Each branch has resistance R + R = 2R, so the overall circuit has resistance 1111 Req 2R 2R R

P31-7 P31-8 (a) The loop through R1 is trivial: i1 = E2=R1 = (5:0 V)=(100 ) = 0:05 A. The loop (b) Vab = E3 + E2 = (5:0 V) + (4:0 V) = 9:0 V.

(a) The three way light-bulb has two laments (or so we are told in the question). There are four ways for these two laments to be wired: either one alone, both in series, or both in parallel. Wiring the laments in series will have the largest total resistance, and since P = V 2=R this arrangement would result in the dimmest light. But we are told the light still operates at the lowest setting, and if a lament burned out in a series arrangement the light would go out. We then conclude that the lowest setting is one lament, the middle setting is another lament, (b) The beauty of parallel settings is that then power is additive (it is also addictive, but that’s a di erent eld.) One lament dissipates 100 W at 120 V; the other lament (the one that burns out) dissipates 200 W at 120 V, and both together dissipate 300 W at 120 V. The resistance of one lament is then (V )2 (120 V)2 R = = = 144 : P (100 W) P31-9

The resistance of the other lament is (V )2 (120 V)2 R = = = 72 : P (200 W) P31-10 We can assume that R \contains” all of the resistance of the resistor, the battery and the ammeter, then R = (1:50 V)=(1:0 m=A) = 1500 : For each of the following parts we apply R + r = V =i, so (a) r = (1:5 V)=(0:1 mA) (1500 ) = 1:35104 , (b) r = (1:5 V)=(0:5 mA) (1500 ) = 1:5103 , (d) R = (1500 ) (18:5 ) = 1482

P31-11 (a) The e ective resistance of the parallel branches on the middle and the right is R2R3 : R2 + R3

The e ective resistance of the circuit as seen by the battery is then R2R3 R1R2 + R1R3 + R2R3 R2 + R3 R2 + R3 The current through the battery is R2 + R3 R1R2 + R1R3 + R2R3 The potential di erence across R1 is then R2 + R3 R1R2 + R1R3 + R2R3 while V3 = E V1, or R2R3 R1R2 + R1R3 + R2R3 so the current through the ammeter is V3 R2 R3 R1R2 + R1R3 + R2R3 or (4 ) i3 = (5:0 V) = 0:45 A: (2 )(4 ) + (2 )(6 ) + (4 )(6 ) (b) Changing the locations of the battery and the ammeter is equivalent to swapping R1 and R3. But since the expression for the current doesn’t change, then the current is the same.

P31-12 V1 + V2 = VS + VX; if Va = Vb, then V1 = VS. Using the rst expression, ia(R1 + R2) = ib(RS + RX );

using the second, iaR1 = ibR2: Dividing the rst by the second, 1 + R2=R1 = 1 + RX=RS;

P31-13 P31-14 Lv = Q=m and Q=t = P = iV , so iV (5:2 A)(12 V) Lv = = = 2:97106J=kg: m=t (21106kg=s)

P31-15 P = i2R. W = pV , where V is volume. p = mg=A and V = Ay, where y is the height of the piston. Then P = dW=dt = mgv. Combining all of this, i2R (0:240 A)2(550 ) v = = = 0:274 m=s: mg (11:8 kg)(9:8 m=s2)

P31-16 (a) Since q = CV , then q = (32106F) (6 V) + (4 V=s)(0:5 s) (2 V=s2)(0:5 s)2 4 = 2:410 C: (b) Since i = dq=dt = C dV =dt, then i = (32106F) (4 V=s) 2(2 V=s2)(0:5 s) = 6:4105A:

(c) Since P = iV , P = [(4 V=s) 2(2 V=s2)(0:5 s) (6 V) + (4 V=s)(0:5 s) (2 V=s2)(0:5 s)2 = 4:8104W:

P31-17 (a) We have P = 30P0 and i = 4i0. Then P 30P0 30 R = = = R0: i2 (4i0)2 16 We don’t really care what happened with the potential di erence, since knowing the change in The volume of the wire is a constant, even upon drawing the wire out, so LA = L0A0; the Resistance is given by R = L=A, but A = L0A0=L, so the length of the wire is ss A0L0R 30 A0L0R0 L = = = 1:37L0: 16

(b) We know that A = L0A0=L, so L A0 A = A0 = = 0:73A0: L0 1:37 P31-18 (a) The capacitor charge as a function of time is given by Eq. 31-27,

while the current through the circuit (and the resistor) is given by Eq. 31-28,

E i = et=RC : R The energy supplied by the emf is ZZ (b) Integrating, Z E2 Z E2 Eq UR = i2Rdt = e2t=RC dt = = : R 2C 2

P31-19 The capacitor charge as a function of time is given by Eq. 31-27,

while the current through the circuit (and the resistor) is given by Eq. 31-28,

E i = et=RC : R The energy stored in the capacitor is given by q2 2C so the rate that energy is being stored in the capacitor is dU q dq q PC = = = i: dt C dt C The rate of energy dissipation in the resistor is PR = i2R;

so the time at which the rate of energy dissipation in the resistor is equal to the rate of energy storage in the capacitor can be found by solving q C iRC = q;

E32-1 All of the paths which involve left hand turns are positive particles (path 1); those paths which involve right hand turns are negative particle (path 2 and path 4); and those paths which don’t turn involve neutral particles (path 3).

E32-2 (a) The greatest magnitude of force is F = qvB = (1:61019C)(7:2106m=s)(83103T) = (b) The force on the electron is F = ma; the angle between the velocity and the magnetic eld is , given by ma = qvB sin . Then (9:11031kg)(4:91016m=s2) = arcsin = 28: (1:61019C)(7:2106m=s)(83103T)

E32-4 (a) v = F=qB sin = (6:481017N=(1:601019C)(2:63103T) sin(23:0) = 3:94105m=s: (b) K = mv2=2 = (938 MeV=c2)(3:94105m=s)2=2 = 809 eV: E32-5 The magnetic force on the proton is F = qvB = (1:61019 C)(2:8107 m=s)(30eex6 T) = 1:31016N: B

The gravitational force on the proton is mg = (1:71027kg)(9:8 m=s2) = 1:71026 N: The ratio is then 7:6109. If, however, you carry the number of signi cant digits for the intermediate answers farther you will get the answer which is in the back of the book.

p E32-6 The speed of the electron is given by v = 2qV =m, or p v = 2(1000 eV)=(5:1105 eV=c2) = 0:063c: The electric eld between the plates is E = (100 V)=(0:020 m) = 5000 V=m. The required magnetic eld is then B = E=v = (5000 V=m)=(0:063c) = 2:6104T:

E32-7 Both have the same velocity. Then K =K = m v2=m v2 = mp=me =. pepe

p E32-8 The speed of the ion is given by v = 2qV =m, or v = p2(10:8 keV)=(6:01)(932 MeV=c2) = 1:96103c: The required electric eld is E = vB = (1:96103c)(1:22 T) = 7:17105V=m.

E32-9 mv (9:111031kg)(0:1)(3:00108 m=s) r = = = 3:4104 m: jqjB (1:61019 C)(0:50 T) (b) The (non-relativistic) kinetic energy of the electron is 11 K = mv2 = (0:511 MeV)(0:10c)2 = 2:6103 MeV: 22

pp (b) B = mv=qr = (9:111031kg)(0:0691c)=(1:601019C)(0:247 m) = 4:78104T: (c) f = qB=2m = (1:601019C)(4:78104T)=2(9:111031kg) = 1:33107 Hz: (d) T = 1=f = 1=(1:33107 Hz) = 7:48108s: pp (b) r = mv=qB = (9:111031kg)(0:037c)=(1:601019C)(0:20T) = 3:16104m:

E32-12 The frequency is f = (7:00)=(1:29103s) = 5:43103 Hz. The mass is given by m = qB=2f , or (1:601019C)(45:0103T) m = = 2:111025kg = 127 u: 2(5:43103 Hz) E32-13 (a) Apply Eq. 32-10, but rearrange it as jqjrB 2(1:61019 C)(0:045 m)(1:2 T) v = = = 2:6106 m=s: m 4:0(1:661027kg) (b) The speed is equal to the circumference divided by the period, so 2r 2m 24:0(1:661027kg) T = = = = 1:1107 s: v jqjB 2(1:6 1019 C)(1:2 T) (c) The (non-relativistic) kinetic energy is jqj2r2B (21:61019 C)2(0:045 m)2(1:2 T)2 K = = = 2:241014J: 2m 2(4:01:661027kg)) To change to electron volts we need merely divide this answer by the charge on one electron, so (2:241014J) K = = 140 keV: (1:61019 C) (d) V = K = (140 keV)=(2e) = 70 V: q

E32-14 (a) R = mv=qB = (938 MeV=c2)(0:100c)=e(1:40 T) = 0:223 m: (b) f = qB=2m = e(1:40 T)=2(938 MeV=c2) = 2:13107 Hz:

E32-15 (a) K =Kp = (q2=m )=(qp2=mp) = 22=4 = 1: (b) Kd=Kp = (qd2=md)=(qp2=mp) = 12=2 = 1=2:

pp (b) r = sqrt2mK=qB. Then rd=rp = (2=1)(1=1)=(1=1) = 2: pp (c) r = sqrt2mK=qB. Then r =rp = (4=1)(2=1)=(2=1) = 2: p pp r = 2mK=jqjB = ( m=jqj)( 2K=B): All three particles are traveling with the same kinetic energy in the same magnetic eld. The relevant factors are in front; we just need to compare p (a) The radius of the deuteron path is 2 rp: 1p 2 E32-17

E32-18 The neutron, being neutral, is una ected by the magnetic eld and moves o in a line tangent to the original path. The proton moves at the same original speed as the deuteron and has the same charge, but since it has half the mass it moves in a circle with half the radius.

E32-19 (a) The proton momentum would be pc = qcBR = e(3:0108m=s)(41106T)(6:4106m) = 7:9104 MeV: Since 79000 MeV is much, much greater than 938 MeV the proton is ultra-relativistic. Then E pc, and since = E=mc2 we have = p=mc. Inverting, s r 2c2 v 1 m2c2 m = 1 = 1 1 0:99993: c 2 p2 2p2

p E32-20 (a) Classically, R = 2mK=qB, or R = p2(0:511 MeV=c2)(10:0 MeV)=e(2:20 T) = 4:84103m: (b) This would be an ultra-relativistic electron, so K E pc, then R = p=qB = K=qBc, or R = (10:0 MeV)=e(2:2 T)(3:00108m=s) = 1:52102m: (c) The electron is e ectively traveling at the speed of light, so T = 2R=c, or T = 2(1:52102m)=(3:00108 m=s) = 3:181010s: This result does depend on the speed! E32-21 Use Eq. 32-10, except we rearrange for the mass, jqjrB 2(1:601019 C)(4:72 m)(1:33 T) m = = = 9:431027 kg v 0:710(3:00108 m=s) However, if it is moving at this velocity then the \mass” which we have here is not the true mass, but a relativistic correction. For a particle moving at 0:710c we have 11 1 v2=c2 1 (0:710)2 so the true mass of the particle is (9:431027 kg)=(1:42) = 6:641027kg: The number of nucleons present in this particle is then (6:641027kg)=(1:671027 kg) = 3:97 4: The charge was +2, which implies two protons, the other two nucleons would be neutrons, so this must be an alpha particle.

E32-22 (a) Since 950 GeV is much, much greater than 938 MeV the proton is ultra-relativistic. = E=mc2, so r m2c4 r v 1 m2c4 = 1 = 1 1 0:9999995: c 2 E2 2E2 (b) Ultra-relativistic motion requires pc E, so B = pc=qRc = (950 GeV)=e(750 m)(3:00108m=s) = 4:44 T:

E32-23 First use 2f = qB=m. The use K = q2B2R2=2m = mR2(2f )2=2. The number of p turns is n = K=2qV , on average the particle is located at a distance R= 2 from the center, so the pp distance traveled is x = n2R= 2 = n 2R. Combining, pp 23R3mf2 23(0:53m)3(2932103keV=c2)(12106=s)2 x = = = 240 m: qV e(80 kV) E32-24 The particle moves in a circle. x = R sin !t and y = R cos !t.

We will use Eq. 32-20, EH = vdB, except we will not take the derivation through to Eq. 32-21. Instead, we will set the drift velocity equal to the speed of the strip. We will, however, set EH = V H=w. Then E V =w (3:9106V)=(0:88102m) H H 1 v = = = = 3:710 m=s: B B (1:2103T) E32-26 (a) v = E=B = (40106V)=(1:2102m)=(1:4 T) = 2:4103m=s: (b) n = (3:2 A)(1:4 T)=(1:61019C)(9:5106m)(40106V) = 7:41028=m3:; Silver.

(b) (0:65 T)=(8:491028=m3)(1:601019C)(1:69108 m) = 0:0028: E32-25

E32-29 Since L~ is perpendicular to B~ can use FB = iLB: Equating the two forces, mg (0:0130 kg)(9:81 m=s2) i = = = 0:467 A: LB (0:620 m)(0:440 T) Use of an appropriate right hand rule will indicate that the current must be directed to the right in order to have a magnetic force directed upward.

E32-30 F = iLB sin = (5:12 103A)(100 m)(58 106T) sin(70) = 27:9 N. The direction is horizontally west.

(a) We use Eq. 32-26 again, and since the (horizontal) axle is perpendicular to the vertical component of the magnetic eld, F (10; 000 N) i = = = 3:3108A: BL (10 T)(3:0 m) (b) The power lost per ohm of resistance in the rails is given by P=r = i2 = (3:3108A)2 = 1:11017 W: (c) If such a train were to be developed the rails would melt well before the train left the station.

E32-32 F = idB, so a = F=m = idB=m. Since a is constant, v = at = idBt=m. The direction is to the left.

Only the ^j component of B~ is of interest. R R E32-33 Then F = dF = i By dx, or Z 3:2 F = (5:0 A)(8103T=m2) x2 dx = 0:414 N: 1:2

E32-34 The magnetic force will have two components: one will lift vertically (Fy = F sin ), the other push horizontally (Fx = F cos ). The rod will move when Fx > (W Fy). We are interested in the minimum value for F as a function of . This occurs when dF d W = = 0: d d cos + sin This happens when = tan . Then = arctan(0:58) = 30, and (0:58)(1:15 kg)(9:81 m=s2) F = = 5:66 N cos(30) + (0:58) sin(30) is the minimum force. Then B = (5:66 N)=(53:2 A)(0:95 m) = 0:112 T.

We choose that the eld points from the shorter side to the longer side. (a) The magnetic eld is parallel to the 130 cm side so there is no magnetic force on that side. The magnetic force on the 50 cm side has magnitude FB = iLB sin ;

where is the angle between the 50 cm side and the magnetic eld. This angle is larger than 90, but the sine can be found directly from the triangle, (120 cm) (130 cm) and then the force on the 50 cm side can be found by (120 cm) (130 cm) The magnetic force on the 120 cm side has magnitude FB = iLB sin ;

where is the angle between the 1200 cm side and the magnetic eld. This angle is larger than 180, but the sine can be found directly from the triangle, (50 cm) (130 cm) and then the force on the 50 cm side can be found by (50 cm) (130 cm) (b) Look at the three numbers above.

E32-36 = N iAB sin , so = (20)(0:1 A)(0:12 m)(0:05 m)(0:5 T) sin(90 33) = 5:0103N m:

E32-37 The external magnetic eld must be in the plane of the clock/wire loop. The clockwise current produces a magnetic dipole moment directed into the plane of the clock. (a) Since the magnetic eld points along the 1 pm line and the torque is perpendicular to both the external eld and the dipole, then the torque must point along either the 4 pm or the 10 pm line. Applying Eq. 32-35, the direction is along the 4 pm line. It will take the minute hand 20 minutes (b) = (6)(2:0 A)(0:15 m)2(0:07 T) = 0:059 N m:

P32-1 Since F~ must be perpendicular to B~ then B~ must be along k^. The magnitude of v is pp (40)2 + (35)2 km/s = 53:1 km/s; the magnitude of F is (4:2)2 + (4:8)2fN = 6:38 fN: Then B = F=qv = (6:381015N)=(1:61019C)(53:1103m=s) = 0:75 T: or B~ = 0:75 T k^:

P32-2 ~a = (q=m)(E~ + ~v B~ ). For the initial velocity given, ~v B~ = (15:0103m=s)(400106T)^j (12:0103m=s)(400106T)k^: But since there is no acceleration in the ^j or k^ direction this must be o set by the electric eld. Consequently, two of the electric eld components are Ey = 6:00 V=m and Ez = 4:80 V=m. The third component of the electric eld is the source of the acceleration, so E = ma =q = (9:111031kg)(2:001012m=s2)=(1:601019C) = 11:4 V=m: xx

(a) Consider rst the cross product, ~v B~ . The electron moves horizontally, there is a component of the B~ which is down, so the cross product results in a vector which points to the left But the force on the electron is given by F~ = q~v B~ , and since the electron has a negative charge the force on the electron would be directed to the right of the electron’s path. (b) The kinetic energy of the electrons is much less than the rest mass energy, so this is non- p relativistic motion. The speed of the electron is then v = 2K=m, and the magnetic force on the electron is FB = qvB, where we are assuming sin = 1 because the electron moves horizontally through a magnetic eld with a vertical component. We can ignore the e ect of the magnetic eld’s horizontal component because the electron is moving parallel to this component. The acceleration of the electron because of the magnetic force is then P32-3

r qvB qB 2K mmm s (1:601019C)(55:0106T) 2(1:921015J) = = 6:271014 m=s2: (9:111031kg) (9:111031kg) (c) The electron travels a horizontal distance of 20.0 cm in a time of (20:0 cm) (20:0 cm) t = p = p = 3:08109 s: 2K=m 2(1:921015J)=(9:111031kg) In this time the electron is accelerated to the side through a distance of 11 d = at2 = (6:271014 m=s2)(3:08109 s)2 = 2:98 mm: 22

P32-4 (a) d needs to be larger than the turn radius, so R d; but 2mK=q2B2 = R2 d2, or p B 2mK=q2d2: (b) Out of the page.

P32-5 Only unde ected ions emerge from the velocity selector, so v = E=B. The ions are then de ected by B0 with a radius of curvature of r = mv=qB; combining and rearranging, q=m = E=rBB0.

P32-6 The ions are given a kinetic energy K = qV ; they are then de ected with a radius of curvature given by R2 = 2mK=q2B2. But x = 2R. Combine all of the above, and m = B2qx2=8V: P32-7 (a) Start with the equation in Problem 6, and take the square root of both sides to get

1 p2 Bq 2 8V and then take the derivative of x with respect to m,

1 2 1 dm B q 2 2 m 8V and then consider nite di erences instead of di erential quantities,

1 2 mB q 2 2V (b) Invert the above expression, 1 2V 2 mB2q and then put in the given values,

1 3 2(7:3310 V) 2 x = (2:0)(1:661027kg); (35:0)(1:661027kg)(0:520 T)2(1:601019C) = 8:02 mm: Note that we used 35.0 u for the mass; if we had used 37.0 u the result would have been closer to the answer in the back of the book.

p p = 5:23107 P32-8 (a) B = 2V m=qr2 = 2(0:105 MV)(238)(932 MeV=c2)=2e(0:973 m)2 T: (b) The number of atoms in a gram is 6:021023=238 = 2:531021. The current is then (0:090)(2:531021)(2)(1:61019C)=(3600 s) = 20:2 mA:

(b) Regardless of speed, the orbital period is T = 2m=qB. But they collide halfway around a complete orbit, so t = m=qB.

P32-11 (a) The period of motion can be found from the reciprocal of Eq. 32-12, 2m 2(9:111031kg) T = = = 7:86108s: jqjB (1:601019C)(455106T) (b) We need to nd the velocity of the electron from the kinetic energy, pp v = 2K=m = 2(22:5 eV)(1:601019 J/eV)=(9:111031kg) = 2:81106 m=s: The velocity can written in terms of components which are parallel and perpendicular to the magnetic eld. Then vjj = v cos and v? = v sin : The pitch is the parallel distance traveled by the electron in one revolution, so p = v T = (2:81106m=s) cos(65:5)(7:86108s) = 9:16 cm: jj (c) The radius of the helical path is given by Eq. 32-10, except that we use the perpendicular velocity component, so mv? (9:111031kg)(2:81106m=s) sin(65:5) R = = = 3:20 cm jqjB (1:601019C)(455106T)

P32-12 F~ = i b d~l B~ . d~l has two components, those R parallel to the path, say d~x and those a perpendicular, say d~y. Then the integral can be written as Zb Zb F~ = d~x B~ + d~y B~ : aa

But B~ is constant, and can be removed from the integral. R b d~x = ~l, a vector that points from a to a R a

P32-13 qvyB = Fx = m dvx=dt; qvxB = Fy = m dvy=dt. Taking the time derivative of the second expression and inserting into the rst we get m d2vy

qB dt2 which has solution vy = v sin(mt=qB), where v is a constant. Using the second equation we nd that there is a similar solution for vx, except that it is out of phase, and so vx = v cos(mt=qB): Integrating, ZZ qBv x = vxdt = v cos(mt=qB) = sin(mt=qB): m Similarly, ZZ qBv y = vydt = v sin(mt=qB) = cos(mt=qB): m This is the equation of a circle.

P32-14 dL~ = ^idx + ^jdy + k^dz. B~ is uniform, so that the integral can be written as I III F~ = i (^idx + ^jdy + k^dz) B~ = i^i B~ dx + i^j B~ dy + ik^ B~ dz;

P32-15 The current pulse provides an impulse which is equal to ZZZ F dt = BiL dt = BL i dt = BLq:

This gives an initial velocity of v0 = BLq=m, which will cause the rod to hop to a height of h = v2=2g = B2L2q2=2m2g: 0

Solving for q, m p (0:013 kg) p q = 2gh = 2(9:8 m=s2)(3:1 m) = 4:2 C: BL (0:12 T)(0:20 m) P32-16

The torque on a current carrying loop depends on the orientation of the loop; the maximum torque occurs when the plane of the loop is parallel to the magnetic eld. In this case the magnitude of the torque is from Eq. 32-34 with sin = 1| P32-17

= NiAB: The area of a circular loop is A = r2 where r is the radius, but since the circumference is C = 2r, we can write C2 A= : 4 The circumference is not the length of the wire, because there may be more than one turn. Instead, Finally, we can write the torque as L2 iL2B 4N 2 4N which is a maximum when N is a minimum, or N = 1.

P32-18 dF~ = i dL~ B~ ; the direction of dF~ will be upward and somewhat toward the center. L~ and B~ are a right angles, but only the upward component of dF~ will survive the integration as the central components will cancel out by symmetry. Hence Z F = iB sin dL = 2riB sin :

P32-19 The torque on the cylinder from gravity is where r is the radius of the cylinder. The torque from magnetism needs to balance this, so mgr sin = N iAB sin = N i2rLB sin ;

or mg (0:262 kg)(9:8 m=s2) i = = = 1:63 A: 2N LB 2(13)(0:127 m)(0:477 T)

(a) The magnetic eld from a moving charge is given by Eq. 33-5. If the protons are moving side by side then the angle is = =2, so E33-1

0 qv B= 4 r2 and we are interested is a distance r = d. The electric eld at that distance is 1q 40 r2 where in both of the above expressions q is the charge of the source proton. On the receiving end is the other proton, and the force on that proton is given by F~ = q(E~ + ~v B~ ): The velocity is the same as that of the rst proton (otherwise they wouldn’t be moving side by side.) This velocity is then perpendicular to the magnetic eld, and the resulting direction for the cross product will be opposite to the direction of E~ . Then for balance, 1 q 0 qv 40 r2 4 r2 1 = v2: 00 (b) This is clearly a relativistic speed!

E33-2B=i=2d=(4107Tm=A)(120A)=2(6:3m)=3:8106T:Thiswilldeectthe 0 compass needle by as much as one degree. However, there is unlikely to be a place on the Earth’s surface where the magnetic eld is 210 T. This was likely a typo, and should probably have been 21.0 T. The de ection would then be some ten degrees, and that is signi cant.

E33-3 B=0i=2d=(4107Tm=A)(50A)=2(1:3103m)=37:7103T: E33-4 (a)i=2dB=0=2(8:13102m)(39:0106T)=(4107Tm=A)=15:9A: (b) Due East.

E33-5 Use 0i (4107N=A2)(1:6 1019 C)(5:6 1014 s1) B = = = 1:2108T: 2d 2(0:0015 m) E33-6 Zero, by symmetry. Any contributions from the top wire are exactly canceled by contribu- tions from the bottom wire.

E33-7 B=i=2d=(4107Tm=A)(48:8A)=2(5:2102m)=1:88104T: 0 F~ = q~v B~ . All cases are either parallel or perpendicular, so either F = 0 or F = qvB. (a) F = qvB = (1:601019C)(1:08107m=s)(1:88104T) = 3:241016N. The direction of (b) F = qvB = (1:601019C)(1:08107m=s)(1:88104T) = 3:241016N. The direction of (c) F = 0.

E33-8 We want B1 = B2, but with opposite directions. Then i1=d1 = i2=d2, since all constants cancel out. Then i2 = (6:6 A)(1:5 cm)=(2:25 cm) = 4:4 A, directed out of the page.

For a single long straight wire, B = 0i=2d but we need a factor of \2″ since there are two wires, then i = dB=0: Finally E33-9

dB (0:0405 m)(296; T) i = = = 30 A 0 (4107N=A2)

E33-10 (a) The semi-circle part contributes half of Eq. 33-21, or 0i=4R. Each long straight wire contributes half of Eq. 33-13, or 0i=4R. Add the three contributions and get i 2 (4107N=A2)(11:5A)

0 2 3 Ba = + 1 = + 1 = 1:1410 T: 4R 4(5:20103m) (b) Each long straight wire contributes Eq. 33-13, or 0i=2R. Add the two contributions and get 0i (4107N=A )(11:5A) 2 Ba = = = 8:85104T: R (5:20103m) The direction is out of the page.

E33-11 z3=0iR2=2B=(4107N=A2)(320)(4:20A)(2:40102m)2=2(5:0106T)=9:73 102m3. Then z = 0:46 m.

E33-12 The circular part contributes a fraction of Eq. 33-21, or 0i=4R. Each long straight wire contributes half of Eq. 33-13, or 0i=4R. Add the three contributions and get 0i B = ( 2): 4R The goal is to get B = 0 that will happen if = 2 radians.

E33-13 There are four current segments that could contribute to the magnetic eld. The straight segments, however, contribute nothing because the straight segments carry currents either directly That leaves the two rounded segments. Each contribution to B~ can be found by starting with There is also a contribution from the top arc; the calculations are almost identical except that this is pointing into the page and r = a, so 0i=4a. The net magnetic eld at P is then

0i 1 1 B = B1 + B2 = : 4 b a E33-14 For each straight wire segment use Eq. 33-12. When the length of wire is L, the distance to the center is W=2; when the length of wire is W the distance to the center is L=2. There are four terms, but they are equal in pairs, so ! 0i 4L 4W 4 W L2=4 + W 2=4 L L2=4 + W 2=4 p L2 2 20i L2 2 20i W + W =p += : L2 + W 2 W L W L W L

We imagine the ribbon conductor to be a collection of thin wires, each of thickness dx and carrying a current di. di and dx are related by di=dx = i=w. The contribution of one of these thin wires to the magnetic eld at P is dB = 0 di=2x; where x is the distance from this thin wire to the point P . We want to change variables to x and integrate, so ZZ Z 0i dx 0i dx B = dB = = : 2wx 2w x E33-15

The limits of integration are from d to d + w, so 0i d + w B = ln : 2w d E33-16 The elds from each wire are perpendicular at P . Each contributes an amount B0 = pp i=2d, but since they are perpendicular there is a net eld of magnitude B = 2B02 = 2 i=2d: 0p 0 (b) Same numerical result, except the direction is up.

Reversing the direction of the second wire (so that now both currents are directed out of the page) will also reverse the direction of B2. Then 0i 1 1 2 b + x b x 0i (b x) (b + x) 2 b2 x2 0i x =: x2 b2

E33-18 (b) By symmetry, only the horizontal component of B~ survives, and must point to the (a) The horizontal component of the eld contributed by the top wire is given by 0i 0i b=2 0ib 2h 2h h (4R2 + b2) p since h is the hypotenuse, or h = R2 + b2=4. But there are two such components, one from the top wire, and an identical component from the bottom wire.

E33-19 (a) We can use Eq. 33-21 to nd the magnetic eld strength at the center of the large loop, i (4107Tm=A)(13A) B = 0 = = 6:8105T: 2R 2(0:12 m) (b) The torque on the smaller loop in the center is given by Eq. 32-34, ~ = NiA~ B~ ;

but since the magnetic eld from the large loop is perpendicular to the plane of the large loop, and the plane of the small loop is also perpendicular to the plane of the large loop, the magnetic eld is in the plane of the small loop. This means that jA~ B~ j = AB. Consequently, the magnitude of the torque on the small loop is =NiAB=(50)(1:3A)()(8:2103m)2(6:8105T)=9:3107Nm:

E33-20 (a) There are two contributions to the eld. One is from the circular loop, and is given by 0i=2R. The other is from the long straight wire, and is given by 0i=2R. The two elds are out of the page and parallel, so 0i B = (1 + 1=): 2R (b) The two components are now at right angles, so 0i p B = 1 + 1=2: 2R The direction is given by tan = 1= or = 18.

E33-21 The force per meter for any pair of parallel currents is given by Eq. 33-25, F=L = 0i2=2d, where d is the separation. The direction of the force is along the line connecting the intersection of the currents with the perpendicular plane. Each current experiences three forces; two are at right p angles and equal in magnitude, so jF~ + F~ j=L = F 2 + F 2 =L = 2 i2=2a. The third force p p12 14 12 14 0 points parallel to this sum, but d = a, so the resultant force is p F 20i2 0i2 4 107N=A2(18:7 A)2 p p = + p = ( 2 + 1= 2) = 6:06104N=m: L 2a 2 2a 2(0:245 m) It is directed toward the center of the square.

E33-22 By symmetry we expect the middle wire to have a net force of zero; the two on the outside will each be attracted toward the center, but the answers will be symmetrically distributed. For the wire which is the farthest left, 2 4 107N=A2(3:22 A)2 F 0i 1 1 1 1 1 1 1 = + + + = 1 + + + = 5:21105N=m: L 2 a 2a 3a 4a 2(0:083 m) 2 3 4 For the second wire over, the contributions from the two adjacent wires should cancel. This leaves F 0i2 1 1 4 107N=A2(3:22 A)2 11 = + + = + = 2:08105N=m: L 2 2a 3a 2(0:083 m) 2 3 E33-23 (a) The force on the projectile is given by the integral of dF~ = i d~l B~ over the length of the projectile (which is w). The magnetic eld strength can be found from adding together the contributions from each rail. If the rails are circular and the distance between them is small compared to the length of the wire we can use Eq. 33-13, 0i 2x where x is the distance from the center of the rail. There is one problem, however, because these are not wires of in nite length. Since the current stops traveling along the rail when it reaches the projectile we have a rod that is only half of an in nite rod, so we need to multiply by a factor of 1/2. But there are two rails, and each will contribute to the eld, so the net magnetic eld strength between the rails is 0i 0i B= + : 4x 4(2r + w x)

In that last term we have an expression that is a measure of the distance from the center of the lower rail in terms of the distance x from the center of the upper rail. The magnitude of the force on the projectile is then Z r+w F = i B dx;

r 2 r+w 1 1 Z 0i 4 r x 2r + w x 0i2 r+w = 2 ln 4 r The current through the projectile is down the page; the magnetic eld through the projectile is into the page; so the force on the projectile, according to F~ = i~l B~ ; is to the right. (b) Numerically the magnitude of the force on the rail is 3A)2 7N=A2 (45010 (410 ) (0:067m)+(0:012m) F = ln = 6:65103 N 2 (0:067 m) The speed of the rail can be found from either energy conservation so we rst nd the work done on the projectile, W = F d = (6:65103 N)(4:0 m) = 2:66104 J: This work results in a change in the kinetic energy, so the nal speed is pp v = 2K=m = 2(2:66104 J)=(0:010 kg) = 2:31103 m=s:

E33-24 The contributions from the left end and the right end of the square cancel out. This leaves the top and the bottom. The net force is the di erence, or 7N=A2 (410 )(28:6A)(21:8A)(0:323m) 1 1 F= ; 2 (1:10102m) (10:30102m) = 3:27103N:

E33-25 The magnetic force on the upper wire near the point d is 0iaibL 0iaibL 0iaibL 2(d + x) 2d 2d2 where x is the distance from the equilibrium point d. The equilibrium magnetic force is equal to the force of gravity mg, so near the equilibrium point we can write x FB = mg mg : d

There is then a restoring force against small perturbations of magnitude mgx=d which corresponds to a spring constant of k = mg=d. This would give a frequency of oscillation of 1p 1p 2 2 which is identical to the pendulum.

E33-27 The magnetic eld inside an ideal solenoid is given by Eq. 33-28 B = 0in; where n is the turns per unit length. Solving for n, B (0:0224 T) n = = = 1:00103=m1: 0i (4107N=A2)(17:8A) The solenoid has a length of 1.33 m, so the total number of turns is N = nL = (1:00103=m1)(1:33 m) = 1330;

and since each turn has a length of one circumference, then the total length of the wire which makes up the solenoid is (1330)(0:026 m) = 109 m:

E33-28 From the solenoid we have Bs = 0nis = 0(11500=m)(1:94 mA) = 0(22:3A=m):

From the wire we have 0iw 0(6:3 A) 0 Bw = = = (1:002 A) 2r 2r r These elds are at right angles, so we are interested in when tan(40) = Bw=Bs, or

(1:002 A) r = = 5:35102m: tan(40)(22:3 A=m) E33-29 Let u = z d. Then 0niR2 Z d+L=2 du 2 [R2 + u2]3=2 dL=2 2 d+L=2 0niR u 2 R2 R2 + u2 dL=2 ! 0ni d + L=2 d L=2 = p p : 2 R2 + (d + L=2)2 R2 + (d L=2)2

If L is much, much greater than R and d then jL=2 dj >> R, and R can be ignored in the denominator of the above expressions, which then simplify to ! 0ni d + L=2 d L=2 B = p p : 2 R2 + (d + L=2)2 R2 + (d L=2)2 ! 0ni d + L=2 d L=2 = p p : 2 (d + L=2)2 (d L=2)2 = 0in:

It is important that we consider the relative size of L=2 and d!

B~ d~s = 50 H E33-30 The net current in the loop is 1i0 + 3i0 + 7i0 6i0 = 5i0. Then i0:

E33-31 (a) The path is clockwise, so a positive current is into page. The net current is 2.0 A out, H

E33-32 Let R0 be the radius of the wire. On the surface of the wire B0 = 0i=2R0. 0

E33-33 (a) We don’t want to reinvent the wheel. The answer is found from Eq. 33-34, except it looks like 0ir B= : 2c2 (b) In the region between the wires the magnetic eld looks like Eq. 33-13, 0i B= : 2r (c) Ampere’s law (Eq. 33-29) is B~ d~s = 0i; where i is the current enclosed. Our Amperian H

loop will still be a circle centered on the axis of the problem, so the left hand side of the above equation will reduce to 2rB, just like in Eq. 33-32. The right hand side, however, depends on the net current enclosed which is the current i in the center wire minus the fraction of the current enclosed in the outer conductor. The cross sectional area of the outer conductor is (a2 b2); so the fraction of the outer current enclosed in the Amperian loop is (r2 b2) r2 b2 i =i : (a2 b2) a2 b2

The net current in the loop is then r2 b2 a2 r2 a2 b2 a2 b2 so the magnetic eld in this region is 0i a2 r2 B= : 2r a2 b2 (d) This part is easy since the net current is zero; consequently B = 0.

E33-34 (a) Ampere’s law (Eq. 33-29) is B~ d~s = 0i; where i is the current enclosed. Our H

Amperian loop will still be a circle centered on the axis of the problem, so the left hand side of the above equation will reduce to 2rB, just like in Eq. 33-32. The right hand side, however, depends on the net current enclosed which is the fraction of the current enclosed in the conductor. The cross sectional area of the conductor is (a2 b2); so the fraction of the current enclosed in the Amperian loop is (r2 b2) r2 b2 i =i : (a2 b2) a2 b2 The magnetic eld in this region is i r2 b2 0 B= : 2r a2 b2 (b) If r = a, then i a2 b2 i 00 2a a2 b2 2a which is what we expect.

If r = b, then 0i b2 b2 2b a2 b2 If b = 0, then 0i r2 02 0ir B= = 2r a2 02 2a2 which is what I expected.

E33-35 The magnitude of the magnetic eld due to the cylinder will be zero at the center of the cylinder and 0i0=2(2R) at point P . The magnitude of the magnetic eld eld due to the wire will be 0i=2(3R) at the center of the cylinder but 0i=2R at P . In order for the net eld to have di erent directions in the two locations the currents in the wire and pipe must be in di erent direction. The net eld at the center of the pipe is 0i=2(3R), while that at P is then 0i0=2(2R) 0i=2R. Set these equal and solve for i;

(a) A positive particle would experience a magnetic force directed to the right for a magnetic eld out of the page. This particle is going the other way, so it must be negative. (b) The magnetic eld of a toroid is given by Eq. 33-36, 0iN 2r while the radius of curvature of a charged particle in a magnetic eld is given by Eq. 32-10 mv R= : jqjB We use the R to distinguish it from r. Combining, 2mv 0iN jqj so the two radii are directly proportional. This means E33-37

The eld from one coil is given by Eq. 33-19 0iR2 B= : 2(R2 + z2)3=2 There are N turns in the coil, so we need a factor of N . There are two coils and we are interested in the magnetic eld at P , a distance R=2 from each coil. The magnetic eld strength will be twice the above expression but with z = R=2, so 20N iR2 80N i B= = : 2(R2 + (R=2)2)3=2 (5)3=2R P33-1

P33-2 (a) Change the limits of integration that lead to Eq. 33-12:

ZL 0id dz 4 0 (z2 + d2)3=2 L 0id z 4 (z2 + d2)1=2 0

0id L =: 4 (L2 + d2)1=2 P33-3 This problem is the all important derivation of the Helmholtz coil properties. (a) The magnetic eld from one coil is 0N iR2 B1 = : 2(R2 + z2)3=2 The magnetic eld from the other coil, located a distance s away, but for points measured from the rst coil, is 0N iR2 B2 = : 2(R2 + (z s)2)3=2 The magnetic eld on the axis between the coils is the sum, 0N iR2 0N iR2 B= + : 2(R2 + z2)3=2 2(R2 + (z s)2)3=2 Take the derivative with respect to z and get dB 30N iR2 30N iR2 = z (z s): dz 2(R2 + z2)5=2 2(R2 + (z s)2)5=2 At z = s=2 this expression vanishes! We expect this by symmetry, because the magnetic eld will be strongest in the plane of either coil, so the mid-point should be a local minimum. (b) Take the derivative again and d2B 30N iR2 150N iR2 = + z2 dz2 2(R2 + z2)5=2 2(R2 + z2)5=2 30N iR2 150N iR2 + (z s)2: 2(R2 + (z s)2)5=2 2(R2 + (z s)2)5=2 We could try and simplify this, but we don’t really want to; we instead want to set it equal to zero, then let z = s=2, and then solve for s. The second derivative will equal zero when 3(R2 + z2) + 15z2 3(R2 + (z s)2) + 15(z s)2 = 0;

and is z = s=2 this expression will simplify to s = R:

P33-4 (a) Each of the side of the square is a straight wire segment of length a which contributes a eld strength of 0i a 4r a2=4 + r2 where r is the distance to the point on the axis of the loop, so p r = a2=4 + z2:

This eld is not parallel to the z axis; the z component is Bz = B(a=2)=r. There are four of these contributions. The o axis components cancel. Consequently, the eld for the square is 0i a a=2 4r a2=4 + r2 r 0i a2 2p 2r a2=4 + r2 0i a2 2(a2 2 p =4 + z ) a2=2 + z2 40i a2 =p: (a2 + 4z2) 2a2 + 4z2 (b) When z = 0 this reduces to 40i a2 40i B= p =p : (a2) 2a2 2 a P33-5 (a) The polygon has n sides. A perpendicular bisector of each side can be drawn to the center and has length x where x=a = cos(=n): Each side has a length L = 2a sin(=n). Each of the side of the polygon is a straight wire segment which contributes a eld strength of 0i L 4x L2=4 + x2 This eld is parallel to the z axis. There are n of these contributions. The o axis components cancel. Consequently, the eld for the polygon 0i L 4x L2=4 + x2 0i 2 4 L2=4 + x2 0i 1 2 a (b) Evaluate: lim n tan(=n) = lim n sin(=n) n=n = : n!1 n!1 Then the answer to part (a) simpli es to 0i B= : 2a

P33-6 For a square loop of wire we have four nite length segments each contributing a term which looks like Eq. 33-12, except that L is replaced by L=4 and d is replaced by L=8. Then at the center, 0i L=4 160i B=4 p =p : 4L=8 L2=64 + L2=64 2L For a circular loop R = L=2 so 0i 0 B= = : 2R L p Since 16= 2 > , the square wins. But only by some 7%!

We want to use the di erential expression in Eq. 33-11, except that the limits of integra- tion are going to be di erent. We have four wire segments. From the top segment, 3L=4 0i d 4 z2 + d2 L=4 ! 0i 3L=4 L=4 = p p : 4d (3L=4)2 + d2 (L=4)2 + d2 P33-7

For the top segment d = L=4, so this simpli es even further to i p p 0 B1 = 2(3 5 + 5) : 10L The bottom segment has the same integral, but d = 3L=4, so i p p 0 B3 = 2( 5 + 5) : 30L By symmetry, the contribution from the right hand side is the same as the bottom, so B2 = B3, and the contribution from the left hand side is the same as that from the top, so B4 = B1. Adding all four terms, 2 i p p p p 0 30L 2 i p p 0 = (2 2 + 10): 3L P33-8 Assume a current ring has a radius r and a width dr, the charge on the ring is dq = 2r dr, where = q=R2. The current in the ring is di = ! dq=2 = !r dr. The ring contributes a eld dB = 0 di=2r. Integrate over all the rings: ZR B = 0!r dr=2r = 0!R=2 = !q=2R: 0

P33-9 B = 0in and mv = qBr. Combine, and mv (5:11105eV=c2)(0:046c) i = = = 0:271 A: 0qrn (4107N=A2)e(0:023m)(10000=m) P33-10 This shape is a triangle with area A = (4d)(3d)=2 = 6d2. The enclosed current is then i = jA = (15 A=m2)6(0:23 m)2 = 4:76 A The line integral is then 0i = 6:0106T m:

P33-11 Assume that B does vary as the picture implies. Then the line integral along the path shown must be nonzero, since B~ ~l on the right is not zero, while it is along the three other sides. Hence B~ d~l is non zero, implying some current passes through the dotted path. But it doesn’t, H

P33-12 (a) Sketch an Amperian loop which is a rectangle which enclosed N wires, has a vertical sides with height h, and horizontal sides with length L. Then B~ d~l = 0N i: Evaluate the integral H

along the four sides. The vertical side contribute nothing, since B~ is perpendicular to ~h, and then B~ ~h = 0. If the integral is performed in a counterclockwise direction (it must, since the sense of integration was determined by assuming the current is positive), we get BL for each horizontal section. Then 0iN 1 B = = 0in: 2L 2 (b) As a ! 1 then tan1(a=2R) ! =2. Then B ! 0i=2a. If we assume that i is made up of several wires, each with current i0, then i=a = i0n.

P33-13 Apply Ampere’s law with an Amperian loop that is a circle centered on the center of the wire. Then III B~ d~s = B ds = B ds = 2rB;

because B~ is tangent to the path and B is uniform along the path by symmetry. The current enclosed is Z ienc = j dA:

This integral is best done in polar coordinates, so dA = (dr)(r d), and then Z r Z 2 00 Zr = 2j0=a r2dr;

0 2j0 3 = r: 3a When r = a the current enclosed is i, so 2j0a2 3i i = or j0 = : 3 2a2 The magnetic eld strength inside the wire is found by gluing together the two parts of Ampere’s law, 2j0 3a 0j0r2 3a 0ir2=: 2a3

P33-14 (a) According to Eq. 33-34, the magnetic eld inside the wire without a hole has magnitude B = 0ir=2R2 = 0jr=2 and is directed radially. If we superimpose a second current to create the hole, the additional eld at the center of the hole is zero, so B = 0jb=2. But the current in the remaining wire is i = jA = j(R2 a2);

E34-1 = B~ A~ = (42106T)(2:5 m2) cos(57) = 5:7105Wb: B

E34-2 jEj=jd B=dtj=AdB=dt=(=4)(0:112m)2(0:157T=s)=1:55mV:

E34-3 (a) The magnitude of the emf induced in a loop is given by Eq. 34-4, dB dt = N (12 mWb=s2)t + (7 mWb=s) There is only one loop, and we want to evaluate this expression for t = 2:0 s, so jEj=(1)(12mWb=s2)(2:0s)+(7mWb=s)=31mV: (b) This part isn’t harder. The magnetic ux through the loop is increasing when t = 2:0 s. The induced current needs to ow in such a direction to create a second magnetic eld to oppose this increase. The original magnetic eld is out of the page and we oppose the increase by pointing the other way, so the second eld will point into the page (inside the loop). By the right hand rule this means the induced current is clockwise through the loop, or to the left through the resistor.

(a) E = (0:16 m)2(0:5 T)=(2 s) = 2:0102V: (b) E = (0:16 m)2(0:0 T)=(2 s) = 0:0102V: (c) E = (0:16 m)2(0:5 T)=(4 s) = 1:0102V:

E34-5 (a) R = L=A = (1:69108 m)[()(0:104 m)]=[(=4)(2:50103m)2] = 1:12103 : (b) E = iR = (9:66 A)(1:12103 ) = 1:08102V. The required dB=dt is then given by dB E = = (1:08102 V)=(=4)(0:104 m)2 = 1:27 T=s: dt A E34-6 E=AB=t=AB=t.ThepowerisP=iE=E2=R.Theenergydissipatedis E2t A2B2 E = P t = = : R Rt E34-7 (a) We could re-derive the steps in the sample problem, or we could start with the end result. We’ll start with the result, di dt except that we have gone ahead and used the derivative instead of the . The rate of change in the current is di dt so the induced emf is E = (130)(3:46104m2)(4107Tm=A)(2:2104=m) (3:0A=s) + (2:0A=s2 )t ;

= (3:73103 V) + (2:48103 V=s)t: (b) When t = 2:0 s the induced emf is 8:69103 V, so the induced current is i = (8:69103 V)=(0:15 ) = 5:8102 A:

E34-8 (a) i = E=R = NAdB=dt. Note that A refers to the area enclosed by the outer solenoid where B is non-zero. This A is then the cross sectional area of the inner solenoid! Then 1 di (120)(=4)(0:032m)2(4107N=A2)(220102=m)(1:5A) i = NA0n = = 4:7103A: R dt (5:3 ) (0:16 s)

E34-9 P = Ei = E2=R = (AdB=dt)2=(L=a), where A is the area of the loop and a is the cross sectional area of the wire. But a = d2=4 and A = L2=4, so L3 2 2 (0:525 m)3 3 2 d dB (1:110 m) P = = (9:82103T=s)2 = 4:97106W: 64 dt 64(1:69108 m)

E34-10 B=BA=B(2:3m)2=2.EB=dB=dt=AdB=dt,or (2:3 m)2 2 E34-11 (a) The induced emf, as a function of time, is given by Eq. 34-5, E(t) = d B(t)=dt This emf drives a current through the loop which obeys E (t) = i(t)R Combining, 1 d B(t) i(t) = : R dt Since the current is de ned by i = dq=dt we can write dq(t) 1 d B (t) = : dt R dt Factor out the dt from both sides, and then integrate:

1 R ZZ 1 R 1 q(t) q(0) = ( B (0) B (t)) R (b) No. The induced current could have increased from zero to some positive value, then decreased to zero and became negative, so that the net charge to ow through the resistor was zero. This would be like sloshing the charge back and forth through the loop.

E34-12 P hiB = 2 B = 2N BA. Then the charge to ow through is

q = 2(125)(1:57 T)(12:2104m2)=(13:3 ) = 3:60102C: E34-13 The part above the long straight wire (a distance b a above it) cancels out contributions below the wire (a distance b a beneath it). The ux through the loop is then a Z 0i 0ib a B = b dr = ln : 2ab 2r 2 2a b

The emf in the loop is then d B 0b a 2 E = = ln [2(4:5 A=s )t (10 A=s)]: dt 2 2a b Evaluating, 4107N=A2(0:16m) (0:12m)

E = ln [2(4:5 A=s2)(3:0 s)(10 A=s)] = 2:20107V: 2 2(0:12 m) (0:16 m)

E34-14 Use Eq. 34-6: E = BDv = (55106T)(1:10 m)(25 m=s) = 1:5103V:

E34-15 If the angle doesn’t vary then the ux, given by = B~ A~ is constant, so there is no emf.

E34-16 (a) Use Eq. 34-6: E = BDv = (1:18T)(0:108 m)(4:86 m=s) = 0:619 V: (b) i = (0:619 V)=(0:415 ) = 1:49 A: (e) P = F v = (0:190 V)(4:86 m=s) = 0:923 W.

The magnetic eld is out of the page, and the current through the rod is down. Then Eq. 32-26 F~ = iL~ B~ shows that the direction of the magnetic force is to the right; furthermore, since everything is perpendicular to everything else, we can get rid of the vector nature of the problem and write F = iLB. Newton’s second law gives F = ma, and the acceleration of an object from rest results in a velocity given by v = at. Combining, E34-17

iLB v(t) = t: m E34-18 (b) The rod will accelerate as long as there is a net force on it. This net force comes from F = iLB. The current is given by iR = E BLv, so as v increases i decreases. When i = 0 the rod (a) E = BLv will give the terminal velocity. In this case, v = E=BL.

E34-19 E34-20 The acceleration is a = R!2; since E = B!R2=2, we can nd a=4E2=B2R3=4(1:4V)2=(1:2T)2(5:3102m)3=3:7104m=s2: E34-21 We will use the results of Exercise 11 that were worked out above. All we need to do is nd the initial ux; ipping the coil up-side-down will simply change the sign of the ux. So (0) = B~ A~ = (59 T)()(0:13 m)2 sin(20) = 1:1106 Wb: B

Then using the results of Exercise 11 we have N R 950 85 = 2:5105 C:

E34-22 (a) The ux through the loop is Z vt Z a+L 0i 0ivt a + L B = dx dr = ln : 0 a 2r 2 a The emf is then d B 0iv a + L E = = ln : dt 2 a Putting in the numbers, (4107N=A2)(110A)(4:86m=s) (0:0102m)+(0:0983m) E = ln = 2:53104V: 2 (0:0102 m) R (d) F = Bil dl, or Z a+L 0i 0i a + L F = il dr = il ln : a 2r 2 a

Putting in the numbers, (4107N=A2)(110A) (0:0102m)+(0:0983m) F = (6:10104A) ln = 3:17108N: 2 (0:0102 m) (e) P = F v = (3:17108N)(4:86 m=s) = 1:54107W.

E34-23 (a) Starting from the beginning, Eq. 33-13 gives 0i B= : 2y The ux through the loop is given by Z B = B~ dA~ ;

but since the magnetic eld from the long straight wire goes through the loop perpendicular to the plane of the loop this expression simpli es to a scalar integral. The loop is a rectangular, so use Combining the above, D+b a i ZZ 0 D 0 2y 0i D+b dy Z 2 D y 0i D + b = a ln 2 D

(b) The ux through the loop is a function of the distance D from the wire. If the loop moves away from the wire at a constant speed v, then the distance D varies as vt. The induced emf is then dB dt 0i b =a: 2 t(vt + b) The current will be this emf divided by the resistance R. The \back-of-the-book” answer is somewhat di erent; the answer is expressed in terms of D instead if t. The two answers are otherwise identical.

E34-24 (a) The area of the triangle is A = x2 tan =2. In this case x = vt, so B = B(vt)2 tan =2;

and then (b)t=E=2Bv2tan=2;so (56:8 V) t = = 2:08 s: 2(0:352 T)(5:21 m=s)2 tan(55) E34-25 E = N BA!, so (24 V) ! = = 39:4 rad=s: (97)(0:33 T)(0:0190 m2)

E34-26 (a) The frequency of the emf is the same as the frequency of rotation, f . (b) The ux changes by BA = Ba2 during a half a revolution. This is a sinusoidal change, so the amplitude of the sinusoidal variation in the emf is E = !=2. Then E = B2a2f . B

We can use Eq. 34-10; the emf is E = BA! sin !t; This will be a maximum when sin !t = 1. The angular frequency, ! is equal to ! = (1000)(2)=(60) rad/s = 105 rad/s The maximum emf is then E34-27

E = (3:5 T) [(100)(0:5 m)(0:3 m)] (105 rad/s) = 5:5 kV: E34-28 (a) The amplitude of the emf is E = BA!, so A=E=2fB=(150V)=2(60=s)(0:50T)=0:798m2:

(a) For this path I E~d~s=dB=dt=(0:212m)2(8:50mT=s)=1:20mV: (b) For this path I E~d~s=dB=dt=(0:323m)2(8:50mT=s)=2:79mV:

(c) For this path I E~d~s=dB=dt=(0:323m)2(8:50mT=s)(0:323m)2(8:50mT=s)=1:59mV:

H E34-30 d B=dt=AdB=dt=A(6:51mT=s),while (a) The path of integration is inside the solenoid, so r2(6:51 mT=s) (0:022 m)(6:51 mT=s) E = = = 7:16105V=m: 2r 2 (b) The path of integration is outside the solenoid, so r2(6:51 mT=s) (0:063 m)2(6:51 mT=s) E = = = 1:58104V=m 2R 2(0:082 m) E34-31 The induced electric eld can be found from applying Eq. 34-13, I E~ d~s = d B : dt We start with the left hand side of this expression. The problem has cylindrical symmetry, so the induced electric eld lines should be circles centered on the axis of the cylindrical volume. If we choose the path of integration to lie along an electric eld line, then the electric eld E~ will be parallel to d~s, and E will be uniform along this path, so III E~ d~s = E ds = E ds = 2rE;

Now for the right hand side. The ux is contained in the path of integration, so B = Br2. All of the time dependence of the ux is contained in B, so we can immediately write dB r dB 2rE = r2 or E = : dt 2 dt What does the negative sign mean? The path of integration is chosen so that if our right hand ngers curl around the path our thumb gives the direction of the magnetic eld which cuts through the path. Since the eld points into the page a positive electric eld would have a clockwise orientation. Since B is decreasing the derivative is negative, but we get another negative from the equation Now for the magnitude.

E = (4:82102 m)(10:7103 T =s)=2 = 2:58104 N=C: The acceleration of the electron at either a or c then has magnitude a = Eq=m = (2:58104 N=C)(1:601019 C)=(9:111031 kg) = 4:53107m=s2:

The induced current is given by i = E =R. The resistance of the loop is given by R = L=A, where A is the cross sectional area. Combining, and writing in terms of the radius of the wire, we have P34-1

The length of the wire is related to the radius of the wire because we have a xed mass. The total volume of the wire is r2L, and this is related to the mass and density by m = r2L. Eliminating r we have mE i= : L2 The length of the wire loop is the same as the circumference, which is related to the radius R of the loop by L = 2R. The emf is related to the changing ux by E = d B=dt, but if the shape of the loop is xed this becomes E = A dB=dt. Combining all of this, mA dB i= : (2R)2 dt We dropped the negative sign because we are only interested in absolute values here. Now A = R2, so this expression can also be written as mR2 dB m dB i= = : (2R)2 dt 4 dt

P34-2 For the lower surface B~ A~ = (76103T)(=2)(0:037 m)2 cos(62) = 7:67105Wb: For the upper surface B~ A~ = (76103T)(=2)(0:037 m)2 cos(28) = 1:44104Wb:. The induced emf is then E = (7:67105Wb + 1:44104Wb)=(4:5103s) = 4:9102V:

P34-3 (a) We are only interested in the portion of the ring in the yz plane. Then E = (3:32 (b) From c to b. Point your right thumb along x to oppose the increasing B~ eld. Your right ngers will curl from c to b.

P34-4 E / N A, but A = r2 and N 2r = L, so E / 1=N . This means use only one loop to maximize the emf.

This is a integral best performed in rectangular coordinates, then dA = (dx)(dy). The magnetic eld is perpendicular to the surface area, so B~ dA~ = B dA. The ux is then P34-5

ZZ Z aZ a 00 1 2 = (2 T=m s2)a3t2: But a = 2:0 cm, so this becomes B = (2 T=m s2)(0:02 m)3t2 = (1:6105Wb=s2)t2: The emf around the square is given by dB dt and at t = 2:5 s this is 8:0105V. Since the magnetic eld is directed out of the page, a positive emf would be counterclockwise (hold your right thumb in the direction of the magnetic eld and your ngers will give a counter clockwise sense around the loop). But the answer was negative, so the emf must be clockwise.

P34-6 (a) Far from the plane of the large loop we can approximate the large loop as a dipole, and then 0iR2 B= : 2×3 The ux through the small loop is then 0i2r2R2 B = r2B = : 2×3 (b) E = d B=dt, so 30i2r2R2 E = v: 2×4 (c) Anti-clockwise when viewed from above.

P34-7 The magnetic eld is perpendicular to the surface area, so B~ dA~ = B dA. The ux is then ZZ B = B~ dA~ = B dA = BA;

since the magnetic eld is uniform. The area is A = r2, where r is the radius of the loop. The induced emf is d B dr E = = 2rB : dt dt It is given that B = 0:785 T, r = 1:23 m, and dr=dt = 7:50102m=s. The negative sign indicate a decreasing radius. Then E = 2(1:23 m)(0:785 T)(7:50102m=s) = 0:455 V:

P34-8 (a) d B=dt = B dA=dt, but dA=dt is A=t, where A is the area swept out during one rotation and t = 1=f . But the area swept out is R2, so

d jEj = B = fBR2: dt (b) If the output current is i then the power is P = E i. But P = ! = 2f , so

P = = iBR2=2: 2f P34-9 (a) E = d B=dt, and B = B~ A~ ,so

E = BLv cos : The component of the force of gravity on the rod which pulls it down the incline is FG = mg sin . The component of the magnetic force on the rod which pulls it up the incline is FB = BiL cos . Equating, BiL cos = mg sin ;

and since E = iR, E mgR sin v= = : BL cos B2L2 cos2 (b)P=iE=E2=R=B2L2v2cos2=R=mgvsin.Thisisidenticaltotherateofchangeof gravitational potential energy.

(a) R = L=a = (r + 2r)=a; with numbers, R = (3:4103 )(2 + ): (b) B = Br2=2; with numbers,

B = (4:32103Wb): (c) i = E=R = B!r2=2R = Ba!r=2( + 2), or Ba tr i= : ( t2 + 4) Take the derivative and set it equal to zero, 4 at2 ( t2 + 4)2

p2 (d) ! = 2 , so q (0:15 T)(1:2106m2) 2(12 rad/s2)(2 rad)(0:24 m) i = = 2:2 A: (1:7108 m)(6 rad)

It does say approximate, so we will be making some rather bold assumptions here. First we will nd an expression for the emf. Since B is constant, the emf must be caused by a change in the area; in this case a shift in position. The small square where B 6= 0 has a width a and sweeps around the disk with a speed r!. An approximation for the emf is then E = Bar!: This emf causes a current. We don’t know exactly where the current ows, but we can reasonably assume that it occurs near the location of the magnetic eld. Let us assume that it is constrained to that region of the disk. The resistance of this portion of the disk is the approximately P34-11

1L 1a 1 A at t where we have assumed that the current is owing radially when de ning the cross sectional area of the \resistor”. The induced current is then (on the order of ) E Bar! = = Bar!t: R 1=(t) This current experiences a breaking force according to F = BIl, so F = B2a2r!t;

Finally we can nd the torque from = rF , and = B2a2r2!t:

P34-12 The induced electric eld in the ring is given by Eq. 34-11: 2RE = jd B=dtj. This electric eld will result in a force on the free charge carries (electrons?), given by F = Ee. The acceleration of the electrons is then a = Ee=me. Then e dB a= : 2Rme dt Integrate both sides with respect to time to nd the speed of the electrons.

ZZ e dB 2Rme dt Z e dB v= e = B: 2Rme The current density is given by j = nev, and the current by iA = ia2. Combining, ne2a2 i = PhiB: 2Rme Actually, it should be pointed out that P hiB refers to the change in ux from external sources. The current induced in the wire will produce a ux which will exactly o set P hiB so that the net ux through the superconducting ring is xed at the value present when the ring became superconducting.

P34-13 Assume that E does vary as the picture implies. Then the line integral along the path shown must be nonzero, since E~ ~l on the right is not zero, while it is along the three other sides. Hence E~ d~l is non zero, implying a change in the magnetic ux through the dotted path. But it H

P34-14 The electric eld a distance r from the center is given by r2dB=dT r dB E= = : 2r 2 dt De ne h to be the distance from the center of the circle to the center of the rod, and evaluate E = E~ d~s, R Z dB r h dt 2 r dB L = h: dt 2 But h2 = R2 (L=2)2, so dB Lp E = R2 (L=2)2: dt 2 P34-15 (a) B = r2Bav, so E (0:32 m) E = = 2(0:28 T)(120 ) = 34 V=m: 2r 2 (b) a = F=m = Eq=m = (33:8 V=m)(1:61019C)=(9:11031kg) = 6:01012m=s2.

E35-1 If the Earth’s magnetic dipole moment were produced by a single current around the core, then that current would be (8:0 1022J=T) i = = = 2:1109 A A (3:5 106m)2

(b) = B = (2:33 A m2)(0:0346 T) = 8:06102N m: E35-3 (a) Using the right hand rule a clockwise current would generate a magnetic moment which would be into the page. Both currents are clockwise, so add the moments: = (7:00 A)(0:20 m)2 + (7:00 A)(0:30 m)2 = 2:86 A m2:

(b) Reversing the current reverses the moment, so = (7:00 A)(0:20 m)2 (7:00 A)(0:30 m)2 = 1:10 A m2:

(b) = B sin = (0:207 A m2)(1:20 T) sin(41) = 0:163 N m: E35-5 (a) The result from Problem 33-4 for a square loop of wire was 40ia2 B(z) = : (4z2 + a2)(4z2 + 2a2)1=2 For z much, much larger than a we can ignore any a terms which are added to or subtracted from z terms. This means that 4z2 + a2 ! 4z2 and (4z2 + 2a2)1=2 ! 2z;

The expression for B then simpli es to 0ia2 2z3 (b) We can rearrange this expression and get 0 2z3 where it is rather evident that ia2 must correspond to ~ , the dipole moment, in Eq. 35-4. So that must be the answer.

(a) For the torque, ~ = ~ B~ = (9:65104N m)^i + (7:24104N m)^j + (8:08104N m)k^:

(b) For the magnetic potential energy, U = ~ B~ = [(0:60)(0:25 T)] = 0:603103J:

E35-7 = iA = i(a2 + b2=2) = i(a2 + b2)=2: E35-8 If the distance to P is very large compared to a or b we can write the Law of Biot and Savart as i ~s ~r B~ = 0 : 4 r3 ~s is perpendicular to ~r for the left and right sides, so the left side contributes 0i b 4 (x + a=2)2 and the right side contributes 0i b B3 = : 4 (x a=2)2 The top and bottom sides each contribute an equal amount 0i a sin 0i a(b=2) B2 = B4 = : 4 x2 + b2=4 4 x3 Add the four terms, expand the denominators, and keep only terms in x3,

0i ab 0 B= = : 4 x3 4 x3 The negative sign indicates that it is into the page.

E35-9 (a) The electric eld at this distance from the proton is 1 (1:601019C) E = = 5:141011N=C: 4(8:851012C2=N m2) (5:291011m)2 (b) The magnetic eld at this from the proton is given by the dipole approximation, 0 2z3 (4107N=A2)(1:411026A=m2) =;2(5:291011m)3 = 1:90102 T

E35-10 1:50 g of water has (2)(6:02 1023)(1:5)=(18) = 1:00 1023 hydrogen nuclei. If all are aligned the net magnetic moment would be = (1:001023)(1:411026J=T) = 1:41103J=T. The eld strength is then 0 (1:41103J=T) B = = (1:00107N=A2) = 9:31013T: 4 x3 (5:33 m)3

E35-11 (a) There is e ectively a current of i = f q = q!=2. The dipole moment is then = iA = (q!=2)(r2) = 1 q!r2: 2 (b) The rotational inertia of the ring is mr2 so L = I! = mr2!. Then (1=2)q!r2 q = =: L mr2! 2m

E35-12 The mass of the bar is 32 m = V = (7:87 g/cm )(4:86 cm)(1:31 cm ) = 50:1 g: The number of atoms in the bar is N = (6:021023)(50:1 g)=(55:8 g) = 5:411023:

The dipole moment of the bar is then = (5:411023)(2:22)(9:271024J=T) = 11:6 J=T: (b) The torque on the magnet is = (11:6 J=T)(1:53 T) = 17:7 N m: E35-13 The magnetic dipole moment is given by = M V , Eq. 35-13. Then = (5; 300 A=m)(0:048 m)(0:0055 m)2 = 0:024 A m2:

E35-14 (a) The original eld is B0 = 0in. The eld will increase to B = mB0, so the increase is

B=( 1)0in=(3:3104)(4107N=A2)(1:3A)(1600=m)=8:6107T: 1 E35-15 The energy to ip the dipoles is given by U = 2B. The temperature is then 2B 4(1:21023J=T)(0:5 T) T = = = 0:58 K: 3k=2 3(1:381023J=K) E35-16 The Curie temperature of iron is 770C, which is 750C higher than the surface temper- ature. This occurs at a depth of (750C)=(30 C=km) = 25 km.

(a) Look at the gure. At 50% (which is 0.5 on the vertical axis), the curve is at (b) Same gure, but now look at the 90% mark. B0=T 1:60 T=K, so B0 480 T. (c) Good question. I think both elds are far beyond our current abilities.

E35-17 E35-18 (a) Look at the gure. At 50% (which is 0.5 on the vertical axis), the curve is at B0=T (b) Same gure, but now look at the 90% mark. B0=T 1:60 T=K, so T (1:8 T)=(1:60 T=K) = 1:1 K.

E35-19 Since (0:5 T)=(10 K) = 0:05 T=K, and all higher temperatures have lower values of the ratio, and this puts all points in the region near where Curie’s Law (the straight line) is valid, then the answer is yes.

E35-20 Using Eq. 35-19, V M MrM (108g/mol)(511103A=m) n = = = = 8:741021A=m2 N A (10490 kg=m3)(6:021023/mol)

E35-21 (a) B = 0=2z3, so (4107N=A2)(1:51023J=T) B = = 9:4106T: 2(10109m)3 (b) U = 2B = 2(1:51023J=T)(9:4106T) = 2:821028J.

E35-22 B = (43106T)(295; 000106m2) = 1:3107Wb: E35-23 (a) We’ll assume, however, that all of the iron atoms are perfectly aligned. Then the dipole moment of the earth will be related to the dipole moment of one atom by Earth = N Fe;

where N is the number of iron atoms in the magnetized sphere. If mA is the relative atomic mass of iron, then the total mass is N mA mA Earth A A Fe where A is Avogadro’s number. Next, the volume of a sphere of mass m is m mA Earth A Fe And nally, the radius of a sphere with this volume would be 1=3 1=3 3V 3EarthmA r= = : 4 4FeA Now we nd the radius by substituting in the known values, !1=3 3(8:01022J=T)(56 g/mol) r = = 1:8105m: 4(14106g/m3)(2:11023J=T)(6:01023/mol) (b) The fractional volume is the cube of the fractional radius, so the answer is (1:8105 m=6:4106)3 = 2:2105:

E35-24 (a) At magnetic equator Lm = 0, so 0 (1:00107N=A2)(8:01022J=T) B = = = 31T: 4r3 (6:37106m)3 (b) Here Lm = 60, so (1:00107N=A2)(8:01022J=T) q q B = 0 1 + 3 sin2 Lm = 1 + 3 sin2(60) = 56T: 4r3 (6:37106m)3 The inclination is given by arctan(Bv=Bh) = arctan(2 tan Lm) = 74: (c) At magnetic north pole Lm = 90, so 2(1:00107N=A2)(8:01022J=T) 0 B = = = 62T: 2r3 (6:37106m)3 There is no horizontal component, so the inclination is 90.

E35-25 This problem is e ectively solving 1=r3 = 1=2 for r measured in Earth radii. Then r = 1:26rE, and the altitude above the Earth is (0:26)(6:37106m) = 1:66106m:

E35-26 The radial distance from the center is r = (6:37106m) (2900103m) = 3:47106m: The eld strength is 2(1:00107N=A2)(8:01022J=T) 0 B = = = 380T: 2r3 (3:47106m)3 E35-27 Here Lm = 90 11:5 = 78:5, so q (1:00107N=A2)(8:01022J=T) q B = 0 1 + 3 sin2 Lm = 1 + 3 sin2(78:5) = 61T: 4r3 (6:37106m)3 The inclination is given by arctan(Bv=Bh) = arctan(2 tan Lm) = 84:

E35-28 The ux out the \other” end is (1:6103T)(0:13 m)2 = 85Wb. The net ux through the surface is zero, so the ux through the curved surface is 0 (85Wb) (25Wb) = 60Wb:. The negative indicates inward.

The total magnetic ux through a closed surface is zero. There is inward ux on faces one, three, and ve for a total of -9 Wb. There is outward ux on faces two and four for a total of +6 Wb. The di erence is +3 Wb; consequently the outward ux on the sixth face must be +3 Wb.

E35-29 E35-31 The eld on the x axis between the wires is 0i 1 1 B= + : 2 2r + x 2r x Since B~ dA~ = 0, we can assume the ux through the curved surface is equal to the H ux through the xz plane within the cylinder. This ux is r 0i 1 1 Z r 2 2r + x 2r x 0i 3r r 2 r 3r 0i = L ln 3:

We can imagine the rotating disk as being composed of a number of rotating rings of radius r, width dr, and circumference 2r. The surface charge density on the disk is = q=R2, and consequently the (di erential) charge on any ring is P35-1

2qr dq = (2r)(dr) = dr R2 The rings \rotate” with angular frequency !, or period T = 2=!. The e ective (di erential) current for each ring is then dq qr! di = = dr: T R2

Each ring contributes to the magnetic moment, and we can glue all of this together as Z = d;

Z ZR 3 qr ! 0 R2 qR2! =: 4 P35-2 (a) The sphere can be sliced into disks. The disks can be sliced into rings. Each ring has some charge qi, radius ri, and mass mi; the period of rotation for a ring is T = 2=!, so the current in the ring is qi=T = !qi=2. The magnetic moment is = (!q =2)r2 = !q r2=2: i i i ii

Note that this is closely related to the expression for angular momentum of a ring: li = !mir2. i Equating, i = qili=2mi: If both mass density and charge density are uniform then we can write qi=mi = q=m, ZZ = d = (q=2m) dl = qL=2m

For a solid sphere L = !I = 2!mR2=5, so = q!R2=5: (b) See part (a)

P35-3 (a) The orbital speed is given by K = mv2=2: The orbital radius is given by mv = qBr, or r = mv=qB. The frequency of revolution is f = v=2r. The e ective current is i = qf . Combining all of the above to nd the dipole moment, v vr mv2 K = iA = q r2 = q = q = : 2r 2 2qB B (b) Since q and m cancel out of the above expression the answer is the same! (c) Work it out: (5:281021=m3)(6:211020 J) (5:281021=m3)(7:581021 J) M = = + = 312 A=m: V (1:18 T) (1:18 T) P35-4 (b) Point the thumb or your right hand to the right. Your ngers curl in the direction of (c) In the vicinity of the wire of the loop B~ has a component which is directed radially outward. Then B~ d~s has a component directed to the left. Hence, the net force is directed to the left.

P35-5 (b) Point the thumb or your right hand to the left. Your ngers curl in the direction of the (c) In the vicinity of the wire of the loop B~ has a component which is directed radially outward. Then B~ d~s has a component directed to the right. Hence, the net force is directed to the right.

P35-6 (a) Let x = B=kT . Adopt the convention that N+ refers to the atoms which have parallel alignment and N those which are anti-parallel. Then N+ + N = N , so +

and x x Note that the denominators are necessary so that N+ + N = N . Finally, ex ex M = (N+ N) = N ex + ex : (b) If B kT then x is very small and ex 1 x. The above expression reduces to (1 + x) (1 x) 2B M = N = Nx = : (1 + x) + (1 x) kT (c) If B kT then x is very large and ex ! 1 while ex ! 0. The above expression reduces to N = N:

(a) Centripetal acceleration is given by a = r!2. Then 0 2r!0!: (b) The change in centripetal acceleration is caused by the additional magnetic force, which has magnitude FB = qvB = er!B: Then a a0 eB ! = = : 2r!0 2m Note that we boldly canceled ! against !0 in this last expression; we are assuming that ! is small, and for these problems it is.

P35-7 P35-8 (a) i = =A = (8:01022J=T)=(6:37106m)2 = 6:3108A: (b) Far enough away both elds act like perfect dipoles, and can then cancel. (c) Close enough neither eld acts like a perfect dipole and the elds will not cancel.

p P35-9 (a) B = B 2 + B 2, so hv 0 qq B = cos2 Lm + 4 sin2 Lm = 0 1 + 3 sin2 Lm: 4r3 4r3 (b) tan i = Bv=Bh = 2 sin Lm= cos Lm = 2 tan Lm:

E36-1 The important relationship is Eq. 36-4, written as iL (5:0 mA)(8:0 mH) B = = = 1:0107Wb N (400)

E36-2 (a) = (34)(2:62103T)(0:103 m)2 = 2:97103Wb: (b) L = =i = (2:97103Wb)=(3:77 A) = 7:88104H:

E36-3 n = 1=d, where d is the diameter of the wire. Then L 0A (4107H=m)(=4)(4:10102m)2 = 0n2A = = = 2:61104H=m: l d2 (2:52103m)2

E36-4 (a) The emf supports the current, so the current must be decreasing. (b)L=E=(di=dt)=(17V)=(25103A=s)=6:8104H: E36-5 (a) Eq. 36-1 can be used to nd the inductance of the coil.

EL (3:0 mV) L = = = 6:0104H: di=dt (5:0 A=s) (b) Eq. 36-4 can then be used to nd the number of turns in the coil.

iL (8:0 A)(6:0104H) N = = = 120 B (40Wb)

(4107H=m)(0:81A)(536)(5:2102m) (5:2102m)+(15:3102m) B = ln ; 2 (15:3102m) = 1:32106Wb:

E36-7 L = m0n2Al = m0N 2A=l, or L=(968)(4107H=m)(1870)2(=4)(5:45102m)2=(1:26m)=7:88H:

(a) E = (4:6 H)(7 A)=(2103s) = 1:6104V: (b) E = (4:6 H)(2 A)=(3103s) = 3:1103V: (c) E = (4:6 H)(5 A)=(1103s) = 2:3104V:

(a) If two inductors are connected in parallel then the current through each inductor will add to the total current through the circuit, i = i1 + i2; Take the derivative of the current with The potential di erence across each inductor is the same, so if we divide by E and apply we get E36-9

so the previous expression can also be written as 111 =+: Leq L1 L2

(b) If the inductors are close enough together then the magnetic eld from one coil will induce currents in the other coil. Then we will need to consider mutual induction e ects, but that is a topic not covered in this text.

E36-10 (a) If two inductors are connected in series then the emf across each inductor will add to Then the current through each inductor is the same, so if we divide by di=dt and apply we get E E1 E2 di=dt di=dt di=dt But E di=dt so the previous expression can also be written as Leq = L1 + L2: (b) If the inductors are close enough together then the magnetic eld from one coil will induce currents in the other coil. Then we will need to consider mutual induction e ects, but that is a topic not covered in this text.

E36-11 Use Eq. 36-17, but rearrange: t (1:50 s) = = = 0:317 s: L ln[i0=i] ln[(1:16 A)=(10:2103A)] Then R = L= L = (9:44 H)=(0:317 s) = 29:8 :

E36-12 (a) There is no current through the resistor, so ER = 0 and then EL = E . (c)n=ln(EL=E)=ln(1=2)=0:693.

E36-13 (a) From Eq. 36-4 we nd the inductance to be N B (26:2103Wb) L = = = 4:78103H: i (5:48 A) Note that B is the ux, while the quantity N B is the number of ux linkages. (b) We can nd the time constant from Eq. 36-14, L = L=R = (4:78103H)=(0:745 ) = 6:42103 s: The we can invert Eq. 36-13 to get

Ri(t) E (0:745 A)(2:53 A) = (6:42103 s) ln 1 = 2:42103 s: (6:00 V)

E36-14 (a) Rearrange: di dt E L di R R dt R di dt = : L E=R i

(b) Integrate: Zt Zi R di 0 L 0 i E=R R i + E=R L E=R E R E 1 et= L = i: R

E36-15 di=dt = (5:0 A=s). Then di E = iR + L = (3:0 A)(4:0 ) + (5:0 A=s)t(4:0 ) + (6:0 H)(5:0 A=s) = 42 V + (20 V=s)t: dt E36-16 (1=3) = (1 et= L ), so

t (5:22 s) L = = = 12:9 s: ln(2=3) ln(2=3)

We want to take the derivative of the current in Eq. 36-13 with respect to time, di E 1 E = et= L = et= L : dt R L L Then L = (5:0102H)=(180 ) = 2:78104s. Using this we nd the rate of change in the current when t = 1:2 ms to be E36-17

di (45 V) 3s)=(2:78104 = e(1:210 s) = 12 A=s: dt ((5:0102H)

E36-18 (b) Consider some time ti: EL(t ) = Eeti= L: i Taking a ratio for two di erent times, EL(t1) EL(t2)

or t2 t1 (2 ms) (1 ms) L = = = 3:58 ms ln[EL(t1)=EL(t2)] ln[(18:24 V)=(13:8 V)] (a) Choose any time, and E = ELet= L = (18:24 V)e(1 ms)=(3:58 ms) = 24 V:

E36-19 (a) When the switch is just closed there is no current through the inductor. So i1 = i2 is given by E (100 V) i1 = = = 3:33 A: R1 + R2 (10 ) + (20 ) (b) A long time later there is current through the inductor, but it is as if the inductor has no e ect on the circuit. Then the e ective resistance of the circuit is found by rst nding the equivalent resistance of the parallel part 1=(30 ) + 1=(20 ) = 1=(12 );

and then nding the equivalent resistance of the circuit (10 ) + (12 ) = 22 : Finally, i1 = (100 V)=(22 ) = 4:55 A and

consequently, i2 = (54:5 V)=(20 ) = 2:73 A: It didn’t ask, but i2 = (4:55 A) (2:73 A) = 1:82 A: (c) After the switch is just opened the current through the battery stops, while that through the (d) All go to zero.

E36-20 (a) For toroids L = 0N 2h ln(b=a)=2. The number of turns is limited by the inner radius: N d = 2a. In this case, N = 2(0:10 m)=(0:00096 m) = 654: The inductance is then (4107H=m)(654)2(0:02m) (0:12m) L = ln = 3:1104H: 2 (0:10 m) (b) Each turn has a length of 4(0:02 m) = 0:08 m. The resistance is then R = N (0:08 m)(0:021 =m) = 1:10 The time constant is L = L=R = (3:1104H)=(1:10 ) = 2:8104s:

(I) When the switch is just closed there is no current through the inductor or R2, so the potential di erence across the inductor must be 10 V. The potential di erence across R1 is always 10 V when the switch is closed, regardless of the amount of time elapsed since closing. (c) The current through the switch is the sum of the above two currents, or 2:0 A. (f) Look at the results of Exercise 36-17. When t = 0 the rate of change of the current is di=dt = E=L. Then di=dt = (10 V)=(5:0 H) = 2:0 A=s: (II) After the switch has been closed for a long period of time the currents are stable and the inductor no longer has an e ect on the circuit. Then the circuit is a simple two resistor parallel network, each resistor has a potential di erence of 10 V across it.

(c) Add the two currents and the current through the switch will be 3.0 A. (f) Zero, since the current is no longer changing.

E36-22 U = (71 J=m3)(0:022 m3) = 1:56 J. Then using U = i2L=2 we get pp i = 2U=L = 2(1:56 J)=(0:092 H) = 5:8 A:

(b) Since the current is squared in the energy expression, doubling the current would quadruple the energy. Then i0 = 2i0 = 2(0:062 A) = 0:124 A.

E36-24 (a) B = 0in and u = B2=20, or u=0i2n2=2=(4107N=A2)(6:57A)2(950=0:853m)2=2=33:6J=m3: (b) U = uAL = (33:6 J=m3)(17:2104m2)(0:853 m) = 4:93102J:

E36-25 uB = B2=20, and from Sample Problem 33-2 we know B, hence (12:6 T)2 u = = 6:32107J=m3: B 2(4107N=A2)

E36-26 (a) uB = B2=20, so (1001012T)2 1 u = = 2:5102 eV/cm3: B 2(4107N=A2) (1:61019J/eV) (b) x = (10)(9:461015m) = 9:461016m. Using the results from part (a) expressed in J/m3 we nd the energy contained is U = (3:981015J=m3)(9:461016m)3 = 3:41036J

The energy density of an electric eld is given by Eq. 36-23; that of a magnetic eld is given by Eq. 36-22. Equating, E= 2 20 B E=p: 00 E36-27

The answer is then p E=(0:50T)= (8:851012C2=Nm2)(4107N=A2)=1:5108V=m:

E36-28 The rate of internal energy increase in the resistor is given by P = iVR. The rate of energy storage in the inductor is dU=dt = Li di=dt = iVL. Since the current is the same through both we want to nd the time when VR = VL. Using Eq. 36-15 we nd 1 et= L = et= L ;

E36-29 (a) Start with Eq. 36-13: iR E t ln(1 iR=E ) (5:20103s) ln[1 (1:96103A)(10:4103 )=(55:0 V)] = 1:12102s: Then L = R = (1:12102s)(10:4103 ) = 116 H: L (b) U = (1=2)(116 H)(1:96103A)2 = 2:23104J: R E36-30 (a) U = Eq; q = idt.

E Z R E2 2 R0 E2 E2L = t + (et= L 1): R R2 Using the numbers provided, L = (5:48 H)=(7:34 ) = 0:7466 s:

Then (12:2 V)2 h i U = (2 s) + (0:7466 s)(e(2 s)=0:7466 s) 1) = 26:4 J (7:34 ) (b) The energy stored in the inductor is UL = Li2=2, or LE2 Z 2

2R2 = 6:57 J: E36-31 This shell has a volume of 4 V = (RE + a)3 RE3 :

3 Since a << RE we can expand the polynomials but keep only the terms which are linear in a. Then V 4R 2a = 4(6:37106m)2(1:6104m) = 8:21018m3: E

The magnetic energy density is found from Eq. 36-22, 1 (60106 T)2 u = B2 = = 1:43103J=m3: B 2 2(4107N=A2) 0

E36-32 (a) B = 0i=2r and u = B2=20 = 0i2=82r2, or B u =(4107H=m)(10A)2=82(1:25103m)2=1:0J=m3: B

(b) E = V=l = iR=l and u = 0E2=2 = 0i2(R=l)2=2. Then E u = (8:851012F=m)(10 A)2(3:3103 =m)2=2 = 4:81015J=m3: E

pp E36-33 i = 2U=L = 2(11:2106J)=(1:48103H) = 0:123 A:

E36-34 C = q2=2U = (1:63106C)2=2(142106J) = 9:36109F: p E36-35 1=2f = LC so L = 1=42f2C, or L = 1=42(10103Hz)2(6:7106F) = 3:8105H:

E36-36 qmax2=2C = Limax2=2, or pp imax = qmax= LC = (2:94106C)= (1:13103H)(3:88106F) = 4:44102A:

Closing a switch has the e ect of \shorting” out the relevant circuit element, which e ectively removes it from the circuit. If S1 is closed we have C = RC or C = C =R, if instead S2 is closed we have L = L=R or L = R L, but if instead S3 is closed we have a LC circuit which will oscillate with period 2 p T = = 2 LC: ! E36-37

Substituting from the expressions above, 2 p T = = 2 L C: !

E36-38 The capacitors can be used individually, or in series, or in parallel. The four possible capacitances are then 2:00F, 5:00F, 2:00F + 5:00F = 7:00F, and (2:00F )(5:00F)(2:00F + 5:00F) = 1:43F. The possible resonant frequencies are then r 11 2 LC s 11 2 (10:0 mH)(1:43F ) s 11 2 (10:0 mH)(2:00F ) s 11 2 (10:0 mH)(5:00F ) s 11 = 602 Hz: 2 (10:0 mH)(7:00F )

pp E36-39 (a) k = (8:13 N)=(0:0021 m) = 3:87103N=m. ! = k=m = (3870 N=m)=(0:485 kg) = 89:3 rad/s: (c) LC = 1=!2, so C = 1=(89:3 rad/s)2(5:20 H) = 2:41105F: pp E36-40 The period of oscillation is T = 2 LC = 2 (52:2mH)(4:21F) = 2:95 ms. It requires one-quarter period for the capacitor to charge, or 0:736 ms.

(a) An LC circuit oscillates so that the energy is converted from all magnetic to all electrical twice each cycle. It occurs twice because once the energy is magnetic with the current owing in one direction through the inductor, and later the energy is magnetic with the current (c) Since it occurs twice during each oscillation it is equal to half a period, or 3.04s.

E36-41 (c) U = q2=2C = (3:24109C)2=2(1:13109F) = 4:64109J: p p 3 (b) i = 2U=L = 2(4:64109J)=(3:17103H) = 1:7110 A:

E36-43 (a) im = qm! and qm = CV m. Multiplying the second expression by L we get Lqm = V m=!2. Combining, Lim! = V m. Then ! (50 V) f = = = 6:1103=s: 2 2(0:042 H)(0:031 A) (c) C = 1=!2L = 1=(26:1103=s)2(0:042 H) = 1:6108F.

pp E36-44 (a) f = 1=2 LC = 1=2 (6:2106F)(54103H) = 275 Hz: (b) Note that from Eq. 36-32 we can deduce imax = !qmax. The capacitor starts with a charge q = CV = (6:2106F)(34 V) = 2:11104C. Then the current amplitude is pp imax = qmax= LC = (2:11104C)= (6:210 F)(5410 H) = 0:365 A: 6 3

pp E36-45 (a) ! = 1= LC = 1= (10106F)(3:0103H) = 5800 rad/s: (b) T = 2=! = 2=(5800 rad/s) = 1:1103s:

E36-46 f = (2105Hz)(1 + =180): C = 42=f2L, so 42 (9:9107F) C= = : (2105Hz)2(1 + =180)2(1 mH) (1 + =180)2

p E36-47 (a)UE=UB=2andUE+UB=U,so3UE=U,or3(q2=2C)=q2 =2C,soq=qmax= 3. max (b) Solve q = qmax cos !t, or Tp t = arccos 1= 3 = 0:152T: 2

E36-48 (a) Add the contribution from the inductor and the capacitor, (24:8103H)(9:16103A)2 (3:83106C)2 U = + = 1:99106J: 2 2(7:73106F) (c) im = p2(1:99106J)=(24:8103H) = 1:27102A.

p (a) The frequency of such a system is given by Eq. 36-26, f = 1=2 LC: Note that maximum frequency occurs with minimum capacitance. Then E36-49

s r f1 C2 (365 pF) = = = 6:04: f2 C1 (10 pF) (b) The desired ratio is 1:60=0:54 = 2:96 Adding a capacitor in parallel will result in an e ective capacitance given by C1;e = C1 + Cadd;

with a similar expression for C2. We want to choose Cadd so that s f1 C 2;e = = 2:96: f2 C 1;e Solving, C2 8:76C1 7:76 (365 pF) 8:76(10 pF) = = 36 pF: 7:76 The necessary inductance is then 11 L = = = 2:2104H: 42f2C 42(0:54106Hz)2(4011012F)

E36-50 The key here is that UE = C(V )2=2. We want to charge a capacitor with one-ninth the capacitance to have three times the potential di erence. Since 32 = 9, it is reasonable to assume that we want to take all of the energy from the rst capacitor and give it to the second. Closing S1 and S2 will not work, because the energy will be shared. Instead, close S2 until the capacitor has completely discharged into the inductor, then simultaneously open S2 while closing S1. The inductor will then discharge into the second capacitor. Open S1 when it is \full”.

p im 3 6 4 p qm = = (2:0 A) (3:010 H)(2:710 F) = 1:8010 C ! (b)dUC=dt=qi=C.Sinceq=qmsin!tandi=imcos!tthendUC=dtisamaximumwhen sin !t cos !t is a maximum, or when sin 2!t is a maximum. This happens when 2!t = =2, or (c) dUC=dt = qmim=2C, or dUC=dt=(1:80104C)(2:0A)=2(2:7106F)=67W:

E36-52 After only complete cycles q = qmaxe . Not only that, but t = N , where = 2=!0. Rt=2L Finally, !0 = p(1=LC) (R=2L)2: Since the rst term under the square root is so much larger than p the second, we can ignore the e ect of damping on the frequency, and simply use !0 ! = 1= LC. Then p p q = qmaxeNR =2L = qmaxeNR LC=L = q eNR C=L: max p p 2 Finally, R C=L = (7:22 ) (3:18 F)=(12:3 H) = 1:1510 . Then N = 5 : q = (6:31C)e100(0:0115) = 1:99C:

E36-53 Use Eq. 36-40, and since U / q2, we want to solve eRt=L = 1=2, then

L t = ln 2: R E36-54 Start by assuming that the presence of the resistance does not signi cantly change the p frequency. Then ! = 1= LC, q = qmaxeRt=2L, t = N , and = 2=!. Combining, pp q = q eNR =2L = q eNR LC=L = q eNR C=L: max max max

Then pp L=C (220mH)=(12F) R = ln(q=qmax) = ln(0:99) = 8700 : N (50) It remains to be veri ed that 1=LC (R=2L)2.

E36-55 The damped (angular) frequency is given by Eq. 36-41; the fractional change would then be ! !0 p p = 1 1 (R=2L!)2 = 1 1 (R2C=4L): ! Setting this equal to 0.01% and then solving for R, s r 4L 4(12:6103H) R = (1 (1 0:0001)2) = (1:9999104) = 2:96 : C (1:15106F) P36-1 The inductance of a toroid is 0N 2h b L = ln : 2 a If the toroid is very large and thin then we can write b = a + , where << a. The natural log then can be approximated as b ln = ln 1 + : a aa The product of and h is the cross sectional area of the toroid, while 2a is the circumference, which we will call l. The inductance of the toroid then reduces to N 2 N 2A 00 L = : 2 a l But N is the number of turns, which can also be written as N = nl, where n is the turns per unit length. Substitute this in and we arrive at Eq. 36-7.

P36-2 (a) Since ni is the net current per unit length and is this case i=W , we can simply write (b) There is only one loop of wire, so L = =i = BA=i = 0iR2=W i = 0R2=W: B

P36-3 Choose the y axis so that it is parallel to the wires and directly between them. The eld in the plane between the wires is

0i 1 1 B= + : 2 d=2 + x d=2 x The ux per length l of the wires is Z d=2a Z d=2a 0i 1 1 B = l B dx = l + dx; d=2+a 2 d=2+a d=2 + x d=2 x Z d=2a 0i 1 2 d=2+a d=2 + x 0i d a = 2l ln : 2 a The inductance is then B 0l d a L = = ln : ia P36-4 (a) Choose the y axis so that it is parallel to the wires and directly between them. The eld in the plane between the wires is

0i 1 1 B= + : 2 d=2 + x d=2 x The ux per length l between the wires is Z d=2a Z d=2a 0i 1 1 1 = B dx = + dx; d=2+a 2 d=2+a d=2 + x d=2 x Z d=2a 0i 1 2 d=2+a d=2 + x 0i d a = 2 ln : 2 a The eld in the plane inside one of the wires, but still between the centers is

0i 1 d=2 x B= + : 2 d=2 + x a2 The additional ux is then d=2 0i d=2 d=2 x ZZ 1 2 = 2 B dx = 2 + dx; d=2a 2 d=2a d=2 + x a2 0i d 1 = 2 ln + : 2 d a 2

The ux per meter between the axes of the wire is the sum, or

0i d 1 a2 (4107H=m)(11:3A) (21:8;mm) 1 = ln + ; (1:3 mm) 2 = 1:5105Wb=m: (b) The fraction f inside the wires is

d1d1 da 2 a 2 (21:8; mm) 1 (21:8; mm) 1 = += +; (21:8; mm) (1:3 mm) 2 (1:3 mm) 2 = 0:09: (c) The net ux is zero for parallel currents.

P36-5 The magnetic eld in the region between the conductors of a coaxial cable is given by 0i 2r so the ux through an area of length l, width b a, and perpendicular to B~ is ZZ B = B~ dA~ = B dA;

b l 0i ZZ a 0 2r 0il b = ln : 2 a We evaluated this integral is cylindrical coordinates: dA = (dr)(dz). As you have been warned so many times before, learn these di erentials! The inductance is then B 0l b L = = ln : i 2 a P36-6 (a) So long as the fuse is not blown there is e ectively no resistance in the circuit. Then the equation for the current is E = L di=dt, but since E is constant, this has a solution i = E t=L. (b) Note that once the fuse blows the maximum steady state current is 2=3 A, so there must be an exponential approach to that current.

P36-7 The initial rate of increase is di=dt = E =L. Since the steady state current is E =R, the current will reach the steady state value in a time given by E =R = i = E t=L, or t = L=R. But that’s L.

P36-8 (a) U = 1 Li2 = (152 H)(32 A)2=2 = 7:8104J: 2 (b) If the coil develops at nite resistance then all of the energy in the eld will be dissipated as heat. The mass of Helium that will boil o is m = Q=Lv = (7:8104J)=(85 J=mol)=(4:00g/mol) = 3:7 kg:

P36-9 (a) B = (0N i)=(2r), so B2 0N 2i2 u= = : 20 82r2 RR (b) U = u dV = urdr d dz. The eld inside the toroid is uniform in z and , so Zb 22 0N i 82r2 a

h0N 2i2 b = ln : 4 a (c) The answers are the same!

P36-10 The energy in the inductor is originally U = Li2=2. The internal energy in the resistor 0 increases at a rate P = i2R. Then Z 1 Z 1 Ri2 Li2 Pdt=R i2e2t=Ldt= 0L= 0: 0 00 2 2

P36-11 (a) In Chapter 33 we found the magnetic eld inside a wire carrying a uniform current density is 0ir B= : 2R2 The magnetic energy density in this wire is 1 0i2r2 uB = B2 = : 20 82R4

We want to integrate in cylindrical coordinates over the volume of the wire. Then the volume element is dV = (dr)(r d)(dz), so Z UB = uBdV;

R l 2 0i2r2 ZZZ 82R4 000 0i2l R Z 4R4 0 0i2l =: 16 (b) Solve L UB = i2 2 for L, and 2UB 0l L= = : i2 8 P36-12 1=C = 1=C1 + 1=C2 and L = L1 + L2. Then

s 1 p C1C2 C2=!2 + C1=!2 1 r = LC = (L1 + L2) = 0 0 = : ! C1 + C2 C1 + C2 !0

P36-13 (a) There is no current in the middle inductor; the loop equation becomes d2q q d2q q L + + L + = 0: dt2 C dt2 C Try q = qm cos !t as a solution: 11 CC p which requires ! = 1= LC: (b) Look at only the left hand loop; the loop equation becomes d2q q d2q L + + 2L = 0: dt2 C dt2 Try q = qm cos !t as a solution: 1 C p which requires ! = 1= 3LC:

P36-14 (b) (!0 !)=! is the fractional shift; this can also be written as !0 p !1 p s (100 )2(7:3106F) = 1 1 = 2:1103: 4(4:4 H) P36-15 We start by focusing on the charge on the capacitor, given by Eq. 36-40 as q = q eRt=2L cos(!0t + ): m

After one oscillation the cosine term has returned to the original value but the exponential term has attenuated the charge on the capacitor according to q = qmeRT=2L;

where T is the period. The fractional energy loss on the capacitor is then U0 U q2 = 1 = 1 eRT=L: U0 q2 m

For small enough damping we can expand the exponent. Not only that, but T = 2=!, so U 2R=!L: U

P36-16 We are given 1=2 = et=2 L when t = 2n=!0. Then 2n 2n nR !0 = = = : t 2(L=R) ln 2 L ln 2 From Eq. 36-41, ! !0 (R=2L)2 ! 2!02 (ln 2)2 82n2 0:0061 =: n2

E37-1 The frequency, f , is related to the angular frequency ! by ! = 2f = 2(60 Hz) = 377 rad/s The current is alternating because that is what the generator is designed to produce. It does this through the con guration of the magnets and coils of wire. One complete turn of the generator will (could?) produce one \cycle”; hence, the generator is turning 60 times per second. Not only does this set the frequency, it also sets the emf, since the emf is proportional to the speed at which the coils move through the magnetic eld.

E37-2 (a) XL = !L, so f = X =2L = (1:28103 )=2(0:0452 H) = 4:51103=s: L

(b) XC = 1=!C, so C = 1=2fX = 1=2(4:51103=s)(1:28103 ) = 2:76108F: C

p E37-3 (a) XL = XC implies !L = 1=!C or ! = 1= LC, so p ! = 1= (6:23103H)(11:4106F) = 3750 rad/s: (b) XL = !L = (3750 rad/s)(6:23103H) = 23:4 (c) See (a) above.

E37-4 (a) im = E=XL = E=!L, so im = (25:0 V)=(377 rad/s)(12:7 H) = 5:22103A:

(b) The current and emf are 90 out of phase. When the current is a maximum, E = 0. (c) !t = arcsin[E(t)=Em], so (13:8 V) !t = arcsin = 0:585 rad: (25:0 V) and i = (5:22103A) cos(0:585) = 4:35103A: (d) Taking energy.

(a) The reactance of the capacitor is from Eq. 37-11, XC = 1=!C. The AC generator from Exercise 4 has E = (25:0 V) sin(377 rad/s)t. So the reactance is E37-5

11 XC = = = 647 : !C (377 rad/s)(4:1F) The maximum value of the current is found from Eq. 37-13, (VC)max) (25:0V) im = = = 3:86102A: XC (647 )

(b) The generator emf is 90 out of phase with the current, so when the current is a maximum the emf is zero.

(c) The emf is -13.8 V when (13:8 V) !t = arcsin = 0:585 rad: (25:0 V) The current leads the voltage by 90 = =2, so i = im sin(!t ) = (3:86102A) sin(0:585 =2) = 3:22102A:

(d) Since both the current and the emf are negative the product is positive and the generator is supplying energy to the circuit.

E37-6 R = (!L 1=omegaC)= tan and ! = 2f = 2(941=s) = 5910 rad/s , so (5910 rad/s)(88:3103H) 1=(5910 rad/s)(937109F) R = = 91:5 : tan(75) E37-7

p (e) tan = (87 37:9 )=(160 ) = 0:3069, so = arctan(0:3069) = 17:

A circuit is considered inductive if XL > XC, this happens when im lags Em. If, on the other hand, XL < XC, and im leads Em, we refer to the circuit as capacitive. This is discussed on (a) At resonance, XL = XC . Since XL = !L and XC = 1=!C we expect that XL grows with Consequently, at frequencies above the resonant frequency XL > XC and the circuit is predomi- nantly inductive. But what does this really mean? It means that the inductor plays a major role in the current through the circuit while the capacitor plays a minor role. The more inductive a circuit is, the less signi cant any capacitance is on the behavior of the circuit. (b) Right at the resonant frequency the inductive e ects are exactly canceled by the capacitive e ects. The impedance is equal to the resistance, and it is (almost) as if neither the capacitor or inductor are even in the circuit.

E37-9 E37-10 The net y component is XC XL. The net x component is R. The magnitude of the resultant is p Z = R2 + (XC XL)2;

while the phase angle is (XC XL) tan = : R

p Atresonance!=1=(1:2H)(1:3106F)=800rad/sandZ=R.Thenim=E=Z= (10 V)=(9:6 ) = 1:04 A, so [VL]m = imXL = (1:08 A)(800 rad/s)(1:2 H) = 1000 V:

E37-12 (a) Let O = X X and A = R, then H2 = A2 + O2 = Z2, so LC

sin = (XL XC )=Z and cos = R=Z: E37-13 (a) The voltage across the generator is the generator emf, so when it is a maximum from (b) The current through the circuit is given by i = im sin(!t ). We found in Sample Problem For a resistive load we apply Eq. 37-3, VR = imR sin(!t ) = (0:196 A)(160 ) sin((=2) (0:513)) = 27:3 V: (c) For a capacitive load we apply Eq. 37-12, VC = imXC sin(!t =2) = (0:196 A)(177 ) sin((0:513)) = 17:0 V: (d) For an inductive load we apply Eq. 37-7, VL = imXL sin(!t + =2) = (0:196 A)(87 ) sin( (0:513)) = 8:4 V: (e) (27:3 V) + (17:0 V) + (8:4 V) = 35:9 V.

E37-14 If circuit 1 and 2 have the same resonant frequency then L1C1 = L2C2. The series combination for the inductors is The series combination for the capacitors is 1=C = 1=C1 + 1=C2;

so C1C2 L1C1C2 + L2C2C1 LC = (L1 + L2) = = L1C1; C1 + C2 C1 + C2 which is the same as both circuit 1 and 2.

(b) Let O = XL XC and A = R, then H2 = A2 + O2 = Z2, so cos = R=Z: Using this relation, R = (39:1 ) cos(56:3) = 21:7 : (c) If the current leads the emf then the circuit is capacitive.

E37-16 (a) Integrating over a single cycle, 1ZT ZT 11 T0 T02 11 = T= : 2T 2 (b) Integrating over a single cycle, 1T T ZZ 11 T0 T02 = 0:

The resistance would be given by Eq. 37-32, P av (0:10)(746 W) R = = = 177 : i 2 (0:650 A)2 rms

This would not be the same as the direct current resistance of the coils of a stopped motor, because there would be no inductive e ects.

E37-17 E37-18 Since irms = Erms=Z, then E2rmsR P av = i2rmsR = : Z2 p E37-19 (a) Z = (160 )2 + (177 )2 = 239 ; then 1 (36 V)2(160 ) P av = = 1:82 W: 2 (239 )2 p (b) Z = (160 )2 + (87 )2 = 182 ; then 1 (36 V)2(160 ) P av = = 3:13 W: 2 (182 )2

p E37-20 (a) Z = (12:2 )2 + (2:30 )2 = 12:4 (b) P av = (120 V)2(12:2 )=(12:4 )2 = 1140 W: (c) irms = (120 V)=(12:4 ) = 9:67 A.

p The rms valuepof any sinusoidal quantity is related to the maximum value by rms 2v = vmax. Since this factor of 2 appears in all of the expressions, we can conclude that if the rms values are equal then so are the maximum values. This means that (VR)max = (VC)max = (VL)max or imR = imXC = imXL or, with one last simpli cation, R = XL = XC: Focus on the right hand side of the last equality. If XC = XL then we have a resonance condition, and the impedance (see Eq. 37-20) is a minimum, and is equal to R. Then, according to Eq. 37-21, Em R

which has the immediate consequence that the rms voltage across the resistor is the same as the rms voltage across the generator. So everything is 100 V.

p E37-22 (a) The antenna is \in-tune” when the impedance is a minimum, or ! = 1= LC. So p f = !=2 = 1=2 (8:22106H)(0:2701012F) = 1:07108Hz: (c) XC = 1=2fC, so V = iX = (1:22107A)=2(1:07108Hz)(0:270 1012F) = 6:72104V: CC E37-21

E37-23 Assuming no inductors or capacitors in the circuit, then the circuit e ectively behaves as a DC circuit. The current through the circuit is i = E =(r + R). The power delivered to R is then P = iV = i2R = E2R=(r + R)2: Evaluate dP=dR and set it equal to zero to nd the maximum. Then dP r R dR (r + R)3 which has the solution r = R.

E37-24 (a) Since P = i 2R=2 = E 2R=2Z2, then P is a maximum when Z is a minimum, and av m m av p vise-versa. Z is a minimum at resonance, when Z = R and f = 1=2 LC. When Z is a minimum C=1=42f2L=1=42(60Hz)2(60mH)=1:2107F: (b) Z is a maximum when XC is a maximum, which occurs when C is very small, like zero. (c) When XC is a maximum P = 0. When P is a maximum Z = R so P = (30 V)2=2(5:0 ) = 90 W: (d) The phase angle is zero for resonance; it is 90 for in nite XC or XL. (e) The power factor is zero for a system which has no power. The power factor is one for a system in resonance.

E37-25 (a) The resistance is R = 15:0 . The inductive reactance is 11 XC = = = 61:3 : !C 2(550 s1)(4:72F) The inductive reactance is given by XL = !L = 2(550 s1)(25:3 mH) = 87:4 :

The impedance is then q Z = (15:0 )2 + ((87:4 ) (61:3 ))2 = 30:1 : Finally, the rms current is E rms (75:0 V) irms = = = 2:49 A: Z (30:1 ) (b) The rms voltages between any two points is given by where Z is not the impedance of the circuit but instead the impedance between the two points in question. When only one device is between the two points the impedance is equal to the reactance We’re not going to show all of the work here, but we will put together a nice table for you PointsImpedance Expression Impedance Value (V )rms, ab bc cd bd ac Z = R Z = 15:0 37.4 V, Z = XC Z = 61:3 153 V, Z = XL Z = 87:4 218 V, Z = jXL XC j Z = 26:1 65 V, p Z = R2 + X2 Z = 63:1 157 V, C

Note that this last one was Vac, and not Vad, because it is more entertaining. You probably (c) The average power dissipated from a capacitor or inductor is zero; that of the resistor is P = [(V ) ]2=R = (37:4 V)2=(15:0 ) = 93:3 W: R R rms

E37-26 (a) The energy stored in the capacitor is a function of the charge on the capacitor; although the charge does vary with time it varies periodically and at the end of the cycle has returned to the original values. As such, the energy stored in the capacitor doesn’t change from one period to the (b) The energy stored in the inductor is a function of the current in the inductor; although the current does vary with time it varies periodically and at the end of the cycle has returned to the original values. As such, the energy stored in the inductor doesn’t change from one period to the (c)P=Ei=Emimsin(!t)sin(!t),sotheenergygeneratedinonecycleis

ZT ZT 00 ZT 0 T = Emim cos : 2 (d) P = i 2R sin2(!t ), so the energy dissipated in one cycle is m

ZT ZT m 00 ZT m 0 T = im2R: 2 E37-27 Apply Eq. 37-41, N s (780) V s = V p = (150 V) = 1:8103 V: N p (65)

E37-28 (a) Apply Eq. 37-41, N s (10) V s = V p = (120 V) = 2:4 V: N p (500)

N s (10) ip = is = (0:16 A) = 3:2103A: N p (500)

The autotransformer could have a primary connected between taps T1 and T2 (200 turns), The same possibilities are true for the secondary connections. Ignoring the one-to-one connections there are 6 choices| three are step up, and three are step down. The step up ratios are 1000=200 = 5, 800=200 = 4, and 1000=800 = 1:25. The step down ratios are the reciprocals of these three values.

E37-30 = (1:69108 m)[1 (4:3103=C)(14:6C)] = 1:58108 m. The resistance of the two wires is L (1:58108 m)2(1:2103m) R = = = 14:9 : A (0:9103m)2 P = i2R = (3:8 A)2(14:9 ) = 220 W.

E37-31 The supply current is p ip = (0:270 A)(74103V= 2)=(220 V) = 64:2 A:

The potential drop across the supply lines is V = (64:2 A)(0:62 ) = 40 V: This is the amount by which the supply voltage must be increased.

E37-32 Use Eq. 37-46: p N p=N s = (1000 )=(10 ) = 10:

P37-1 (a) The emf is a maximum when !t =4 = =2, so t = 3=4! = 3=4(350 rad/s) = (b) The current is a maximum when !t 3=4 = =2, so t = 5=4! = 5=4(350 rad/s) = (d) XL = Em=im and XL = !L, so E m (31:4 V) L = = = 0:144 H: im! (0:622 A)(350 rad/s)

P37-2 (a) The emf is a maximum when !t =4 = =2, so t = 3=4! = 3=4(350 rad/s) = (b) The current is a maximum when !t+=4 = =2, so t = =4! = =4(350 rad/s) = 2:24103s. (d) XC = Em=im and XC = 1=!C, so

im (0:622 A) C = = = 5:66105F: Em! (31:4V)(350rad/s)

P37-3 (a) Since the maximum values for the voltages across the individual devices are propor- tional to the reactances (or resistances) for devices in series (the constant of proportionality is the From Eq. 37-18, XL XC 2R R RR (b) The impedance of the circuit, in terms of the resistive element, is ppp Z = R2 + (X X )2 = R2 + (2R R)2 = 2 R: LC

ButEm=imZ,soZ=(34:4V)=(0:320A)=108.Thenwecanuseourpreviousworktondsthat R = 76 .

P37-4 When the switch is open the circuit is an LRC circuit. In position 1 the circuit is an RLC circuit, but the capacitance is equal to the two capacitors of C in parallel, or 2C. In position 2 the The impedance when the switch is in position 2 is Z2 = j!L 1=!Cj. But

Z2 = (170 V)=(2:82 A) = 60:3 : The phase angle when the switch is open is 0 = 20. But

!L 1=!C Z2 RR The phase angle when the switch is in position 1 is !L 1=!2C R

so !L 1=!2C = (166 ) tan(10) = 29:2 . Equating the !L part, C = 1=2(377 rad/s)[(60:3 ) + (29:2 )] = 1:48105F: Finally, (60:3 ) + 1=(377 rad/s)(1:48105F) L = = 0:315 H: (377 rad/s)

All three wires have emfs which vary sinusoidally in time; if we choose any two wires the phase di erence will have an absolute value of 120. We can then choose any two wires and expect (by symmetry) to get the same result. We choose 1 and 2. The potential di erence is then P37-5

V1 V2 = V m (sin !t sin(!t 120)) : We need to add these two sine functions to get just one. We use 11 sin sin = 2 sin ( ) cos ( + ): 22 Then 11 22 p 3 2 p = 3V m sin(!t + 30):

pp (c) The current amplitude will be halved when the impedance is doubled, or when Z = 2R. This occurs when 3R2 = (!L 1=!C)2, or 3R2!2 = !4L2 2!2L=C + 1=C2: The solution to this quadratic is p 2L + 3CR2 9C2R4 + 12CR2L 2L2C (d) !=! = (8:9 rad/s)=(229 rad/s) = 0:039.

P37-8 (a) The current amplitude will be halved when the impedance is doubled, or when Z = 2R. This occurs when 3R2 = (!L 1=!C)2, or 3R2!2 = !4L2 2!2L=C + 1=C2: The solution to this quadratic is p 2L + 3CR2 9C2R4 + 12CR2L 2L2C Note that ! = !+ !; with a wee bit of algebra, !(! + ! ) = !2 !2 : ++

Also, !+ + ! 2!. Hence, p 9C2R4 + 12CR2L 2L2C p !2R 9C2R2 + 12LC 2L p !R 9!2C2R2 + 12 2L p ! R 9CR2=L + 12 ! 2L! p 3R !L assuming that CR2 4L=3.

P37-9 P37-11 (a) The resistance of this bulb is (V )2 (120 V)2 R = = = 14:4 : P (1000 W)

The power is directly related to the brightness; if the bulb is to be varied in brightness by a factor of 5 then it would have a minimum power of 200 W. The rms current through the bulb at this power would be pp irms = P=R = (200 W)=(14:4 ) = 3:73 A: The impedance of the circuit must have been E rms (120 V) Z = = = 32:2 : irms (3:73 A)

The inductive reactance would then be p p X = Z2 R2 = (32:2 )2 (14:4 )2 = 28:8 : L

Finally, the inductance would be L = XL=! = (28:8 )=(2(60:0 s1)) = 7:64 H:

(b) One could use a variable resistor, and since it would be in series with the lamp a value of 32:2 14:4 = 17:8 would work. But the resistor would get hot, while on average there is no power radiated from a pure inductor.

E38-1 The maximum value occurs where r = R; there Bmax = 1 00R dE=dt. For r < R B is 2 half of Bmax when r = R=2. For r > R B is half of Bmax when r = 2R. Then the two values of r are 2:5 cm and 10:0 cm.

E38-2 For a parallel plate capacitor E = =0 and the ux is then E = A=0 = q=0. Then

d E dq d dV id = 0 = = CV = C : dt dt dt dt E38-3 Use the results of Exercise 2, and change the potential di erence across the plates of the capacitor at a rate dV id (1:0 mA) = = = 1:0 kV/s: dt C (1:0F) Provide a constant current to the capacitor dQ d dV i = = CV = C = id: dt dt dt E38-4 Since E is uniform between the plates E = EA, regardless of the size of the region of interest. Since jd = id=A, id 1 d E dE jd = = 0 = 0 : A A dt dt (b) Since E = q=0A, dE=dt = i=0A, or

dE=dt = (1:84 A)=(8:851012F=m)(1:22 m)2 = 1:401011V=m: (c) id = 0d E=dt = 0adE=dt. a here refers to the area of the smaller square. Combine this with the results of part (b) and id = ia=A = (1:84 A)(0:61 m=1:22 m)2 = 0:46 A:

(d) B~d~s=0id=(4107H=m)(0:46A)=5:78107Tm: H E38-6 Substitute Eq. 38-8 into the results obtained in Sample Problem 38-1. Outside the capacitor E = R2E, so 0 0R2dE 0 B = = id: 2r dt 2r Inside the capacitor the enclosed ux is E = r2E; but we want instead to de ne id in terms of the total E inside the capacitor as was done above. Consequently, inside the conductor

0r 0R2dE 0r B = = id: 2R2 dt 2R2 E38-7 Since the electric eld is uniform in the area and perpendicular to the surface area we have ZZZ E = E~ dA~ = E dA = E dA = EA:

The displacement current is then dE dE id = 0A = (8:851012F=m)(1:9 m2) : dt dt

(a) In the rst region the electric eld decreases by 0.2 MV/m in 4s, so (0:2106V=m) id = (8:851012F=m)(1:9 m2) = 0:84 A: (4106s) (b) The electric eld is constant so there is no change in the electric ux, and hence there is no (c) In the last region the electric eld decreases by 0.4 MV/m in 5s, so (0:4106V=m) id = (8:851012F=m)(1:9 m2) = 1:3 A: (5106s)

E38-8 (a) Because of the circular symmetry B~ d~s = 2rB, where r is the distance from the H

center of the circular plates. Not only that, but id = jdA = r2jd. Equate these two expressions, and B=0rjd=2=(4107H=m)(0:053m)(1:87101A=m)=2=6:23107T: (b) dE=dt = id=0A = jd=0 = (1:87101A=m)=(8:851012F=m) = 2:111012V=m:

E38-9 The magnitude of E is given by (162 V) (4:8103m) Using the results from Sample Problem 38-1, 00R dE 2 dt t=0 (4107H=m)(8:851012F=m)(0:0321m) (162V) = 2(60=s) ; 2 (4:8103m) = 2:271012T:

(b) Eq. 27-11 from page 618 and the equation from Ex. 27-25 on page 630. (d) Eqs. 34-16 and 34-17 from page 785.

(a) Consider the path abef a. The closed line integral consists of two parts: b ! e and e ! f ! a ! b. Then E38-11 I d E~ d~s = dt can be written as ZZ d E~ d~s + E~ d~s = abef : b!e e!f !a!b dt Now consider the path bcdeb. The closed line integral consists of two parts: b ! c ! d ! e and e ! b. Then I d E~ d~s = dt can be written as ZZ d E~ d~s + E~ d~s = bcde: b!c!d!e e!b dt

These two expressions can be added together, and since ZZ E~ d~s = E~ d~s e!b b!e we get ZZ d E~ d~s + E~ d~s = ( abef + bcde) : e!f !a!b b!c!d!e dt The left hand side of this is just the line integral over the closed path ef adcde; the right hand side is the net change in ux through the two surfaces. Then we can simplify this expression as I d E~ d~s = : dt (c) If the equations were not self consistent we would arrive at di erent values of E and B depending on how we de ned our surfaces. This multi-valued result would be quite unphysical.

E38-12 (a) Consider the part on the left. It has a shared surface s, and the other surfaces l. Applying Eq. I, IZZ ql=0 = E~ dA~ = E~ dA~ + E~ dA~ : sl Consider the part on the right. It has a shared surface s, and the other surfaces r. Applying Eq. I, IZZ qr=0 = E~ dA~ = E~ dA~ + E~ dA~ : sr Adding these two expressions will result in a canceling out of the part Z E~ dA~ s

since one is oriented opposite the other. We are left with ZZI qr + ql = E~ dA~ + E~ dA~ = E~ dA~ : 0 r l E38-13

E38-14 (a) Electric dipole is because the charges are separating like an electric dipole. Magnetic dipole because the current loop acts like a magnetic dipole.

E38-15 A series LC circuit will oscillate naturally at a frequency !1 f= = p 2 2 LC We will need to combine this with v = f , where v = c is the speed of EM waves. We want to know the inductance required to produce an EM wave of wavelength = 550109m, so 2 (550 109m)2 L = = = 5:01 1021 H: 42c2C 42(3:00 108m=s)2(17 1012 F) This is a small inductance!

E38-16 (a) B = E=c, and B must be pointing in the negative y direction in order that the wave be propagating in the positive x direction. Then Bx = Bz = 0, and B = E =c = (2:34104V=m)=(3:00108m=s) = (7:801013T) sin k(x ct): yz

(b) = 2=k = 2=(9:72106=m) = 6:46107m: E38-17 The electric and magnetic eld of an electromagnetic wave are related by Eqs. 38-15 and 38-16, E (321V=m) B = = = 1:07 pT: c (3:00 108m=s) E38-18 Take the partial of Eq. 38-14 with respect to x, @ @E @ @B @x @x @x @t @2E @2B = : @x2 @x@t Take the partial of Eq. 38-17 with respect to t, @ @B @ @E @t @x @t @t @2B @2E = 00 : @t@x @t2 Equate, and let 00 = 1=c2, then @2E 1 @2E =: @x2 c2 @t2 Repeat, except now take the partial of Eq. 38-14 with respect to t, and then take the partial of Eq. 38-17 with respect to x.

E38-19 (a) Since sin(kx !t) is of the form f (kx !t), then we only need do part (b). (b) The constant Em drops out of the wave equation, so we need only concern ourselves with f (kx !t). Letting g = kx !t, @2f @2f @t2 @x2 2 2 @2f 2 @ f @g @g @g2 @t @g2 @x @g @g @t @x ! = ck:

Intensity is given by Eq. 38-28, which is simply an expression of power divided by surface area. To nd the intensity of the TV signal at -Centauri we need to nd the distance in meters;

r = (4:30 light-years)(3:00108 m=s)(3:15107 s/year) = 4:06 1016 m: E38-23

The intensity of the signal when it has arrived at out nearest neighbor is then P (960 kW) 2 I = = = 4:63 1029 W/m 4r2 4(4:06 1016 m)2

E38-24 (a) From Eq. 38-22, S = cB2=0. B = Bm sin !t. The time average is de ned as 1T 2T 2 ZZ cBm cBm S dt = cos2 !t dt = : T 0 0T 0 20 (b) S = (3:0108m=s)(1:0104T)2=2(4107H=m) = 1:2106W=m2: av

E38-25 I = P=4r2, so pp r = P=4I = (1:0103W)=4(130 W=m2) = 0:78 m:

E38-26 uE = 0E2=2 = 0(cB)2=2 = B2=20 = uB: (a) Intensity is related to distance by Eq. 38-28. If r1 is the original distance from the street lamp and I1 the intensity at that distance, then

P I1 = : 4r2 1 There is a similar expression for the closer distance r2 = r1 162 m and the intensity at that distance I2 = 1:50I1. We can combine the two expression for intensity, PP 4r2 4r2 21 1 p2 r1 = 1:50 (r1 162 m): (b) No, we can’t nd the power output from the lamp, because we were never provided with an absolute intensity reference.

p E38-28 (a) Em = 20cI, so p Em = 2(4107H=m)(3:00108m=s)(1:38103W=m2) = 1:02103V=m: (b) Bm = Em=c = (1:02103V=m)=(3:00108m=s) = 3:40106T: (b)I=E2=2c=(1:96V)2=2(4107H=m)(3:00108m=s)=5:10103W=m2: m0 (c) P = 4r2I = 4(11:2 m)2(5:10103W=m2) = 8:04 W: E38-27

E38-30 (a) The intensity is P (11012W) I = = = 1:961027W=m2: A 4(6:37106m)2 The power received by the Arecibo antenna is P = IA = (1:961027W=m2)(305 m)2=4 = 1:41022W: (b) The power of the transmitter at the center of the galaxy would be P = IA = (1:961027W)(2:3104ly)2(9:461015m/ly)2 = 2:91014W: E38-31 (a) The electric eld amplitude is related to the intensity by Eq. 38-26, E2 20c or Em = p20cI = p2(4107H=m)(3:00108m=s)(7:83W=m2) = 7:68102 V=m: (b) The magnetic eld amplitude is given by Em (7:68 102 V=m) Bm = = = 2:56 1010 T c (3:00 108m=s) (c) The power radiated by the transmitter can be found from Eq. 38-28, P = 4r2I = 4(11:3 km)2(7:83W=m2) = 12:6 kW:

E38-32 (a) The power incident on (and then re ected by) the target craft is P1 = I1A = P0A=2r2. The intensity of the re ected beam is I2 = P1=2r2 = P A=42r4. Then 0

I = (183103W)(0:222 m2)=42(88:2103m)4 = 1:701017W=m2: 2 (b) Use Eq. 38-26: p p 7 Em = 20cI = 2(4107H=m)(3:00108m=s)(1:701017W=m2) = 1:1310 V=m: pp (c) Brms = Em= 2c = (1:13107 V=m)= 2(3:00108m=s) = 2:661016T:

Radiation pressure for absorption is given by Eq. 38-34, but we need to nd the energy absorbed before we can apply that. We are given an intensity, a surface area, and a time, so U = (1:1103W=m2)(1:3 m2)(9:0103s) = 1:3107J: The momentum delivered is p = (U)=c = (1:3107J)=(3:00108m=s) = 4:3102kg m=s: E38-34 (a) F=A = I=c = (1:38103W=m2)=(3:00108m=s) = 4:60106Pa: (b) (4:60106Pa)=(101105Pa) = 4:551011: E38-35 F=A = 2P=Ac = 2(1:5109W)=(1:3106m2)(3:0108m=s) = 7:7106Pa: E38-33

E38-36 F=A = P=4r2c, so F=A = (500 W)=4(1:50 m)2(3:00108m=s) = 5:89108Pa:

E38-37 (a) F = IA=c, so (1:38103W=m2)(6:37106m)2 F = = 5:86108N: (3:00108m=s)

E38-38 (a) Assuming MKSA, the units are m F V N m C V sN Ns = =: s m m Am s Vm m Cm m2s

(b) Assuming MKSA, the units are A2 V N A2 J N 1 J J ===: N m Am N Cm Am sm m m2s

We can treat the object as having two surfaces, one completely re ecting and the other completely absorbing. If the entire surface has an area A then the absorbing part has an area f A while the re ecting part has area (1 f )A. The average force is then the sum of the force on each part, E38-39

I 2I cc which can be written in terms of pressure as F av I = (2 f ): Ac E38-40 We can treat the object as having two surfaces, one completely re ecting and the other completely absorbing. If the entire surface has an area A then the absorbing part has an area f A while the re ecting part has area (1 f )A. The average force is then the sum of the force on each part, I 2I cc which can be written in terms of pressure as F av I = (2 f ): Ac The intensity I is that of the incident beam; the re ected beam will have an intensity (1 f )I . Each beam will contribute to the energy density| I =c and (1 f )I =c, respectively. Add these two energy densities to get the net energy density outside the surface. The result is (2 f )I =c, which is the left hand side of the pressure relation above.

E38-41 The bullet density is = N m=V . Let V = Ah; the kinetic energy density is K=V = 1 N mv2=Ah: h=v, however, is the time taken for N balls to strike the surface, so that 2

F Nmv Nmv2 2K P= = = = : A At Ah V

E38-42 F = IA=c; P = IA; a = F=m; and v = at. Combine: v = P t=mc = (10103W)(86400 s)=(1500 kg)(3108m=s) = 1:9103m=s:

E38-43 The force of radiation on the bottom of the cylinder is F = 2IA=c. The force of gravity on the cylinder is W = mg = HAg: Equating, 2I=c = Hg. The intensity of the beam is given by I = 4P=d2: Solving for H, 8P 8(4:6 W) H = = = 4:9107m: cgd2 (3:0108m=s)(1200 kg=m3)(9:8 m=s2)(2:6103m)2 E38-44 F = 2IA=c. The value for I is in Ex. 38-37, among other places. Then F = (1:38103W=m2)(3:1106m2)=(3:00108m=s) = 29 N:

P38-1 For the two outer circles use Eq. 33-13. For the inner circle use E = V=d, Q = CV , C = 0A=d, and i = dQ=dt. Then dQ 0A dV dE i = = = 0A : dt d dt dt ThechangeinuxisdE=dt=AdE=dt.Then I B~ d~l = 0 d E dt so B = 0i=2r.

P38-2 (a) id = i. Assuming V = (174103V) sin !t, then q = CV and i = dq=dt = Cd(V )=dt. Combine, and use ! = 2(50:0=s), id = (1001012F)(174103V)2(50:0=s) = 5:47103A:

(b) d E=dt = id=0 = (7:63A)=(8:851012F=m) = 8:6210 V=m: 5 (c) i = dq=dt = Cd(V )=dt; C = 0A=d; [d(V )=dt]m = Em!. Combine, and 0A 0AEm! (8:851012F=m)(0:182m)2(225V)(128rad/s) d = = = = 3:48103m: C i (7:63A)

RR (d) 2rB = 00r2dE=dt, so B = 0r(dE=dt)=2 = 0 rt=2R2: (e) Check Exercise 38-10!

P38-5 (a) E~ = E^j and B~ = Bk^. Then ~S = E~ B~ =0, or ~S = EB=mu0 ^i: Energy only passes through the yz faces; it goes in one face and out the other. The rate is P = (b) The net change is zero.

P38-6 (a) For a sinusoidal time dependence jdE=dtjm = !Em = 2fEm. Then jdE=dtjm = 2(2:4109=s)(13103V=m) = 1:961014V=m s: (b) Using the result of part (b) of Sample Problem 38-1, 11 B = (4107H=m)(8:91012F=m)(2:4102m) (1:961014V=m s) = 1:3105T: 22

Look back to Chapter 14 for a discussion on the elliptic orbit. On page 312 it is pointed out that the closest distance to the sun is Rp = a(1 e) while the farthest distance is Ra = a(1 + e), The fractional variation in intensity is I I p I a I Ia Ip Ia Ra2 Rp2 (1 + e)2 = 1: (1 e)2 We need to expand this expression for small e using (1 + e)2 1 + 2e; and (1 e)2 1 + 2e; and nally (1 + 2e)2 1 + 4e: Combining, I (1 + 2e)2 1 4e: I

P38-8 The beam radius grows as r = (0:440 rad)R, where R is the distance from the origin. The beam intensity is P (3850 W) I = = = 4:3102W: r2 (0:440 rad)2(3:82108m)2 P38-7

P38-9 Eq. 38-14 requires @E @B @x @t Emk = Bm!:

Eq. 38-17 requires @E @B @t @x 00Em! = Bmk:

or !1 =c=p k 00 Not only that, but Em = cBm. You’ve seen an expression similar to this before, and you’ll see (b) We’ll assume that Eq. 38-21 is applicable here. Then 1 EmBm 0 0 E2 = m sin 2kx sin 2!t 40c (c) The time averaged power ow across any surface is the value of 1T ZZ T0 where T is the period of the oscillation. We’ll just gloss over any concerns about direction, and assume that the ~S will be constant in direction so that we will, at most, need to concern ourselves about a constant factor cos . We can then deal with a scalar, instead of vector, integral, and we can integrate it in any order we want. We want to do the t integration rst, because an integral over sin !t for a period T = 2=! is zero. Then we are done! (d) There is no energy ow; the energy remains inside the container.

P38-10 (a) The electric eld is parallel to the wire and given by E=V=d=iR=d=(25:0A)(1:00 =300m)=8:33102V=m (b) The magnetic eld is in rings around the wire. Using Eq. 33-13, 0i (4107H=m)(25A) B = = = 4:03103T: 2r 2(1:24103m) (c) S = EB=0, so S=(8:33102V=m)(4:03103T)=(4107H=m)=267W=m2: P38-11 (a) We’ve already calculated B previously. It is 0i E B = where i = : 2r R The electric eld of a long straight wire has the form E = k=r, where k is some constant. But

Z Zb V = E~ d~s = E dr = k ln(b=a): a

In this problem the inner conductor is at the higher potential, so V E ln(b=a) ln(b=a)

and then the electric eld is E E= : r ln(b=a) This is also a vector eld, and if E is positive the electric eld points radially out from the central (b) The Poynting vector is 1 0 E~ is radial while B~ is circular, so they are perpendicular. Assuming that E is positive the direction of ~S is away from the battery. Switching the sign of E (connecting the battery in reverse) will ip the direction of both E~ and B~ , so ~S will pick up two negative signs and therefore still point away The magnitude is EB E2 S= = 0 2R ln(b=a)r2 (c) We want to evaluate a surface integral in polar coordinates and so dA = (dr)(rd). We have already established that ~S is pointing away from the battery parallel to the central axis. Then we can integrate ZZ P = ~S dA~ = S dA;

Z b Z 2 E 2 a 0 2R ln(b=a)r2 Z b E2 a R ln(b=a)r E2 =: R (d) Read part (b) above.

P38-12 (a) B~ is oriented as rings around the cylinder. If the thumb is in the direction of current then the ngers of the right hand grip ion the direction of the magnetic eld lines. E~ is directed parallel to the wire in the direction of the current. ~S is found from the cross product of these two, (b) The magnetic eld on the surface is given by Eq. 33-13: B = 0i=2a: The electric eld on the surface is given by E = V =l = iR=l Then S has magnitude i iR i2R S = EB=0 = = : 2a l 2al R ~S dA~ is only evaluated on the surface of the cylinder, not the end caps. ~S is everywhere parallel to dA~ , so the dot product reduces to S dA; S is uniform, so it can be brought out of the integral; R Hence, Z ~S dA~ = i2R;

(b) B~ must be directed along the z axis. The magnitude is B = E=c = (288 V=m)=(3:00108m=s) = 9:6107T: (c) k = 2= = 2=(3:18 m) = 1:98=m while ! = 2f , so ! = 2(9:43107Hz) = 5:93108 rad/s: (d) I = EmBm=20, so (288 V)(9:6107T) I = = 110 W: 2(4107H=m) (e) P = I=c = (110 W)=(3:00108m=s) = 3:67107Pa.

P38-14 (a) B~ is oriented as rings around the cylinder. If the thumb is in the direction of current then the ngers of the right hand grip ion the direction of the magnetic eld lines. E~ is directed parallel to the wire in the direction of the current. ~S is found from the cross product of these two, (b) The magnitude of the electric eld is V Q Q it E= = = = : d Cd 0A 0A The magnitude of the magnetic eld on the outside of the plates is given by Sample Problem 38-1, 00R dE 00iR 00R B = = = E: 2 dt 20A 2t ~S has magnitude EB 0R S = = E2: 0 2t Integrating, Z2 R 0E ~S dA~ = 0 E22Rd = Ad : 2t t But E is linear in t, so d(E2)=dt = 2E2=t; and then

Z d1 ~S dA~ = Ad 0E2 : dt 2 (b) p = I=c = (3:6109W=m2)=(3:00108m=s) = 12 Pa (d) a = F=m = F=V , so (1:671011N) a = = 2:9103m=s2: 4(4880 kg=m3)(1:05)3(633109)3=3

P38-16 The force from the sun is F = GM m=r2. The force from radiation pressure is 2IA 2PA F= = : c 4r2c

Equating, 4GMm 2P=c so 4(6:671011N m2=kg2)(1:991030kg)(1650 kg) A = = 1:06106m2: 2(3:91026W)=(3:0108m=s) That’s about one square kilometer.

E39-1 Both scales are logarithmic; choose any data point from the right hand side such as c = f (1 Hz)(3108m) = 3108m=s;

and another from the left hand side such as c = f (11021 Hz)(31013m) = 3108m=s:

E39-2 (a) f = v= = (3:0108m=s)=(1:0104)(6:37106m) = 4:7103 Hz: If we assume that this is the data transmission rate in bits per second (a generous assumption), then it would take 140 days to download a web-page which would take only 1 second on a 56K modem! (b) T = 1=f = 212 s = 3:5 min.

E39-3 (a) Apply v = f . Then f = (3:0108m=s)=(0:0671015m) = 4:51024 Hz: (b) = (3:0108m=s)=(30 Hz) = 1:0107m:

E39-4 Don’t simply take reciprocal of linewidth! f = c=, so f = (c=2). Ignore the negative, and f = (3:00108m=s)(0:010109m)=(632:8109m)2 = 7:5109 Hz: E39-5 (a) We refer to Fig. 39-6 to answer this question. The limits are approximately 520 nm (b) The wavelength for which the eye is most sensitive is 550 nm. This corresponds to to a frequency of f = c= = (3:00 108 m=s)=(550 109m) = 5:45 1014 Hz: This frequency corresponds to a period of T = 1=f = 1:83 1015s.

E39-6 f = c=. The number of complete pulses is f t, or ft = ct= = (3:00108m=s)(4301012s)=(520109m) = 2:48105:

(b) The distance traveled by the light is (1:51011m) + 2(3:8108m), so t = (1:511011m)=(3108m=s) = 503 s: (d) 1054 6500 5400 BC.

This is a question of how much time it takes light to travel 4 cm, because the light traveled from the Earth to the moon, bounced o of the re ector, and then traveled back. The time to travel 4 cm is t = (0:04 m)=(3 108 m=s) = 0:13 ns. Note that I interpreted the question di erently than the answer in the back of the book.

E39-10 Consider any incoming ray. The path of the ray can be projected onto the xy plane, the xz plane, or the yz plane. If the projected rays is exactly re ected in all three cases then the three dimensional incoming ray will be re ected exactly reversed. But the problem is symmetric, so it is Now the problem has been reduced to Sample Problem 39-2, so we are done.

We will choose the mirror to lie in the xy plane at z = 0. There is no loss of generality in doing so; we had to de ne our coordinate system somehow. The choice is convenient in that any normal is then parallel to the z axis. Furthermore, we can arbitrarily de ne the incident ray to originate at (0; 0; z1). Lastly, we can rotate the coordinate system about the z axis so that the The point of re ection for this ray is somewhere on the surface of the mirror, say (x2; y2; 0). This distance traveled from the point 1 to the re ection point 2 is E39-11

q p d = (0 x2)2 + (0 y )2 + (z 0)2 = x2 + y2 + z2 12 2 1 2 2 1

and the distance traveled from the re ection point 2 to the nal point 3 is q p d23 = (x2 0)2 + (y2 y3)2 + (0 z3)2 = x2 + (y2 y3)2 + z2: 23

The only point which is free to move is the re ection point, (x2; y2; 0), and that point can only move in the xy plane. Fermat’s principle states that the re ection point will be such to minimize the total distance, qq d12 + d23 = x + y + z + x2 2 2 2 2 + (y y )2 + z2: 221 33

We do this minimization by taking the partial derivative with respect to both x2 and y2. But we can do part by inspection alone. Any non-zero value of x2 can only add to the total distance, regardless of the value of any of the other quantities. Consequently, x2 = 0 is one of the conditions We are done! Although you are invited to nish the minimization process, once we know that x2 = 0 we have that point 1, point 2, and point 3 all lie in the yz plane. The normal is parallel to the z axis, so it also lies in the yz plane. Everything is then in the yz plane.

22 E39-17 The speed of light in a substance with index of refraction n is given by v = c=n. An electron will then emit Cerenkov radiation in this particular liquid if the speed exceeds v = c=n = (3:00 108 m=s)=(1:54) = 1:95108 m=s:

E39-18 Since t = d=v = nd=c, t = n d=c. Then t = (1:00029 1:00000)(1:61103m)=(3:00108m=s) = 1:56109s: E39-19 The angle of the refracted ray is 2 = 90, the angle of the incident ray can be found by trigonometry, (1:14 m) (0:85 m) We can use these two angles, along with the index of refraction of air, to nd that the index of refraction of the liquid from Eq. 39-4, sin 2 (sin 90) n1 = n2 = (1:00) = 1:25: sin 1 (sin 53:3) There are no units attached to this quantity.

E39-20 For an equilateral prism = 60. Then sin[ + ]=2 sin[(37) + (60)]=2 n = = = 1:5: sin[ =2] sin[(60)=2] E39-21 p E39-22 t = d=v; but L=d = cos 2 = 1 sin2 2 and v = c=n. Combining, nL n2L (1:63)2(0:547 m) t = p = p = q = 3:07109s: c 1 sin2 c n2 sin2 2 1 (3108m=s) (1:632) sin2(24)

E39-23 The ray of light from the top of the smokestack to the life ring is 1, where tan 1 = L=h Snell’s law gives n1 sin 1 = n2 sin 2, so = arcsin[(1:33) sin(27)=(1:00)] = 37:1: 1

Then L = h tan 1 = (98 m) tan(37:1) = 74 m: E39-24 The length of the shadow on the surface of the water is x1 = (0:64 m)= tan(55) = 0:448 m: The ray of light which forms the \end” of the shadow has an angle of incidence of 35, so the ray travels into the water at an angle of

(1:00) = arcsin sin(35) = 25:5: 2 (1:33) The ray travels an additional distance x = (2:00 m 0:64 m)= tan(90 25:5) = 0:649 m 2

The total length of the shadow is (0:448 m) + (0:649 m) = 1:10 m:

E39-25 We’ll rely heavily on the gure for our arguments. Let x be the distance between the points on the surface where the vertical ray crosses and the bent ray crosses.

2 x d app 1 d In this exercise we will take advantage of the fact that, for small angles , sin tan In this approximation Snell’s law takes on the particularly simple form n11 = n22 The two angles here are conveniently found from the gure, x d and x 2 tan 2 = : dapp Inserting these two angles into the simpli ed Snell’s law, as well as substituting n1 = n and n2 = 1:0, xx d dapp d dapp = : n

E39-26 (a) You need to address the issue of total internal re ection to answer this question. (b) Rearrange sin[ + ]=2 n = sin[ =2] = and = ( + )=2 to get = arcsin (n sin[ =2]) = arcsin ((1:60) sin[(60)=2]) = 53:1:

E39-27 Use the results of Ex. 39-35. The apparent thickness of the carbon tetrachloride layer, as viewed by an observer in the water, is dc;w = nwdc=nc = (1:33)(41 mm)=(1:46) = 37:5 mm:

The total \thickness” from the water perspective is then (37:5 mm) + (20 mm) = 57:5 mm. The apparent thickness of the entire system as view from the air is then dapp = (57:5 mm)=(1:33) = 43:2 mm:

E39-28 (a) Use the results of Ex. 39-35. dapp = (2:16 m)=(1:33) = 1:62 m. (b) Need a diagram here!

E39-29 (a) n = =n = (612 nm)=(1:51) = 405 nm: (b) L = nLn = (1:51)(1:57 pm) = 2:37 pm. There is actually a typo: the \p” in \pm” was supposed to be a . This makes a huge di erence for part (c)!

E39-30 (a) f = c= = (3:00108m=s)=(589 nm) = 5:091014Hz: (b) n = =n = (589 nm)=(1:53) = 385 nm: (c) v = f = (5:091014Hz)(385 nm) = 1:96108m=s:

E39-31 (a) The second derivative of p p L = a2 + x2 + b2 + (d x)2 is a2(b2 + (d 2)2)3=2 + b2(a2 + x2)3=2 : (b2 + (d 2)2)3=2(a2 + x2)3=2 (a) The second derivative of p p L = n1 a2 + x2 + n2 b + (d x)2 2

is n1a2(b2 + (d 2)2)3=2 + n b2(a2 + x2)3=2 2 : (b2 + (d 2)2)3=2(a2 + x2)3=2 This is always a positive number, so dL=dx = 0 is a minimum.

E39-32 (a) The angle of incidence on the face ac will be 90 . Total internal re ection occurs when sin(90 ) > 1=n, or < 90 arcsin[1=(1:52)] = 48:9: (b) Total internal re ection occurs when sin(90 ) > nw=n, or

< 90 arcsin[(1:33)=(1:52)] = 29:0: E39-33 (a) The critical angle is given by Eq. 39-17, n2 (1:586) c = sin1 = sin1 = 72:07: n1 (1:667)

(b) Critical angles only exist when \attempting” to travel from a medium of higher index of refraction to a medium of lower index of refraction; in this case from A to B.

E39-34 If the re is at the water’s edge then the light travels the water is numerically equivalent to a critical angle, so the sh = arcsin(1=1:33) = 49 with the vertical. That’s the same as 41

178 along the surface, entering the Then the angle of refraction in needs to look up at an angle of with the horizontal.

E39-35 Light can only emerge from the water if it has an angle of incidence less than the critical angle, or < c = arcsin 1=n = arcsin 1=(1:33) = 48:8: The radius of the circle of light is given by r=d = tan c, where d is the depth. The diameter is twice this radius, or 2(0:82 m) tan(48:8) = 1:87 m:

E39-36 The refracted angle is given by n sin 1 = sin(39). This ray strikes the left surface with an angle of incidence of 90 1. Total internal re ection occurs when sin(90 1) = 1=n;

but sin(90 1) = cos 1, so we can combine and get tan = sin(39) with solution 1 = 32:2 : The

index of refraction of the glass is then n = sin(39)= sin(32:2) = 1:18: E39-37 The light strikes the quartz-air interface from the inside; it is originally \white”, so if the re ected ray is to appear \bluish” (reddish) then the refracted ray should have been \reddish” (bluish). Since part of the light undergoes total internal re ection while the other part does not, then the angle of incidence must be approximately equal to the critical angle. (a) Look at Fig. 39-11, the index of refraction of fused quartz is given as a function of the wavelength. As the wavelength increases the index of refraction decreases. The critical angle is a function of the index of refraction; for a substance in air the critical angle is given by sin c = 1=n: As n decreases 1=n increases so c increases. For fused quartz, then, as wavelength increases c also In short, red light has a larger critical angle than blue light. If the angle of incidence is midway between the critical angle of red and the critical angle of blue, then the blue component of the light will experience total internal re ection while the red component will pass through as a refracted ray. (c) Choose an angle of incidence between the two critical angles as described in part (a). Using a value of n = 1:46 from Fig. 39-11, = sin1(1=1:46) = 43:2: c

E39-38 (a) There needs to be an opaque spot in the center of each face so that no refracted ray emerges. The radius of the spot will be large enough to cover rays which meet the surface at less than the critical angle. This means tan c = r=d, where d is the distance from the surface to the spot, or 6.3 mm. Since then r = (6:3 mm) tan(41:1) = 5:50 mm: (b) The circles have an area of a = (5:50 mm)2 = 95:0 mm2. Each side has an area of (12:6 mm)2; the fraction covered is then (95:0 mm2)=(12:6 mm)2 = 0:598:

E39-39 For u c the relativistic Doppler shift simpli es to f = f0u=c = u=0;

E39-40 c = f , so 0 = f + f . Then = = f =f . Furthermore, f0 f , from Eq. 39-21, is f0u=c for small enough u. Then f f0 u = = : f0 c E39-41 The Doppler theory for light gives 1 u=c 1 (0:2) f = f0 p = f0 p = 0:82 f0: 1 u2=c2 1 (0:2)2 The frequency is shifted down to about 80%, which means the wavelength is shifted up by an additional 25%. Blue light (480 nm) would appear yellow/orange (585 nm).

E39-42 Use Eq. 39-20: 1 u=c 1 (0:892) f = f0 p = (100 Mhz) p = 23:9 MHz: 1 u2=c2 1 (0:892)2

E39-43 (a) If the wavelength classical Doppler shift (b) For the relativistic shift,

is three times longer then the frequency is one-third, so for the

1 u=c 1 u2=c2 0 = 10u2 18uc + 8c2: p E39-44 (a) f0=f = =0. This shift is small, so we apply the approximation:

0 (462 nm) u = c 1 = (3108m=s) 1 = 1:9107m=s: (434 nm) (b) A red shift corresponds to objects moving away from us.

E39-45 The sun rotates once every 26 days at the equator, while the radius is 7:0108m. The speed of a point on the equator is then 2R 2(7:0108m) v = = = 2:0103 m=s: T (2:2106s)

This corresponds to a velocity parameter of = u=c = (2:0103 m=s)=(3:0108 m=s) = 6:7106:

This is a case of small numbers, so we’ll use the formula that you derived in Exercise 39-40: = = (553 nm)(6:7106) = 3:7103 nm:

E39-46 Use Eq. 39-23 written as 0 which can be rearranged as 2 2 (540 nm)2 (620 nm)2 u=c = 0 = = 0:137: 2 + 2 (540 nm)2 + (620 nm)2 0

so f c c f c+v cv 2v2 c2 v2 2(8:65105m=s)2 (3:00108m=s)2 (8:65105m=s)2 = 1:66105: p (b) f1 = f (c u)=sqrtc2 u2 and f2 = f (c + u)= c2 u2.

so f 2c f c2 u2 2(3:00108m=s) (3:00108m=s)2 (8:65105m=s)2 = 8:3106:

(b) The Doppler e ect at this speed is 1 u=c 1 (0:6) 1 u2=c2 1 (0:6)2 this means the frequency is one half, so the period is doubled to 12 minutes. (c) If C send the signal at the instant the signal from A passes, then the two signals travel together to C, so C would get B’s signals at the same rate that it gets A’s signals: every six minutes.

E39-49 p E39-50 The transverse Doppler e ect is = 0= 1 u2=c2. Then p = (589:00 nm)= 1 (0:122)2 = 593:43 nm: The shift is (593:43 nm) (589:00 nm) = 4:43 nm.

E39-51 The frequency observed by the detector from the rst source is (Eq. 39-31) p f = f1 1 (0:717)2 = 0:697f1: The frequency observed by the detector from the second source is (Eq. 39-30) p 1 (0:717)2 0:697f2 f = f2 = : 1 + (0:717) cos 1 + (0:717) cos We need to equate these and solve for . Then 0:697f2 1 + 0:717 cos = 101:1: Subtract from 180 to nd the angle with the line of sight.

E39-52 P39-1 Consider the triangle in Fig. 39-45. The true position corresponds to the speed of light, the opposite side corresponds to the velocity of earth in the orbit. Then = arctan(29:8103m=s)=(3:00108m=s) = 20:500:

P39-2 The distance to Jupiter from point x is dx = rj re. The distance to Jupiter from point y is q d = r2 + r2: 2ej The di erence in distance is related to the time according to (d2 d1)=t = c;

so p (778109m)2 + (150109m)2 (778109m) + (150109m) = 2:7108m=s: (600 s) P39-3 sin(30)=(4:0 m=s) = sin =(3:0 m=s). Then = 22: Water waves travel more slowly in shallower water, which means they always bend toward the normal as they approach land.

P39-4 (a) If the ray is normal to the water’s surface then it passes into the water unde ected. Once in the water the problem is identical to Sample Problem 39-2. The re ected ray in the water is parallel to the incident ray in the water, so it also strikes the water normal, and is transmitted (b) Assume the ray strikes the water at an angle 1. It then passes into the water at an angle 2, where nw sin 2 = na sin 1: Once the ray is in the water then the problem is identical to Sample Problem 39-2. The re ected ray in the water is parallel to the incident ray in the water, so it also strikes the water at an angle 2. When the ray travels back into the air it travels with an angle 3, where nw sin 2 = na sin 3: Comparing the two equations yields 1 = 3, so the outgoing ray in the air is parallel to the incoming ray.

P39-5 (a) As was done in Ex. 39-25 above we use the small angle approximation of sin tan The incident angle is ; if the light were to go in a straight line we would expect it to strike a distance y1 beneath the normal on the right hand side. The various distances are related to the angle by tan y1=t: The light, however, does not go in a straight line, it is refracted according to (the small angle approximation to) Snell’s law, n11 = n22; which we will simplify further by letting 1 = , n2 = n, and n1 = 1, = n2: The point where the refracted ray does strike is related to the angle by 2 tan 2 = y2=t: Combining the three expressions,

y1 = ny2: The di erence, y1 y2 is the vertical distance between the displaced ray and the original ray as measured on the plate glass. A little algebra yields n1 = t : n The perpendicular distance x is related to this di erence by cos = x=(y1 y2):

In the small angle approximation cos 1 2=2. If is suciently small we can ignore the square (b) Remember to use radians and not degrees whenever the small angle approximation is applied. Then (1:52) 1 x = (1:0 cm)(0:175 rad) = 0:060 cm: (1:52)

P39-6 (a) At the top layer, at the next layer, at the next layer, n3 sin 3 = n2 sin 2:

Combining all three expressions, n3 sin 3 = sin : (b) = arcsin[sin(50)=(1:00029)] = 49:98. Then shift is (50) (49:98) = 0:02. 3

P39-7 The \big idea” of Problem 6 is that when light travels through layers the angle that it makes in any layer depends only on the incident angle, the index of refraction where that incident That means that light which leaves the surface of the runway at 90 to the normal will make an angle n0 sin 90 = n0(1 + ay) sin

at some height y above the runway. It is mildly entertaining to note that the value of n0 is unim- portant, only the value of a! The expression 1 sin = 1 ay 1 + ay can be used to nd the angle made by the curved path against the normal as a function of y. The slope of the curve at any point is given by dy cos = tan(90 ) = cot = : dx sin Now we need to know cos . It is p cos = 1 sin2 p2ay:

Combining p dy 2ay dx 1 ay and now we integrate. We will ignore the ay term in the denominator because it will always be small compared to 1. Then d Z h dy Z 0 0 2ay s r 2h 2(1:7 m) d = = = 1500 m: a (1:5106m1)

P39-8 The energy of a particle is given by E2 = p2c2 + m2c4. This energy is related to the mass p by E = mc2. is related to the speed by = 1= 1 u2=c2. Rearranging, s r u 1 m2c2 c 2 p2 + m2c2 s p2 =: p2 + m2c2

Since n = c=u we can write this as s s m2c2 mc2 2 n= 1+ = 1+ : p2 pc

For the pion, s 2 (135 MeV) n = 1 + = 1:37: (145 MeV) For the muon, s 2 (106 MeV) n = 1 + = 1:24: (145 MeV)

P39-9 (a) Before adding the drop of liquid project the light ray along the angle so that = 0. Increase slowly until total internal re ection occurs at angle 1. Then ng sin 1 = 1 Now put the liquid on the glass and repeat the above process until total internal re ection occurs at angle 2. Then ng sin 2 = nl: (b) This is not terribly practical.

P39-10 Let the internal angle at Q be Q. Then n sin Q = 1, because it is a critical angle. Let the internal angle at P be P . Then P + Q = 90. Combine this with the other formula and q 1 = n sin(90 P ) = n cos Q = n 1 sin2 P : Not only that, but sin 1 = n sin P , or p 1 = n 1 (sin 1)2=n2;

which can be solved for n to yield q n = 1 + sin2 1: p (a) The fraction of light energy which escapes from the water is dependent on the critical angle. Light radiates in all directions from the source, but only that which strikes the surface at an angle less than the critical angle will escape. This critical angle is sin c = 1=n: We want to nd the solid angle of the light which escapes; this is found by integrating

2 c ZZ = sin d d : 00 This is not a hard integral to do. The result is = 2(1 cos c): There are 4 steradians in a spherical surface, so the fraction which escapes is 1 1q f = (1 cos c) = (1 1 sin2 c): 22 The last substitution is easy enough. We never needed to know the depth h. (b) f = 1 (1 p1 (1=(1:3))2) = 0:18: 2 P39-11

P39-12 (a) The beam of light strikes the face of the ber at an angle and is refracted according to n1 sin 1 = sin : The beam then travels inside the ber until it hits the cladding interface; it does so at an angle of 90 1 to the normal. It will be re ected if it exceeds the critical angle of

or if which can be written as cos 1 n2=n1: but if this is the cosine, then we can use sin2 + cos2 = 1 to nd the sine, and q sin 1 n2=n2: 1 21

Combine this with the rst equation and q arcsin n2 n2: 12

(b) = arcsin p(1:58)2 (1:53)2 = 23:2: Consider the two possible extremes: a ray of light can propagate in a straight line directly down the axis of the ber, or it can re ect o of the sides with the minimum possible angle The minimum angle of incidence that will still involve re ection is the critical angle, so n2 sin c = : n1 This light ray has farther to travel than the ray down the ber axis because it is traveling at an angle. The distance traveled by this ray is 0 n1 n2 The time taken for this bouncing ray to travel a length L down the ber is then L0 L0n L n2 011 t= = = : v c c n2 Now for the easier ray. It travels straight down the ber in a time P39-13

L t = n1: c The di erence is L n2 Ln1 t0 t = t = 1 n = (n n ): 1 12 c n2 cn2 (b) For the numbers in Problem 12 we have (350103 m)(1:58) t = ((1:58) (1:53)) = 6:02105s: (3:00108m=s)(1:53)

P39-14 P39-15 We can assume the airplane speed is small compared to the speed of light, and use Eq. 39-21. f = 990 Hz; so jf j = f0u=c = u=0;

hence u = (990=s)(0:12 m) = 119 m=s: The actual answer for the speed of the airplane is half this because there were two Doppler shifts: once when the microwaves struck the plane, and one when the re ected beam was received by the station. Hence, the plane approaches with a speed of 59:4 m=s.

(a) In order to change from the frame of reference of the mirror to your own frame of reference you need to subtract vo from all velocities. Then your velocity is vo v0 = 0, the mirror is moving with velocity 0 vo = vo and your image is moving with velocity vo vo = 2vo.

E40-2 You are 30 cm from the mirror, the image is 10 cm behind the mirror. You need to focus 40 cm away.

If the mirror rotates through an angle then the angle of incidence will increase by an angle , and so will the angle of re ection. But that means that the angle between the incident angle and the re ected angle has increased by twice.

E40-3 E40-4 Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah can see any image which is located between that line and the mirror. By similar triangles, the image of Bernie will be d=2 = (3:0 m)=2 = 1=5 m from the mirror when it becomes visible. Since i = o, Bernie will also be 1:5 m from the mirror.

The apparent depth of the swimming pool is given by the work done for Exercise 39- 25, dapp = d=n The water then \appears” to be only 186 cm=1:33 = 140 cm deep. The apparent distance between the light and the mirror is then 250 cm + 140 cm = 390 cm; consequently the image of the light is 390 cm beneath the surface of the mirror.

E40-7 E40-8 Three. There is a single direct image in each mirror and one more image of an image in one of the mirrors.

We want to know over what surface area of the mirror are rays of light re ected from the object into the eye. By similar triangles the diameter of the pupil and the diameter of the part of the mirror (d) which re ects light into the eye are related by E40-9

d (5:0 mm) (10 cm) (24 cm) + (10 cm) which has solution d = 1:47 mm The area of the circle on the mirror is A = (1:47 mm)2=4 = 1:7 mm2:

E40-11 Seven. Three images are the ones from Exercise 8. But each image has an image in the ceiling mirror. That would make a total of six, except that you also have an image in the ceiling mirror (look up, eh?). So the total is seven!

E40-12 A point focus is not formed. The envelope of rays is called the caustic. You can see a similar e ect when you allow light to re ect o of a spoon onto a table.

The image is magni ed by a factor of 2.7, so the image distance is 2.7 times farther from the mirror than the object. An important question to ask is whether or not the image is real or virtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could see it. If it were a real image it would be in front of the mirror, and the man, who serves as the object and is therefore closer to the mirror than the image, would not be able to see it. So we shall assume that the image is virtual. The image distance is then a negative number. The focal length is half of the radius of curvature, so we want to solve Eq. 40-6, with f = 17:5 cm and i = 2:7o 1 1 1 0:63 (17:5 cm) o 2:7o o which has solution o = 11 cm.

E40-13 E40-14 The image will be located at a point given by 111111 == = : i f o (10 cm) (15 cm) (30 cm) The vertical scale is three times the horizontal scale in the gure below.

E40-15 This problem requires repeated application of 1=f = 1=o + 1=i, r = 2f , m = i=o, or the properties of plane, convex, or concave mirrors. All dimensioned variables below (f; r; i; o) are (a) Concave mirrors have positive focal lengths, so f = +20; r = 2f = +40;

(b) m = +1 for plane mirrors only; r = 1 for at surface; f = 1=2 = 1; i = o = 10; the (c) If f is positive the mirror is concave; r = 2f = +40;

(d) If m is negative then the image is real and inverted; only Concave mirrors produce real images (from real objects); i = mo = (0:5)(60) = 30;

(f) If m is positive the image is virtual and upright; if m is less than one the image is reduced, but only convex mirrors produce reduced virtual images (from real objects); f = 20 for convex mirrors; r = 2f = 40; let i = mo = o=10, then 1=f = 1=o + 1=i = 1=o 10=o = 9=o;

(g) r is negative for convex mirrors, so r = 40; f = r=2 = 20; convex mirrors produce only virtual upright images (from real objects); so i is negative; and 1=o = 1=f 1=i = 1=(20) 1=(4) = 1=(5);

(h) Inverted images are real; only concave mirrors produce real images (from real objects); inverted images have negative m; i = mo = (0:5)(24) = 12;

E40-16 Use the angle de nitions provided by Eq. 40-8. From triangle OaI we have + = 2;

while from triangle IaC we have += : Combining to eliminate we get = 2 :

Substitute Eq. 40-8 and eliminate s, 112 oir or 112 o i r which is the same as Eq. 40-4 if i ! i and r ! r.

(a) Consider the point A. Light from this point travels along the line ABC and will be parallel to the horizontal center line from the center of the cylinder. Since the tangent to a circle de nes the outer limit of the intersection with a line, this line must describe the apparent size. (b) The angle of incidence of ray AB is given by E40-17

sin 1 = r=R: The angle of refraction of ray BC is given by sin 2 = r=R:

Snell’s law, and a little algebra, yields r r RR nr = r:

In the last line we used the fact that n2 = 1, because it is in the air, and n1 = n, the index of refraction of the glass.

E40-18 This problem requires repeated application of (n2 n1)=r = n1=o + n2=i. All dimensioned (a) (1:5) (1:0) (1:0) (30) (10) (b) (1:0) (1:5) (10) (13) (c) (1:5) (1:0) (1:5) (30) (600)

(d) Rearrange the formula to solve for n2, then 1 11 n2 1i = n1 + : r ro Substituting the numbers,

11 11 (20) (20) (20) (20) (e) (1:5) (1:0) (10) (6) (f ) (1:0) (1:5) (1:0) (30) (7:5) (g) (1:0) (1:5) (1:5) (30) (70) (h) Solving Eq. 40-10 for n2 yields 1=o + 1=r 1=r 1=i so 1=(100) + 1=(30) n2 = (1:5) = 1:0 1=(30) 1=(600) and the image is real.

E40-19 (b) If the beam is small we can use Eq. 40-10. Parallel incoming rays correspond to an object at in nity. Solving for n2 yields 1=o + 1=r 1=r 1=i so if o ! 1 and i = 2r, then 1=1 + 1=r n2 = (1:0) = 2:0 1=r 1=2r (c) There is no solution if i = r!

E40-20 The image will be located at a point given by 111111 == = : i f o (10 cm) (6 cm) (15 cm)

E40-21 The image location can be found from Eq. 40-15, 111 1 1 1 i f o (30 cm) (20 cm) 12 cm so the image is located 12 cm from the thin lens, on the same side as the object.

E40-22 For a double convex lens r1 > 0 and r2 < 0 (see Fig. 40-21 and the accompanying text). Then the problem states that r2 = r1=2. The lens maker's equation can be applied to get 1 1 1 3(n 1) f r1 r2 r1 so r1 = 3(n 1)f = 3(1:5 1)(60 mm) = 90 mm, and r2 = 45 mm:

E40-23 The object distance is essentially o = 1, so 1=f = 1=o + 1=i implies f = i, and the image forms at the focal point. In reality, however, the object distance is not in nite, so the magni cation is given by m = i=o f =o, where o is the Earth/Sun distance. The size of the image is then h=hf=o=2(6:96108m)(0:27m)=(1:501011m)=2:5mm: io

The factor of two is because the sun’s radius is given, and we need the diameter!

E40-24 (a) The at side has r2 = 1, so 1=f = (n 1)=r, where r is the curved side. Then (b) 1=i = 1=f 1=o = 1=(0:40 m) 1=(0:40 m) = 0. Then i is 1.

E40-26 (a) 1=f = (n 1)[1=(r) 1=r], so 1=f = 2(1 n)=r. 1=i = 1=f 1=o so if o = r, then 1=i = 2(1 n)=r 1=r = (1 2n)=r;

(b) For n > 0:5 the image is virtual; the magni cation is m = i=o = r=(1 2n)=r = 1=(2n 1): E40-27 According to the de nitions, o = f + x and i = f + x0. Starting with Eq. 40-15, 11 1 oi f i+o 1 oi f 2f + x + x0 1 (f + x)(f + x0) f f 2 = xx0:

E40-28 (a) You can’t determine r1, r2, or n. i is found from 1111 i +10 +20 +20 the image is real and inverted. m = (20)=(20) = 1: (b) You can’t determine r1, r2, or n. The lens is converging since f is positive. i is found from 1111 i +10 +5 10 the image is virtual and upright. m = (10)=(+5) = 2: (c) You can’t determine r1, r2, or n. Since m is positive and greater than one the lens is converging. Then f is positive. i is found from 1111 i +10 +5 10 the image is virtual and upright. m = (10)=(+5) = 2: (d) You can’t determine r1, r2, or n. Since m is positive and less than one the lens is diverging. Then f is negative. i is found from 1111 i 10 +5 3:3 the image is virtual and upright. m = (3:3)=(+5) = 0:66: (e) f is found from 1111 = (1:5 1) = : f +30 30 +30 The lens is converging. i is found from 1111 i +30 +10 15 the image is virtual and upright. m = (15)=(+10) = 1:5: (f ) f is found from 1111 = (1:5 1) = : f 30 +30 30 The lens is diverging. i is found from 1111 i 30 +10 7:5 the image is virtual and upright. m = (7:5)=(+10) = 0:75: (g) f is found from 1111 = (1:5 1) = : f 30 60 120 The lens is diverging. i is found from 1111 i 120 +10 9:2 the image is virtual and upright. m = (9:2)=(+10) = 0:92: (h) You can’t determine r1, r2, or n. Upright images have positive magni cation. i is found from i = (0:5)(10) = 5;

f is found from 1111 f +10 5 10 (h) You can’t determine r1, r2, or n. Real images have negative magni cation. i is found from

f is found from 1111 f +10 5 +3:33 E40-29 o + i = 0:44 m = L, so 11111L f o i o L o o(L o)

which can also be written as o2 oL + f L = 0. This has solution pp L L2 4f L (0:44 m) (0:44 m) 4(0:11 m)(0:44 m) o = = = 0:22 m: 22 There is only one solution to this problem, but sometimes there are two, and other times there are none!

E40-30 (a) Real images (from real objects) are only produced by converging lenses. (b) Since hi = h0=2, then i = o=2. But d = i+o = o+o=2 = 3o=2, so o = 2(0:40 m)=3 = 0:267 m, (c) 1=f = 1=o + 1=i = 1=(0:267 m) + 1=(0:133 m) = 1=(0:0889 m).

Step through the exercise one lens at a time. The object is 40 cm to the left of a converging lens with a focal length of +20 cm. The image from this rst lens will be located by solving E40-31

111111 i f o (20 cm) (40 cm) 40 cm so i = 40 cm. Since i is positive it is a real image, and it is located to the right of the converging The image from the converging lens is located 40 cm – 10 cm from the diverging lens, but it is located on the wrong side: the diverging lens is \in the way” so the rays which would form the image hit the diverging lens before they have a chance to form the image. That means that the real image from the converging lens is a virtual object in the diverging lens, so that the object distance for the The image formed by the diverging lens is located by solving 111 1 1 1 i f o (15 cm) (30 cm) 30 cm

or i = 30 cm. This would mean the image formed by the diverging lens would be a virtual image, The image is virtual, so it is upright. The magni cation from the rst lens is m1 = i=o = (40 cm)=(40 cm)) = 1;

the magni cation from the second lens is E40-32 (a) The parallel rays of light which strike the lens of focal length f will converge on the focal point. This point will act like an object for the second lens. If the second lens is located a distance L from the rst then the object distance for the second lens will be L f. Note that this will be a negative value for L < f, which means the object is virtual. The image will form at a point 1=i = 1=(f ) 1=(L f ) = L=f (f L): Note that i will be positive if L < f , so the rays really do converge on a point. (b) The same equation applies, except switch the sign of f . Then 1=i = 1=(f ) 1=(L f ) = L=f (L f ): This is negative for L < f , so there is no real image, and no converging of the light rays. (c) If L = 0 then i = 1, which means the rays coming from the second lens are parallel.

E40-33 The image from the converging lens is found from 1111 == i1 (0:58 m) (1:12 m) 1:20 m

This real image is 1:97 m 1:20 m = 0:77 m in front of the plane mirror. It acts as an object for the mirror. The mirror produces a virtual image 0:77 m behind the plane mirror. This image is upright relative to the object which formed it, which was inverted relative to the original object. This second image is 1:97 m + 0:77 m = 2:74 m away from the lens. This second image acts as an object for the lens, the image of which is found from 1111 == i3 (0:58 m) (2:74 m) 0:736 m

so i3 = 0:736 m, and the image is real and inverted relative to the object which formed it, which was inverted relative to the original object. So this image is actually upright.

E40-34 (a) The rst lens forms a real image at a location given by 1=i = 1=f 1=o = 1=(0:1 m) 1=(0:2 m) = 1=(0:2 m): The image and object distance are the same, so the image has a magni cation of 1. This image is 0:3 m 0:2 m = 0:1 m from the second lens. The second lens forms an image at a location given by 1=i = 1=f 1=o = 1=(0:125 m) 1=(0:1 m) = 1=(0:5 m): Note that this puts the nal image at the location of the original object! The image is magni ed by (c) The image is virtual, but inverted.

E40-35 If the two lenses \pass” the same amount of light then the solid angle subtended by each lens as seen from the respective focal points must be the same. If we assume the lenses have the same round shape then we can write this as do=f o = de=f e. Then

de f o do f e E40-36 (a) f = (0:25 m)=(200) 1:3 mm: Then 1=f = (n 1)(2=r) can be used to nd r; (b) The diameter would be twice the radius. In e ect, these were tiny glass balls.

E40-37 (a) In Fig. 40-46(a) the image is at the focal point. This means that in Fig. 40-46(b) i = f = 2:5 cm, even though f 0 6= f . Solving, 1111 =+=: f (36 cm) (2:5 cm) 2:34 cm (b) The e ective radii of curvature must have decreased.

(b) i = (25 cm) (7:7 cm) = 17:3 cm. Then 1111 == o (4:2 cm) (17:3 cm) 5:54 cm: (e) M = mm = 10.

Microscope magni cation is given by Eq. 40-33. We need to rst nd the focal length of the objective lens before we can use this formula. We are told in the text, however, that the microscope is constructed so the at the object is placed just beyond the focal point of the objective lens, then f ob 12:0 mm. Similarly, the intermediate image is formed at the focal point of the eyepiece, so f ey 48:0 mm. The magni cation is then E40-39

s(250 mm) (285 mm)(250 mm) m = = = 124: f obf ey (12:0 mm)(48:0 mm)

A more accurate answer can be found by calculating the real focal length of the objective lens, which is 11:4 mm, but since there is a huge uncertainty in the near point of the eye, I see no point in trying to be more accurate than this.

P40-1 The old intensity is Io = P=4d2, where P is the power of the point source. With the mirror in place there is an additional amount of light which needs to travel a total distance of 3d in order to get to the screen, so it contributes an additional P =4(3d)2 to the intensity. The new intensity is then I = P=4d2 + P=4(3d)2 = (10=9)P=4d2 = (10=9)I : no

P40-2 (a) vi = di=dt; but i = f o=(o f ) and f = r=2 so 2 2 d ro r do r vi = = = vo: dt 2o r 2o r dt 2o r (b) Put in the numbers! 2 (15 cm) vi = (5:0 cm/s) = 6:2102 cm/s: 2(75 cm) (15 cm) (c) Put in the numbers! 2 (15 cm) vi = (5:0 cm/s) = 70 m=s 2(7:7 cm) (15 cm) (d) Put in the numbers! 2 (15 cm) vi = (5:0 cm/s) = 5:2 cm/s: 2(0:15 cm) (15 cm)

(b) There are two ends to the object of length L, one of these ends is a distance o1 from the mirror, and the other is a distance o2 from the mirror. The images of the two ends will be Since we are told that the object has a short length L we will assume that a di erential approach to the problem is in order. Then P40-3

Finding the ratio of L0=L is then reduced to L0 i di =: L o do We can take the derivative of Eq. 40-15 with respect to changes in o and i, di do i2 o2 or L0 di i2 L do o2 (a) Since i is given by 1 1 1 of i f o of the fraction i=o can also be written i of f ==: o o(o f ) o f

P40-4 The left surface produces an image which is found from n=i = (n 1)=R 1=o, but since the incoming rays are parallel we take o = 1 and the expression simpli es to i = nR=(n 1). This image is located a distance o = 2R i = (n 2)R=(n 1) from the right surface, and the image produced by this surface can be found from 1=i = (1 n)=(R) n=o = (n 1)=R n(n 1)=(n 2)R = 2(1 n)=(n 2)R: Then i = (n 2)R=2(n 1):

P40-5 The \1″ in Eq. 40-18 is actually nair; the assumption is that the thin lens is in the air. If that isn’t so, then we need to replace \1″ with n0, so Eq. 40-18 becomes n0 n n n0 =: o ji0j r1 A similar correction happens to Eq. 40-21: n n0 n n0 + = : ji0j i r2

Adding these two equations, n0 n0 1 1

+ = (n n0) : o i r1 r2

This yields a focal length given by 0 1 nn 1 1 = : f n r1 r2

P40-6 Start with Eq. 40-4 11 1 oi f jf j jf j jf j oif 11 y y0 where + is when f is positive and is when f is negative.

(a) The image (which will appear on the screen) and object are a distance D = o + i apart. We can use this information to eliminate one variable from Eq. 40-15, P40-7

11 1 oi f 111 o Do f D1 o(D o) f o2 oD + f D = 0: This last expression is a quadratic, and we would expect to get two solutions for o. These solutions will be of the form \something” plus/minus \something else”; the distance between the two locations for o will evidently be twice the \something else”, which is then pp d = o+ o = (D)2 4(f D) = D(D 4f ):

(b) The ratio of the image sizes is m+=m, or i+o=io+. Now it seems we must nd the actual values of o+ and o. From the quadratic in part (a) we have p D D(D 4f) D d 22 so the ratio is o D d =: o+ D + d But i = o+, and vice-versa, so the ratio of the image sizes is this quantity squared.

P40-8 1=i = 1=f 1=o implies i = f o=(o f ). i is only real if o f . The distance between the image and object is of o2 y=i+o= +o= : of of This quantity is a minimum when dy=do = 0, which occurs when o = 2f . Then i = 2f , and y = 4f .

P40-9 (a) The angular size of each lens is the same when viewed from the shared focal point. This means W1=f1 = W2=f2, or W2 = (f2=f1)W1: (b) Pass the light through the diverging lens rst; choose the separation of the lenses so that the focal point of the converging lens is at the same location as the focal point of the diverging lens (c) Since I / 1=A, where A is the area of the beam, we have I / 1=W 2. Consequently,

I2=I1 = (W1=W2)2 = (f1=f2)2 P40-10 The location of the image in the mirror is given by 111 =: i f a+b

The location of the image in the plate is given by i0 = a, which is located at b a relative to the mirror. Equating, 111 ba b+a f 2b 1 b2 a2 f p p = (7:5 cm)2 2(7:5 cm)(28:2 cm) = 21:9 cm: P40-11 We’ll solve the problem by nding out what happens if you put an object in front of the Let the object distance be o1. The rst lens will create an image at i1, where 111 = i1 f1 o1 If the rst image is real (i1 positive) then the image will be on the \wrong” side of the second lens, and as such the real image will act like a virtual object. In short, o2 = i1 will give the correct sign to the object distance when the image from the rst lens acts like an object for the second lens. The image formed by the second lens will then be at 111 i2 f2 o2 11 f2 i2 111 =+: f2 f1 o1 In this case it appears as if the combination 11 + f2 f1 is equivalent to the reciprocal of a focal length. We will go ahead and make this connection, and 1 1 1 f1 + f2 =+= : f f2 f1 f1f2 The rest is straightforward enough.

P40-12 (a) The image formed by the rst lens can be found from 1111 ==: i1 f1 2f1 2f1 This is a distance o2 = 2(f1 + f2) = 2f2 from the mirror. The image formed by the mirror is at an image distance given by 1111 ==: i2 f2 2f2 2f2 Which is at the same point as i1!. This means it will act as an object o3 in the lens, and, reversing the rst step, produce a nal image at O, the location of the original object. There are then three images formed; each is real, same size, and inverted. Three inversions nets an inverted image. The nal image at O is therefore inverted.

P40-13 (a) Place an object at o. The image will be at a point i0 given by 111 i0 f o

or i0 = f o=(o f ): (b) The lens must be shifted a distance i0 i, or fo i0 i = 1: of (c) The range of motion is (0:05 m)(1:2 m) i = 1 = 5:2 cm: (1:2 m) (0:05 m)

(b) Using the results of Problem 40-11, 111 f f2 f1

We want the maximum linear motion of the train to move no more than 0.75 mm on the lm; this means we want to nd the size of an object on the train that will form a 0.75 mm image. The object distance is much larger than the focal length, so the image distance is approximately equal to the focal length. The magni cation is then m = i=o = (3:6 cm)=(44:5 m) = 0:00081. The size of an object on the train that would produce a 0.75 mm image on the lm is then How much time does it take the train to move that far?

P40-15 (0:93 m) t = = 25 ms: (135 km/hr)(1=3600 hr/s)

P40-16 (a) The derivation leading to Eq. 40-34 depends only on the fact that two converging optical devices are used. Replacing the objective lens with an objective mirror doesn’t change (b) The image will be located very close to the focal point, so jmj f =o, and (16:8 m) hi = (1:0 m) = 8:4103m (2000 m) (c) f e = (5 m)=(200) = 0:025 m. Note that we were given the radius of curvature, not the focal length, of the mirror!

E41-1 In this problem we look for the location of the third-order bright fringe, so 1 m 1 (3)(554 109m) = sin = sin = 12:5 = 0:22 rad: d (7:7 106m)

E41-2 d1 sin = gives the rst maximum; d2 sin = 2 puts the second maximum at the location of the rst. Divide the second expression by the rst and d2 = 2d1. This is a 100% increase in d.

E41-3 y = D=d = (512109m)(5:4 m)=(1:2103m) = 2:3103m: E41-4 d = = sin = (592109m)= sin(1:00) = 3:39105m:

Since the angles are very small, we can assume sin for angles measured in radians. If the interference fringes are 0:23 apart, then the angular position of the rst bright fringe is 0:23 away from the central maximum. Eq. 41-1, written with the small angle approximation in mind, is d = for this rst (m = 1) bright fringe. The goal is to nd the wavelength which increases by 10%. To do this we must increase the right hand side of the equation by 10%, which means increasing by 10%. The new wavelength will be 0 = 1:1 = 1:1(589 nm) = 650 nm E41-5

E41-6 Immersing the apparatus in water will shorten the wavelengths to =n. Start with d sin 0 = ; and then nd from d sin = =n. Combining the two expressions, = arcsin[sin =n] = arcsin[sin(0:20)=(1:33)] = 0:15: 0

E41-7 The third-order fringe for a wavelength will be located at y = 3D=d, where y is measured from the central maximum. Then y is y y = 3( )D=d = 3(612109m 480109m)(1:36 m)=(5:22103m) = 1:03104m: 1212

= d sin = (0:120 m) sin[arctan(0:180 m=2:0 m)] = 1:08102m: Then f = v= = (0:25 m=s)=(1:08102m) = 23 Hz: E41-9 A variation of Eq. 41-3 is in order:

1 D ym = m + 2d We are given the distance (on the screen) between the rst minima (m = 0) and the tenth minima (m = 9). Then (50 cm) (0:15 mm) or = 6104 mm = 600 nm:

E41-10 The \maximum” maxima is given by the integer part of m = d sin(90)= = (2:0 m)=(0:50 m) = 4: Since there is no integer part, the \maximum” maxima occurs at 90. These are point sources radiating in both directions, so there are two central maxima, and four maxima each with m = 1, m = 2, and m = 3. But the m = 4 values overlap at 90, so there are only two. The total is 16.

E41-11 E41-13 Consider Fig. 41-5, and solve it exactly for the information given. For the tenth bright fringe r1 = 10 + r2. There are two important triangles: r2 = D2 + (y d=2)2 2

and r2 = D2 + (y + d=2)2 1 Solving to eliminate r2, pp D2 + (y + d=2)2 = D2 + (y d=2)2 + 10:

This has solution r 4D2 + d2 1002 y = 5 : d2 1002 The solution predicted by Eq. 41-1 is 10 p d or r 4D2 y0 = 5 : d2 1002 The fractional error is y0=y 1, or r 4D2 4D2 + d2 1002 or s 4(40 mm)2 1 = 3:1104: 4(40 mm)2 + (2 mm)2 100(589106mm)2

E41-15 Leading by 90 is the same as leading by a quarter wavelength, since there are 360 in a circle. The distance from A to the detector is 100 m longer than the distance from B to the detector. Since the wavelength is 400 m, 100 m corresponds to a quarter wavelength. So a wave peak starts out from source A and travels to the detector. When it has traveled a quarter wavelength a wave peak leaves source B. But when the wave peak from A has traveled a quarter wavelength it is now located at the same distance from the detector as source B, which They are in phase.

E41-16 The rst dark fringe involves waves radians out of phase. Each dark fringe after that involves an additional 2 radians of phase di erence. So the mth dark fringe has a phase di erence of (2m + 1) radians.

2 2d E41-17 I = 4I0 cos sin , so for this problem we want to plot

2(0:60 mm) I =I0 = cos2 sin = cos2 (6280 sin ) : (600109m)

E41-18 The resultant quantity will be of the form A sin(!t + ). Solve the problem by looking at t = 0; then y1 = 0, but x1 = 10, and y2 = 8 sin 30 = 4 and x2 = 8 cos 30 = 6:93. Then the resultant is of length p A = (4)2 + (10 + 6:93)2 = 17:4;

and has an angle given by = arctan(4=16:93) = 13:3: (a) We want to know the path length di erence of the two sources to the detector. Assume the detector is at x and the second source is at y = d. The distance S D is x; the pp1 distance S2D is x2 + d . The di erence is x2 + d2 x. If this di erence is an integral number 2

of wavelengths then we have a maximum; if instead it is a half integral number of wavelengths we have a minimum. For part (a) we are looking for the maxima, so we set the path length di erence equal to m and solve for xm.

E41-19 p m mm m mm d2 m22 xm = 2m The rst question we need to ask is what happens when m = 0. The right hand side becomes indeterminate, so we need to go back to the rst line in the above derivation. If m = 0 then d2 = 0; In fact, we may have even more troubles. xm needs to be a positive value, so the maximum allowed value for m will be given by m < d= = (4:17 m)=(1:06 m) = 3:93;

The rst three maxima occur at m = 3, m = 2, and m = 1. These maxima are located at (4:17 m)2 (3)2(1:06 m)2 2(3)(1:06 m) (4:17 m)2 (2)2(1:06 m)2 2(2)(1:06 m) (4:17 m)2 (1)2(1:06 m)2 x1 = = 7:67 m: 2(1)(1:06 m) Interestingly enough, as m decreases the maxima get farther away! (b) The closest maxima to the origin occurs at x = 6:94 cm. What then is x = 0? It is a local minimum, but the intensity isn’t zero. It corresponds to a point where the path length di erence is 3.93 wavelengths. It should be half an integer to be a complete minimum.

E41-20 The resultant can be written in the form A sin(!t + ). Consider t = 0. The three components can be written as y = 0 + 6:14 3:08 = 3:06: and x = 10 + 12:6 + 3:55 = 26:2: Then A = p(3:06)2 + (26:2)2 = 26:4 and = arctan(3:06=26:2) = 6:66:

E41-21 The order of the indices of refraction is the same as in Sample Problem 41-4, so d = =4n = (620 nm)=4(1:25) = 124 nm:

2dn 2(410 nm)(1:50) 1230 nm = = = : m 1=2 m 1=2 m 1=2 The result is only in the visible range when m = 3, so = 492 nm.

(a) Light from above the oil slick can be re ected back up from the top of the oil layer or from the bottom of the oil layer. For both re ections the light is re ecting o a substance with a higher index of refraction so both re ected rays pick up a phase change of . Since both waves have this phase the equation for a maxima is 11 2d + n + n = mn: 22 Remember that n = =n, where n is the index of refraction of the thin lm. Then 2nd = (m 1) is the condition for a maxima. We know n = 1:20 and d = 460 nm. We don’t know m or . It might E41-23

seem as if there isn’t enough information to solve the problem, but we can. We need to nd the wavelength in the visible range (400 nm to 700 nm) which has an integer m. Trial and error might work. If = 700 nm, then m is 2nd 2(1:20)(460 nm) m = + 1 = + 1 = 2:58 (700 nm) But m needs to be an integer. If we increase m to 3, then 2(1:20)(460 nm) = = 552 nm (3 1) (b) One of the most profound aspects of thin lm interference is that wavelengths which are maximally re ected are minimally transmitted, and vice versa. Finding the maximally transmitted wavelengths is the same as nding the minimally re ected wavelengths, or looking for values of m The most obvious choice is m = 3:5, and then 2(1:20)(460 nm) = = 442 nm: (3:5 1) E41-24 The condition for constructive interference is 2nd = (m 1=2). Assuming a minimum value of m = 1 one nds d = =4n = (560 nm)=4(2:0) = 70 nm:

E41-25 The top surface contributes a phase di erence of , so the phase di erence because of the thickness is 2, or one complete wavelength. Then 2d = =n, or d = (572 nm)=2(1:33) = 215 nm:

E41-26 The wave re ected from the rst surface picks up a phase shift of . The wave which is re ected o of the second surface travels an additional path di erence of 2d. The interference will be bright if 2d + n=2 = mn results in m being an integer.

As with the oil on the water in Ex. 41-23, both the light which re ects o of the acetone and the light which re ects o of the glass undergoes a phase shift of . Then the maxima for re ection are given by 2nd = (m 1). We don’t know m, but at some integer value of m we have = 700 nm. If m is increased by exactly 1 then we are at a minimum of = 600 nm. Consequently, 2

E41-27 we can set these two expressions equal to each other to nd m, (m 1)(700 nm) = (m 1=2)(600 nm);

so m = 4. Then we can nd the thickness, d = (4 1)(700 nm)=2(1:25) = 840 nm:

E41-28 The wave re ected from the rst surface picks up a phase shift of . The wave which is re ected o of the second surface travels an additional path di erence of 2d. The interference will be bright if 2d + n=2 = mn results in m being an integer. Then 2nd = (m 1=2)1 is bright, and 2nd = m2 is dark. Divide one by the other and (m 1=2)1 = m2, so

then d = m2=2n = (2)(450 nm)=2(1:33) = 338 nm: E41-29 Constructive interference happens when 2d = (m 1=2). The minimum value for m is m = 1; the maximum value is the integer portion of 2d=+1=2 = 2(4:8105m)=(680109m)+1=2 = 141:67, so mmax = 141. There are then 141 bright bands.

E41-30 (a) A half wavelength phase shift occurs for both the air/water interface and the water/oil interface, so if d = 0 the two re ected waves are in phase. It will be bright! (b) 2nd = 3, or d = 3(475 nm)=2(1:20) = 594 nm:

E41-31 There is a phase shift on one surface only, so the bright bands are given by 2nd = (m 1=2). Let the rst band be given by 2nd1 = (m1 1=2). The last bright band is then given by 2nd2 = (m1 + 9 1=2). Subtract the two equations to get the change in thickness:

d = 9=2n = 9(630 nm)=2(1:50) = 1:89 m: E41-32 Apply Eq. 41-21: 2nd = m. In one case we have 2nair = (4001);

in the other, 2nvac = (4000): E41-33 (a) We can start with the last equation from Sample Problem 41-5, r 1 2 and solve for m, r2 1 m= + R 2 In this exercise R = 5:0 m, r = 0:01 m, and = 589 nm. Then (0:01 m)2 m = = 34 (589 nm)(5:0 m) (b) Putting the apparatus in water e ectively changes the wavelength to (589 nm)=(1:33) = 443 nm;

so the number of rings will now be (0:01 m)2 m = = 45: (443 nm)(5:0 m)

qq E41-34 (1:42 cm) = (10 1 )R, while (1:27 cm) = (10 1 )R=n. Divide one expression by 2p 2 the other, and (1:42 cm)=(1:27 cm) = n, or n = 1:25.

qq E41-35 (0:162 cm) = (n 1 )R, while (0:368 cm) = (n + 20 1 )R. Square both expres- 22 sions, the divide one by the other, and nd (n + 19:5)=(n 0:5) = (0:368 cm=0:162 cm)2 = 5:16 which can be rearranged to yield 19:5 + 5:16 0:5 n = = 5:308: 5:16 1 Oops! That should be an integer, shouldn’t it? The above work is correct, which means that there really aren’t bright bands at the speci ed locations. I’m just going to gloss over that fact and solve for R using the value of m = 5:308. Then R = r2=(m 1=2) = (0:162 cm)2=(5:308 0:5)(546 nm) = 1:00 m: Well, at least we got the answer which is in the back of the book…

E41-36 Pretend the ship is a two point source emitter, one h above the water, and one h below the water. The one below the water is out of phase by half a wavelength. Then d sin = , where d = 2h, gives the angle for theta for the rst minimum.

so D = (160 m)=(7:46102) = 2:14 km: E41-37 The phase di erence is 2=n times the path di erence which is 2d, so = 4d=n = 4nd=: (a) = 4(1:38)(100109m)=(450109m) = 3:85. Then I (3:85) = cos2 = 0:12: I0 2 (b) = 4(1:38)(100109m)=(650109m) = 2:67. Then I (2:67) = cos2 = 0:055: I0 2 The re ected ray is diminished by 1 0:055 = 95%.

E41-38 The change in the optical path length is 2(d d=n), so 7=n = 2d(1 1=n), or 7(589109m) d = = 4:9106m: 2(1:42) 2

When M2 moves through a distance of =2 a fringe has will be produced, destroyed, and then produced again. This is because the light travels twice through any change in distance. The wavelength of light is then E41-39

2(0:233 mm) = = 588 nm: 792 E41-40 The change in the optical path length is 2(d d=n), so 60 = 2d(1 1=n), or 11 n = = = 1:00030: 1 60=2d 1 60(500109m)=2(5102m) P41-1 (a) This is a small angle problem, so we use Eq. 41-4. The distance to the screen is 2 20 m, because the light travels to the mirror and back again. Then D (632:8 nm)(40:0 m) d = = = 0:253 mm: y (0:1 m) (b) Placing the cellophane over one slit will cause the interference pattern to shift to the left or right, but not disappear or change size. How does it shift? Since we are picking up 2.5 waves then we are, in e ect, swapping bright fringes for dark fringes.

P41-2 The change in the optical path length is d d=n, so 7=n = d(1 1=n), or 7(550109m) d = = 6:64106m: (1:58) 1 p P41-3 The distance from S1 to P is r1 = (x + d=2)2 + y2. The distance from S2 to P is p r2 = (x d=2)2 + y2. The di erence in distances is xed at some value, say c, so that 122 1 2 12 12 12 4(d2 c2)x2 4c2y2 = c2(d2 c2):

P41-4 The change in the optical path length for each slit is nt t, where n is the corresponding index of refraction. The net change in the path di erence is then n2t n1t. Consequently, m = t(n2 n1), so (5)(480109m) t = = 8:0106m: (1:7) (1:4) P41-5 The intensity is given by Eq. 41-17, which, in the small angle approximation, can be written as d I = 4I0 cos2 :

The intensity will be half of the maximum when 1 d=2 = cos2 2 or d 4 2 which will happen if = =2d:

P41-6 Follow the construction in Fig. 41-10, except that one of the electric eld amplitudes is twice the other. The resultant eld will have a length given by p = E0 5 + 4 cos ;

so squaring this yields 2d sin d sin Im d sin = 1 + 8 cos2 : 9 P41-7 We actually did this problem in Exercise 41-27, although slightly di erently. One maxi- mum is 2(1:32)d = (m 1=2)(679 nm);

the other is 2(1:32)d = (m + 1=2)(485 nm): Set these equations equal to each other, (m 1=2)(679 nm) = (m + 1=2)(485 nm);

and nd m = 3. Then the thickness is d = (3 1=2)(679 nm)=2(1:32) = 643 nm:

P41-8 (a) Since we are concerned with transmission there is a phase shift for two rays, so 2d = mn

The minimum thickness occurs when m = 1; solving for d yields (525109m) d = = = 169109m: 2n 2(1:55) (b) The wavelengths are di erent, so the other parts have di ering phase di erences. (c) The nearest destructive interference wavelength occurs when m = 1:5, or = 2nd = 2(1:55)1:5(169109m) = 393109m: This is blue-violet.

P41-9 It doesn’t matter if we are looking at bright are dark bands. It doesn’t even matter if we concern ourselves with phase shifts. All that cancels out. Consider 2d = m; then d = (10)(480 nm)=2 = 2:4 m:

P41-10 (a) Apply 2d = m. Then d = (7)(600109m)=2 = 2100109m: (b) When water seeps in it introduces an extra phase shift. Point A becomes then a bright fringe, and the equation for the number of bright fringes is 2nd = m. Solving for m, m = 2(1:33)(2100109m)=(600109m) = 9:3;

P41-11 (a) Look back at the work for Sample Problem 41-5 where it was found r 1 2 We can write this as s 1 rm = 1 mR 2m and expand the part in parentheses in a binomial expansion, 11 p

rm 1 mR: 2 2m We will do the same with r 1 2 expanding s 1 rm+1 = 1 + mR 2m to get p 11 rm+1 1 + mR: 2 2m Then 1p 2m or 1p r R=m: 2 (b) The area between adjacent rings is found from the di erence, A = r2 r2 ;

m+1 m and into this expression we will substitute the exact values for rm and rm+1, 11 22 = R: Unlike part (a), we did not need to assume m 1 in order to arrive at this expression; it is exact for all m.

P41-12 The path length shift that occurs when moving the mirror as distance x is 2x. This means = 22x= = 4x=. The intensity is then 2x I = 4I0 cos2

E42-1 = a sin = (0:022 mm) sin(1:8) = 6:91107m: E42-2 a = = sin = (0:10109m)= sin(0:12103 rad=2) = 1:7 m:

(a) This is a valid small angle approximation problem: the distance between the points on the screen is much less than the distance to the screen. Then (0:0162 m) = 7:5 103 rad: (2:16 m) (b) The di raction minima are described by Eq. 42-3, a = 1:18 104 m: E42-3

E42-4 a = = sin = (633109m)= sin(1:97=2) = 36:8 m: (a) We again use Eq. 42-3, but we will need to throw in a few extra subscripts to distinguish between which wavelength we are dealing with. If the angles match, then so will the sine of the angles. We then have sin a;1 = sin b;2 or, using Eq. 42-3, (1)a (2)b aa (b) Will any other minima coincide? We want to solve for the values of ma and mb that will be integers and have the same angle. Using Eq. 42-3 one more time, maa mbb aa and then substituting into this the relationship between the wavelengths, ma = mb=2: whenever mb is an even integer ma is an integer. Then all of the di raction minima from a are overlapped by a minima from b.

E42-5 E42-6 The angle is given by sin = 2=a. This is a small angle, so we can use the small angle approximation of sin = y=D. Then y = 2D=a = 2(0:714 m)(593109m)=(420106m) = 2:02 mm:

E42-7 Small angles, so y=D = sin = =a. Then a = D=y = (0:823 m)(546109m)=(5:20103m=2) = 1:73104m:

E42-8 (b) Small angles, so y=D = m=a. Then a = mD=y = (5 1)(0:413 m)(546109m)=(0:350103m) = 2:58 mm: (a) = arcsin(=a) = arcsin[(546109m)=(2:58 mm)] = 1:21102:

E42-9 Small angles, so y=D = m=a. Then y = mD=a = (2 1)(2:94 m)(589109m)=(1:16103m) = 1:49103m:

E42-10 Doubling the width of the slit results in a narrowing of the di raction pattern. Since the width of the central maximum is e ectively cut in half, then there is twice the energy in half the space, producing four times the intensity.

E42-11 (a) This is a small angle approximation problem, so = (1:13 cm)=(3:48 m) = 3:25 103 rad: (b) A convenient measure of the phase di erence, is related to through Eq. 42-7, a (25:2 106m) = sin = sin(3:25 103 rad) = 0:478 rad (538 109m) (c) The intensity at a point is related to the intensity at the central maximum by Eq. 42-8, 2 2 I sin sin(0:478 rad) = = = 0:926 Im (0:478 rad)

E42-12 Consider Fig. 42-11; the angle with the vertical is given by ( )=2. For Fig. 42-10(d) the circle has wrapped once around onto itself so the angle with the vertical is (3 )=2. Substitute Use the result from Problem 42-3 that tan = for the maxima. The lowest non-zero solution is = 4:49341 rad. The angle against the vertical is then 0:21898 rad, or 12:5.

E42-13 Drawing heavily from Sample Problem 42-4, x 1:39 x = arcsin = arcsin = 2:54: a 10 Finally, = 2x = 5:1.

E42-14 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects have an angular separation of at least 9 1 1:22 1 1:22(540 10 ) R = sin = sin = 1:34 104 rad d (4:90 103m) (b) The linear separation is y = D = (1:34 104 rad)(163103m) = 21:9 m: E42-15 (a) Rayleigh’s criterion for resolving images (Eq. 42-11) requires that two objects have an angular separation of at least 9 1:22 1 1:22(562 10 ) R = sin1 = sin = 1:37 104 rad: d (5:00 103m) (b) Once again, this is a small angle, so we can use the small angle approximation to nd the distance to the car. In that case R = y=D; where y is the headlight separation and D the distance to the car. Solving, R

E42-16 y=D = 1:22=a; or D = (5:20103m)(4:60103=m)=1:22(542109m) = 36:2 m: E42-17 The smallest resolvable angular separation will be given by Eq. 42-11,

1 1:22 1 1:22(565 109m) R = sin = sin = 1:36 107 rad; d (5:08 m) The smallest objects resolvable on the Moon’s surface by this telescope have a size y where y = D = (3:84 108 m)(1:36 107 rad) = 52:2 m R

E42-18 y=D = 1:22=a; or y = 1:22(1:57102m)(6:25103m)=(2:33 m) = 51:4 m

E42-19 y=D = 1:22=a; or D = (4:8102m)(4:3103=m)=1:22(0:12109m) = 1:4106 m:

E42-20 y=D = 1:22=a; or d = 1:22(550109m)(160103m)=(0:30 m) = 0:36 m:

Using Eq. 42-11, we nd the minimum resolvable angular separation is given by

1 1:22 1 1:22(475 109m) R = sin = sin = 1:32 104 rad d (4:4 103m) The dots are 2 mm apart, so we want to stand a distance D away such that D > y= = (2 103m)=(1:32 104 rad) = 15 m: R E42-21

E42-22 y=D = 1:22=a; or y = 1:22(500109m)(354103m)=(9:14 m=2) = 4:73102 m:

E42-23 (a) = v=f . Now use Eq. 42-11: (1450 m=s) = arcsin 1:22 = 6:77: (25103Hz)(0:60 m) (b) Following the same approach,

E42-24 (a) = v=f . Now use Eq. 42-11: (3108 m=s) = arcsin 1:22 = 0:173: (220109Hz)(0:55 m) This is the angle from the central maximum; the angular width is twice this, or 0:35. (b) use Eq. 42-11: (0:0157 m) = arcsin 1:22 = 0:471: (2:33 m) This is the angle from the central maximum; the angular width is twice this, or 0:94.

E42-25 The linear separation of the fringes is given by y D = = or y = Ddd for suciently small d compared to .

E42-26 (a) d sin = 4 gives the location of the fourth interference maximum, while a sin = gives the location of the rst di raction minimum. Hence, if d = 4a there will be no fourth interference maximum! (b) Since d sin m = mi gives the interference maxima and a sin m = md gives the di raction id minima, and d = 4a, then whenever mi = 4md there will be a missing maximum.

E42-27 (a) The central di raction envelope is contained in the range

= arcsin a This angle corresponds to the mth maxima of the interference pattern, where sin = m=d = m=2a: Equating, m = 2, so there are three interference bands, since the m = 2 band is \washed out” by (b) If d = a then = and the expression reduces to sin2 2

sin2(2 ) 22 0 2 sin 0 E42-28 Remember that the central peak has an envelope width twice that of any other peak. Ignoring the central maximum there are (11 1)=2 = 5 fringes in any other peak envelope.

E42-29 (a) The rst di raction minimum is given at an angle such that a sin = ; the order of the interference maximum at that point is given by d sin = m: Dividing one expression by the other we get d=a = m; with solution m = (0:150)=(0:030) = 5. The fact that the answer is exactly 5 implies that the fth interference maximum is squelched by the di raction minimum. Then there are only four complete fringes on either side of the central maximum. Add this to the central maximum (b) For the third fringe m = 3, so d sin = 3. Then is Eq. 42-14 is 3, while in Eq. 42-16 is a 3 a d d so the relative intensity of the third fringe is, from Eq. 42-17, 2 sin(3a=d) (cos 3)2 = 0:255: (3a=d)

P42-1 y = mD=a. Then y = (10)(632:8109m)(2:65 m)=(1:37103m) = 1:224102m: The separation is twice this, or 2.45 cm.

P42-2 If a then the di raction pattern is extremely tight, and there is e ectively no light at P . In the event that either shape produces an interference pattern at P then the other shape must produce an equal but opposite electric eld vector at that point so that when both patterns from But the intensity is the eld vector squared; hence the two patterns look identical.

P42-3 (a) We want to take the derivative of Eq. 42-8 with respect to , so

2 dI d sin dd sin cos sin 2

sin = Im2 ( cos sin ) : 3 This equals zero whenever sin = 0 or cos = sin ; the former is the case for a minima while the latter is the case for the maxima. The maxima case can also be written as tan = : (b) Note that as the order of the maxima increases the solutions get closer and closer to odd integers times =2. The solutions are = 0; 1:43; 2:46; etc.

P42-4 The outgoing beam strikes the moon with a circular spot of radius r = 1:22D=a = 1:22(0:69106m)(3:82108m)=(2 1:3 m) = 123 m: If P0 is the power in the light, then

Z ZR 0 where R is the radius of the central peak and I is the angular intensity. For a we can write ar=D, then

P0 = 2Im d 2Im a 0 a Then the intensity at the center falls o with distance D as 2 =2 sin2 2 Z D D (0:82):

2 Im = 1:9 (a=D) P0 The fraction of light collected by the mirror on the moon is then 2 (2 1:3 m) P1=P0 = 1:9 (0:10 m)2 = 5:6106: (0:69106m)(3:82108m) The fraction of light collected by the mirror on the Earth is then 2 (2 0:10 m) P2=P1 = 1:9 (1:3 m)2 = 5:6106: (0:69106m)(3:82108m) Finally, P2=P0 = 31011.

P42-5 (a) The ring is reddish because it occurs at the blue minimum. (b) Apply Eq. 42-11 for blue light: d = 1:22= sin = 1:22(400 nm)= sin(0:375) = 70 m: (c) Apply Eq. 42-11 for red light: = arcsin (1:22(700 nm)=(70 m)) 0:7;

P42-6 The di raction pattern is a property of the speaker, not the interference between the speak- ers. The di raction pattern should be una ected by the phase shift. The interference pattern, however, should shift up or down as the phase of the second speaker is varied.

(a) The missing fringe at = 5 is a good hint as to what is going on. There should be some sort of interference fringe, unless the di raction pattern has a minimum at that point. This would be the rst minimum, so a sin(5) = (440 109m) would be a good measure of the width of each slit. Then a = 5:05 106m.

(b) If the di raction pattern envelope were not present we could expect that the fourth interfer- ence maxima beyond the central maximum would occur at this point, and then d sin(5) = 4(440 109m) yielding d = 2:02 105m: (c) Apply Eq. 42-17, where = m and a a m a = sin = = m = m=4: d d Then for m = 1 we have 2 sin(=4) (=4) while for m = 2 we have 2 sin(2=4) I2 = (7) = 2:8: (2=4) These are in good agreement with the gure.

(b) There are a number of angles allowed: = arcsin[(5)(589109m)=(3:50106m)] = 57:3:

E43-2 The distance between adjacent rulings is (2)(612109m) d = = 2:235106m: sin(33:2) The number of lines is then N = D=d = (2:86102m)=(2:235106m) = 12; 800:

We want to nd a relationship between the angle and the order number which is linear. We’ll plot the data in this representation, and then use a least squares t to nd the wavelength. The data to be plotted is E43-3

m sin 1 17.6 0.302 2 37.3 0.606 3 65.2 0.908 m sin -1 -17.6 -0.302 -2 -37.1 -0.603 -3 -65.0 -0.906 On my calculator I get the best straight line t as 0:302m + 8:33 104 = sin m;

which means that = (0:302)(1:73 m) = 522 nm: E43-4 Although an approach like the solution to Exercise 3 should be used, we’ll assume that each measurement is perfect and error free. Then randomly choosing the third maximum, d sin (5040109m) sin(20:33) = = = 586109m: m (3) E43-5 (a) The principle maxima occur at points given by Eq. 43-1,

sin m = m : d The di erence of the sine of the angle between any two adjacent orders is

sin m+1 sin m = (m + 1) m = : ddd Using the information provided we can nd d from (600 109) d = = = 6 m: sin m+1 sin m (0:30) (0:20)

It doesn’t take much imagination to recognize that the second and third order maxima were given. (b) If the fourth order maxima is missing it must be because the di raction pattern envelope has a minimum at that point. Any fourth order maxima should have occurred at sin 4 = 0:4. If it is a di raction minima then a sin m = m where sin m = 0:4 We can solve this expression and nd (600 109m) a = m = m = m1:5 m: sin m (0:4)

E43-6 (a) Find the maximum integer value of m = d= = (930 nm)=(615 nm) = 1:5, hence (b) The rst order maximum occurs at = arcsin(615 nm)=(930 nm) = 41:4: The width of the maximum is (615 nm) (1120)(930 nm) cos(41:4)

E43-7 The fth order maxima will be visible if d= 5; this means d (1103m) = = 635109m: 5 (315 rulings)(5)

E43-8 (a) The maximum could be the rst, and then d sin (1103m) sin(28) = = = 2367109m: m (200)(1) That’s not visible. The rst visible wavelength is at m = 4, then d sin (1103m) sin(28) = = = 589109m: m (200)(4) The next is at m = 5, then d sin (1103m) sin(28) = = = 469109m: m (200)(5) (b) Yellow-orange and blue.

E43-9 A grating with 400 rulings/mm has a slit separation of

1 d = = 2:5 103 mm: 400 mm1 To nd the number of orders of the entire visible spectrum that will be present we need only consider the wavelength which will be on the outside of the maxima. That will be the longer wavelengths, so we only need to look at the 700 nm behavior. Using Eq. 43-1, d sin = m;

and using the maximum angle 90, we nd d (2:5 106m) (700 109m) so there can be at most three orders of the entire spectrum.

E43-10 In this case d = 2a. Since interference maxima are given by sin = m=d while di raction minima are given at sin = m0=a = 2m0=d then di raction minima overlap with interference maxima whenever m = 2m0. Consequently, all even m are at di raction minima and therefore E43-11 If the second-order spectra overlaps the third-order, it is because the 700 nm second-order Start with the wavelengths multiplied by the appropriate order parameter, then divide both side by d, and nally apply Eq. 43-1.

2(700 nm) 3(400 nm) dd E43-12 Fig. 32-2 shows the path length di erence for the right hand side of the grating as d sin . If the beam strikes the grating at ang angle then there will be an additional path length di erence of d sin on the right hand side of the gure. The di raction pattern then has two contributions to the path length di erence, these add to give d(sin + sin psi) = m:

E43-13 E43-14 Let d sin i = i and 1 + 20 = 2. Then sin 2 = sin 1 cos(20) + cos 1 sin(20):

Rearranging, q sin 2 = sin 1 cos(20) + 1 sin2 1 sin(20): Substituting the equations together yields a rather nasty expression, 2 1 p = cos(20) + 1 ( =d)2 sin(20): 1 dd

Rearranging, 2 2 2 2 (2 1 cos(20 )) = d 1 sin (20 ): Use 1 = 430 nm and 2 = 680 nm, then solve for d to nd d = 914 nm. This corresponds to 1090 rulings/mm.

E43-15 The shortest wavelength passes through at an angle of 1 = arctan(50 mm)=(300 mm) = 9:46: This corresponds to a wavelength of (1103m) sin(9:46) 1 = = 470109m: (350) The longest wavelength passes through at an angle of 2 = arctan(60 mm)=(300 mm) = 11:3: This corresponds to a wavelength of (1103m) sin(11:3) 2 = = 560109m: (350)

E43-16 (a) = =R = =N m; so = (481 nm)=(620 rulings/mm)(5:05 mm)(3) = 0:0512 nm: (b) mm is the largest integer smaller than d=, or mm 1=(481109m)(620 rulings/mm) = 3:35;

E43-17 The required resolving power of the grating is given by Eq. 43-10 (589:0 nm) R = = = 982: (589:6 nm) (589:0 nm) Using Eq. 43-11 we can nd the number of grating lines required. We are looking at the second- order maxima, so R (1000) N = = = 500: m (2)

E43-18 (a) N = R=m = =m, so (415:5 nm) N = = 23100: (2)(415:496 nm 415:487 nm) (b) d = w=N , where w is the width of the grating. Then m (23100)(2)(415:5109m) = arcsin = arcsin = 27:6: d (4:15102m)

E43-19 N = R=m = =m, so (656:3 nm) N = = 3650 (1)(0:180 nm) E43-20 Start with Eq. 43-9: m d sin = tan D= = = : d cos d cos

(a) We nd the ruling spacing by Eq. 43-1, m (3)(589 nm) d = = = 9:98 m: sin m sin(10:2) (b) The resolving power of the grating needs to be at least R = 1000 for the third-order line; see the work for Ex. 43-17 above. The number of lines required is given by Eq. 43-11, R (1000) m (3) so the width of the grating (or at least the part that is being used) is 333(9:98 m) = 3:3 mm.

E43-21 E43-22 (a) Condition (1) is satis ed if d 2(600 nm)= sin(30) = 2400 nm: (b) To remove the third order requires d = 3a, or a = 800 nm.

E43-23 (a) The angles of the rst three orders are (1)(589109m)(40000) (76103m) (2)(589109m)(40000) (76103m) (3)(589109m)(40000) 3 = arcsin = 68:4: (76103m) The dispersion for each order is (1)(40000) 360 D1 = = 3:2102=nm; (76106nm) cos(18:1) 2 (2)(40000) 360 D2 = = 7:7102=nm; (76106nm) cos(38:3) 2 (3)(40000) 360 D3 = = 2:5101=nm: (76106nm) cos(68:4) 2

E43-24 d = m=2 sin , so E43-25 (2)(0:122 nm) d = = 0:259 nm: 2 sin(28:1)

Bragg re ection is given by Eq. 43-12 where the angles are measured not against the normal, but against the plane. The value of d depends on the family of planes under consideration, but it is at never larger than a0, the unit cell dimension. We are looking for the smallest angle; this will correspond to the largest d and the smallest m. That means m = 1 and d = 0:313 nm. Then the minimum angle is 1 (1)(29:3 1012 m) = sin = 2:68: 2(0:313 109 m)

E43-26 2d= = sin 1 and 2d=2 = sin 2. Then = arcsin[2 sin(3:40)] = 6:81: 2

E43-27 We apply Eq. 43-12 to each of the peaks and nd the product m = 2d sin : The four values are 26 pm, 39 pm, 52 pm, and 78 pm. The last two values are twice the rst two, so the wavelengths are 26 pm and 39 pm.

E43-28 (a) 2d sin = m, so (3)(96:7 pm) d = = 171 pm: 2 sin(58:0) (b) = 2(171 pm) sin(23:2)=(1) = 135 pm:

E43-29 The angle against the face of the crystal is 90 51:3 = 38:7. The wavelength is = 2(39:8 pm) sin(38:7)=(1) = 49:8 pm:

E43-30 If > 2d then =2d > 1. But =2d = sin =m: E43-32 A wavelength will be di racted if m = 2d sin . The possible solutions are 4 = 2(275 pm) sin(47:8)=(4) = 102 pm:

E43-33 m (1)(0:261 109 m) d = = = 1:45 1010 m: 2 sin 2 sin(63:8)

d is the spacing between the planes in Fig. 43-28; it correspond to half of the diagonal distance between two cell centers. Then 00

or p p a0 = 2d = 2(1:45 1010 m) = 0:205 nm:

E43-34 Di raction occurs when 2d sin = m. The angles in this case are then given by (0:125109m) sin = m = (0:248)m: 2(0:252109m) There are four solutions to this equation. They are 14:4, 29:7, 48:1, and 82:7. They involve rotating the crystal from the original orientation (90 42:4 = 47:6) by amounts 47:6 82:7 = 35:1:

Since the slits are so narrow we only need to consider interference e ects, not di raction e ects. There are three waves which contribute at any point. The phase angle between adjacent waves is P43-1

= 2d sin =: We can add the electric eld vectors as was done in the previous chapters, or we can do it in a di erent order as is shown in the gure below.

Then the vectors sum to E(1 + 2 cos ): We need to square this quantity, and then normalize it so that the central maximum is the maximum. Then (1 + 4 cos + 4 cos2 ) I = Im : 9

P43-2 (a) Solve for I = Im=2, this occurs when 3 2 or = 0:976 rad. The corresponding angle x is

(0:976) x = = : 2d 2d 6:44d But = 2x, so

: 3:2d (b) For the two slit pattern the half width was found to be = =2d: The half width in the three slit case is smaller.

(a) and (b) A plot of the intensity quickly reveals that there is an alternation of large maximum, then a smaller maximum, etc. The large maxima are at = 2n, the smaller maxima (c) The intensity at these secondary maxima is then P43-3

(1 + 4 cos + 4 cos2 ) Im I = Im = : 99 Note that the minima are not located half-way between the maxima!

P43-4 Covering up the middle slit will result in a two slit apparatus with a slit separation of 2d. The half width, as found in Problem 41-5, is then = =2(2d); = =4d;

P43-5 (a) If N is large we can treat the phasors as summing to form a exible \line” of length NE. We then assume (incorrectly) that the secondary maxima occur when the loop wraps around on itself as shown in the gures below. Note that the resultant phasor always points straight up. This isn’t right, but it is close to reality.

The length of the resultant depends on how many loops there are. For k = 0 there are none. For k = 1 there are one and a half loops. The circumference of the resulting circle is 2NE=3, the diameter is NE=3. For k = 2 there are two and a half loops. The circumference of the resulting circle is 2NE=5, the diameter is NE=5. The pattern for higher k is similar: the circumference is 2NE=(2k + 1), the diameter is NE=(k + 1=2): The intensity at this \approximate” maxima is proportional to the resultant squared, or (NE)2 Ik / : (k + 1=2)22

but Im is proportional to (NE)2, so 1 Ik = I m : (k + 1=2)22

(b) Near the middle the vectors simply fold back on one another, leaving a resultant of E. Then (NE)2 N2 so Im N2 (c) Let have the values which result in sin = 1, and then the two expressions are identical!

P43-6 (a) v = f , so v = f + f . Assuming v = 0, we have f =f = =. Ignore the negative sign (we don’t need it here). Then fc f f and then cc f = = : R Nm (b) The ray on the top gets there rst, the ray on the bottom must travel an additional distance of N d sin . It takes a time t = N d sin =c (c) Since m = d sin , the two resulting expression can be multiplied together to yield c N d sin (f )(t) = = 1: Nm c This is almost, but not quite, one of Heisenberg’s uncertainty relations!

P43-7 (b) We sketch parallel lines which connect centers to form almost any right triangle similar to the one shown in the Fig. 43-18. The triangle will have two sides which have integer multiple p lengths of the lattice spacing a . The hypotenuse of the triangle will then have length h2 + k2a , 00 where h and k are the integers. In Fig. 43-18 h = 2 while k = 1. The number of planes which cut the diagonal is equal to h2 + k2 if, and only if, h and k are relatively prime. The inter-planar spacing is then p h2 + k2a a 0p0 d= = : h2 + k2 h2 + k2

(a) The next ve spacings are then p p p p p h = 1; k = 4; d = a0= 17: P43-8 The middle layer cells will also di ract a beam, but this beam will be exactly out of phase with the top layer. The two beams will then cancel out exactly because of destructive interference.

(a) The direction of propagation is determined by considering the argument of the sine function. As t increases y must decrease to keep the sine function \looking” the same, so the wave (b) The electric eld is orthogonal (perpendicular) to the magnetic eld (so Ex = 0) and the direction of motion (so Ey = 0); Consequently, the only non-zero term is Ez. The magnitude of E will be equal to the magnitude of B times c. Since ~S = E~ B~ =0, when B~ points in the positive x direction then E~ must point in the negative z direction in order that ~S point in the negative y direction. Then Ez = cB sin(ky + !t): (c) The polarization is given by the direction of the electric eld, so the wave is linearly polarized in the z direction.

E44-1 E44-2 Let one wave be polarized in the x direction and the other in the y direction. Then the net electric eld is given by E2 = E2 + E2, or xy

2 2 sin2(kz !t) + sin2 0 where is the phase di erence. We can consider any point in space, including z = 0, and then average the result over a full cycle. Since merely shifts the integration limits, then the result is independent of . Consequently, there are no interference e ects.

E44-3 (a) The transmitted intensity is I0=2 = 6:1103W=m2. The maximum value of the electric eld is pp E = 20cI = 2(1:26106H=m)(3:00108m=s)(6:1103W=m2) = 2:15 V=m: m

(b) The radiation pressure is caused by the absorbed half of the incident light, so p = I=c = (6:1103W=m2)=(3:00108m=s) = 2:031011Pa:

E44-4 The rst sheet transmits half the original intensity, the second transmits an amount pro- portional to cos2 . Then I = (I =2) cos2 , or 0

p p 35:3 = arccos 2I=I0 = arccos 2(I0=3)=I0 : E44-5 The rst sheet polarizes the un-polarized light, half of the intensity is transmitted, so 2 The second sheet transmits according to Eq. 44-1, 11 24 and the transmitted light is polarized in the direction of the second sheet. The third sheet is 45 to the second sheet, so the intensity of the light which is transmitted through the third sheet is 11 I = I cos2 = I cos2(45) = I : 32 0 0 48

E44-6 The transmitted intensity through the rst sheet is proportional to cos2 , the transmitted intensity through the second sheet is proportional to cos2(90 ) = sin2 . Then I = I0 cos2 sin2 = (I0=4) sin2 2;

or 1p1p = arcsin 4I=I0 = arcsin 4(0:100I0)=I0 = 19:6: 22 Note that 70:4 is also a valid solution!

E44-7 The rst sheet transmits half of the original intensity; each of the remaining sheets transmits an amount proportional to cos2 , where = 30. Then I1316 2 )) = cos = (cos(30 = 0:211 I0 2 2 E44-8 The rst sheet transmits an amount proportional to cos2 , where = 58:8. The second sheet transmits an amount proportional to cos2(90 ) = sin2 . Then I = I0 cos2 sin2 = (43:3 W=m2) cos2(58:8) sin2(58:8) = 8:50 W=m2:

Since the incident beam is unpolarized the rst sheet transmits 1/2 of the original intensity. The transmitted beam then has a polarization set by the rst sheet: 58:8 to the vertical. The second sheet is horizontal, which puts it 31:2 to the rst sheet. Then the second sheet transmits cos2(31:2) of the intensity incident on the second sheet. The nal intensity transmitted by the second sheet can be found from the product of these terms, E44-9

1 2 2(31:2 2 I = (43:3 W=m ) cos ) = 15:8 W=m : 2

(a) The angle for complete polarization of the re ected ray is Brewster’s angle, and is given by Eq. 44-3 (since the rst medium is air) E44-11

p = tan1 n = tan1(1:33) = 53:1: E44-13 The angles are between p = tan1 n = tan1(1:472) = 55:81:

and p = tan1 n = tan1(1:456) = 55:52: E44-14 The smallest possible thickness t will allow for one half a wavelength phase di erence for the o and e waves. Then nt = =2, or t = (525109m)=2(0:022) = 1:2105m:

(a) The incident wave is at 45 to the optical axis. This means that there are two components; assume they originally point in the +y and +z direction. When they travel through the half wave plate they are now out of phase by 180; this means that when one component is in the +y direction the other is in the z direction. In e ect the polarization has been rotated by 90. (b) Since the half wave plate will delay one component so that it emerges 180 \later” than it should, it will in e ect reverse the handedness of the circular polarization. (c) Pretend that an unpolarized beam can be broken into two orthogonal linearly polarized components. Both are then rotated through 90; but when recombined it looks like the original beam. As such, there is no apparent change.

E44-15 E44-16 The quarter wave plate has a thickness of x = =4n, so the number of plates that can be cut is given by N = (0:250103m)4(0:181)=(488109m) = 371:

Intensity is proportional to the electric eld squared, so the original intensity reaching the eye is I0, with components Ih = (2:3)2Iv, and then P44-1

I0 = Ih + Iv = 6:3Iv or Iv = 0:16I0: (a) When the sun-bather is standing only the vertical component passes, while (b) when the sun-bather is lying down only the horizontal component passes.

P44-2 The intensity of the transmitted light which was originally unpolarized is reduced to Iu=2, regardless of the orientation of the polarizing sheet. The intensity of the transmitted light which was originally polarized is between 0 and Ip, depending on the orientation of the polarizing sheet. Then the maximum transmitted intensity is Iu=2 + Ip, while the minimum transmitted intensity is Iu=2. The ratio is 5, so Iu=2 + Ip Ip Iu=2 Iu or Ip=Iu = 2. Then the beam is 1/3 unpolarized and 2/3 polarized.

P44-3 Each sheet transmits a fraction cos2 = cos2 : N

There are N sheets, so the fraction transmitted through the stack is N cos2 : N

As N gets larger we can use a small angle approximation to the cosine function,

1 cos x 1 x2 for x 1 2 The the transmitted intensity is 2 2N 1 1 : 2 N2

This expression can also be expanded in a binomial expansion to get 1 2 2 N2 The stack then transmits all of the light which makes it past the rst lter. Assuming the light is originally unpolarized, then the stack transmits half the original intensity.

P44-4 (a) Stack several polarizing sheets so that the angle between any two sheets is suciently (b) The transmitted intensity fraction needs to be 0:95. Each sheet will transmit a fraction cos2 , where = 90=N , with N the number of sheets. Then we want to solve 2 N 0:95 = cos (90 =N ) for N . For large enough N , will be small, so we can expand the cosine function as cos2 = 1 sin2 1 2;

so 0:95 1 (=2N )2N 1 N (=2N )2 which has solution N = 2=4(0:05) = 49:

Since passing through a quarter wave plate twice can rotate the polarization of a linearly polarized wave by 90, then if the light passes through a polarizer, through the plate, re ects o the coin, then through the plate, and through the polarizer, it would be possible that when it passes through the polarizer the second time it is 90 to the polarizer and no light will pass. You won’t see On the other hand if the light passes rst through the plate, then through the polarizer, then is re ected, the passes again through the polarizer, all the re ected light will pass through he polarizer and eventually work its way out through the plate. So the coin will be visible. Hence, side A must be the polarizing sheet, and that sheet must be at 45 to the optical axis.

P44-5 P44-6 (a) The displacement of a ray is given by so the shift is y = t(tan e tan o):

Solving for each angle, 1 (1:486) 1 o = arcsin sin(38:8) = 22:21: (1:658) The shift is then y = (1:12102m) (tan(24:94) tan(22:21)) = 6:35104m: (c) The rays have polarizations which are perpendicular to each other; the o-wave being polarized (d) One ray, then the other, would disappear.

P44-7 The method is outline in Sample Problem 44-24; use a polarizing sheet to pick out the o-ray or the e-ray.

E45-1 (a) The energy of a photon is given by Eq. 45-1, E = hf , so hc E = hf = :

Putting in \best” numbers (6:626068761034J s) hc = (2:99792458108m=s) = 1:23984106 eV m: (1:6021764621019 C) This means that hc = 1240 eV nm is accurate to almost one part in 8000! (b) E = (1240 eV nm)=(589 nm) = 2:11 eV.

E45-2 Using the results of Exercise 45-1, (1240 eV nm) (0:60 eV) which is in the infrared.

E45-3 Using the results of Exercise 45-1, (1240 eV nm) (375 nm) and (1240 eV nm) (580 nm) The di erence is E = (3:307 eV) (2:138 eV) = 1:17 eV:

E45-4 P = E=t, so, using the result of Exercise 45-1, (1240 eV nm) P = (100=s) = 230 eV/s: (540 nm) That’s a small 3:681017W.

When talking about the regions in the sun’s spectrum it is more common to refer to wavelengths than frequencies. So we will use the results of Exercise 45-1(a), and solve = hc=E = (1240 eV nm)=E: The energies are between E = (1:01018J)=(1:61019C) = 6:25 eV and E = (1:01016J)=(1:6 1019C) = 625 eV. These energies correspond to wavelengths between 198 nm and 1.98 nm; this is the ultraviolet range.

E45-5 E45-6 The energy per photon is E = hf = hc=. The intensity is power per area, which is energy per time per area, so P E nhc hc n I= = = = : A At At A t But R = n=t is the rate of photons per unit time. Since h and c are constants and I and A are equal for the two beams, we have R1=1 = R2=2, or R1=R2 = 1=2:

(a) Since the power is the same, the bulb with the larger energy per photon will emits fewer photons per second. Since longer wavelengths have lower energies, the bulb emitting 700 nm (b) How many more photons per second? If E1 is the energy per photon for one of the bulbs, then N1 = P =E1 is the number of photons per second emitted. The di erence is then E45-7

PPP E1 E2 hc or (130 W) N N = ((700109m) (400109m)) = 1:961020: 1 2 34 (6:6310 Js)(3:00108m=s) E45-8 Using the results of Exercise 45-1, the energy of one photon is (1240 eV nm) (630 nm) The total light energy given o by the bulb is Et = P t = (0:932)(70 W)(730 hr)(3600 s/hr) = 1:71108J:

The number of photons is Et (1:71108J) n = = = 5:431026: E0 (1:968 eV)(1:61019J=eV)

E45-9 Apply Wien’s law, Eq. 45-4, maxT = 2898 m K; so (2898 m K) T = = 91106K: (321012m) Actually, the wavelength was supposed to be 32 m. Then the temperature would be 91 K.

E45-10 Apply Wien’s law, Eq. 45-4, maxT = 2898 m K; so (2898 m K) = = 1:45 m: (0:0020 K) This is in the radio region, near 207 on the FM dial.

E45-11 The wavelength of the maximum spectral radiancy is given by Wien’s law, Eq. 45-4, maxT = 2898 m K:

E45-12 (a) Apply Wien’s law, Eq. 45-4, maxT = 2898 m K:; so (2898 m K) = = 5:00107m: (5800 K) (b) Apply Wien’s law, Eq. 45-4, maxT = 2898 m K:; so (2898 m K) T = = 5270 K: (550109m)

E45-13 I=T4andP=IA.Then P = (5:67108W=m2 K4)(1900 K)4(0:5103m)2 = 0:58 W: E45-14 Since I / T4, doubling T results in a 24 = 16 times increase in I. Then the new power level is (16)(12:0 mW) = 192 mW: E45-15 (a) We want to apply Eq. 45-6, 2c2h 1 R(; T ) = : 5 ehc=kT 1 We know the ratio of the spectral radiancies at two di erent wavelengths. Dividing the above equation at the rst wavelength by the same equation at the second wavelength, 5 ehc=1kT 1 1 5 ehc=2kT 1 2

where 1 = 200 nm and 2 = 400 nm. We can considerably simplify this expression if we let x = ehc=2kT ;

because since 2 = 21 we would have ehc=1kT = e2hc=2kT = x2: Then we get 1 5 x2 1 1 3:5 = = (x + 1): 2 x 1 32 We will use the results of Exercise 45-1 for the exponents and then rearrange to get hc (3:10 eV) T = = = 7640 K: 1k ln(111) (8:62105 eV=K) ln(111) (b) The method is the same, except that instead of 3.5 we have 1/3.5; this means the equation for x is 11 3:5 32 with solution x = 8:14, so then hc (3:10 eV) T = = = 17200 K: 1k ln(8:14) (8:62105 eV=K) ln(8:14)

E45-16 hf = , so (5:32 eV) f = = = 1:281015Hz: h (4:141015eV s) E45-17 We’ll use the results of Exercise 45-1. Visible red light has an energy of (1240 eV nm) E = = 1:9 eV: (650 nm) The substance must have a work function less than this to work with red light. This means that only cesium will work with red light. Visible blue light has an energy of (1240 eV nm) E = = 2:75 eV: (450 nm) This means that barium, lithium, and cesium will work with blue light.

E45-18 Since Km = hf , K = (4:141015eV s)(3:191015Hz) (2:33 eV) = 10:9 eV: m

(a) Use the results of Exercise 45-1 to nd the energy of the corresponding photon, hc (1240 eV nm) E = = = 1:83 eV: (678 nm) Since this energy is less than than the minimum energy required to remove an electron then the (b) The cut-o wavelength is the longest possible wavelength of a photon that will still result in the photo-electric e ect occurring. That wavelength is (1240 eV nm) (1240 eV nm) = = = 544 nm: E (2:28 eV) E45-19

E45-20 (a) Since Km = hc= , (1240 eV nm) Km = (4:2 eV) = 2:0 eV: (200 nm) (b) The minimum kinetic energy is zero; the electron just barely makes it o the surface. (c) V s = Km=q = 2:0 V: (d) The cut-o wavelength is the longest possible wavelength of a photon that will still result in the photo-electric e ect occurring. That wavelength is (1240 eV nm) (1240 eV nm) = = = 295 nm: E (4:2 eV)

E45-21 Km = qV s = 4:92 eV. But Km = hc= , so (1240 eV nm) = = 172 nm: (4:92 eV + 2:28 eV)

E45-22 (a) Km = qV s and Km = hc= . We have two di erent values for qV s and , so subtracting this equation from itself yields q(V s;1 V s;2) = hc=1 hc=2: Solving for 2, hc hc=1 q(V s;1 V s;2) (1240 eV nm) (1240 eV nm)=(491 nm) (0:710 eV) + (1:43 eV) = 382 nm: (b) Km = qV s and Km = hc= , so = (1240 eV nm)=(491 nm) (0:710 eV) = 1:82 eV:

(a) The stopping potential is given by Eq. 45-11, h ee so (1240 eV nm) (1:85 eV V0 = = 1:17 V: e(410 nm e E45-23

(b) These are not relativistic electrons, so v = 2K=m = c 2K=mc2 = c 2(1:17 eV)=(0:511106 or v = 64200 m=s.

E45-24 It will have become the stopping potential, or h ee so (4:141015eV m) (2:49 eV) V0 = (6:331014=s) = 0:131 V: (1:0e) (1:0e) E45-25

E45-26 (a) Using the results of Exercise 45-1, (1240 eV nm) = = 62 pm: (20103 eV) (b) This is in the x-ray region.

E45-27 (a) Using the results of Exercise 45-1, (1240 eV nm) E = = 29; 800 eV: (41:6103 nm) (b) f = c= = (3108m=s)=(41:6 pm) = 7:211018=s: (c) p = E=c = 29; 800 eV=c = 2:98104 eV=c:

E45-28 (a) E = hf , so (0:511106eV) f = = 1:231020=s: (4:141015eV s) (b) = c=f = (3108m=s)=(1:231020=s) = 2:43 pm: (c) p = E=c = (0:511106eV)=c.

The initial momentum of the system is the momentum of the photon, p = h=. This momentum is imparted to the sodium atom, so the nal speed of the sodium is v = p=m, where m is the mass of the sodium. Then h (6:631034J s) v = = = 2:9 cm/s: m (589109m)(23)(1:71027kg) E45-29

E45-30 (a) C = h=mc = hc=mc2, so (1240 eV nm) C = = 2:43 pm: (0:511106eV) (c) Since E = hf = hc=, and = h=mc = hc=mc2, then E = hc= = mc2:

E45-31 The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17, h 0 = + (1 cos ): mc We’ll use the results of Exercise 45-30 to save some time, and let h=mc = C, which is 2.43 pm. (a) For = 35, 0 = (2:17 pm) + (2:43 pm)(1 cos 35) = 2:61 pm: (b) For = 115, 0 = (2:17 pm) + (2:43 pm)(1 cos 115) = 5:63 pm:

E45-32 (a) We’ll use the results of Exercise 45-1: (1240 eV nm) = = 2:43 pm: (0:511106eV) (b) The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17,

h 0 = + (1 cos ): mc 0 = (2:43 pm) + (2:43 pm)(1 cos 72) = 4:11 pm: (c) We’ll use the results of Exercise 45-1: (1240 eV nm) E = = 302 keV: (4:11 pm)

E45-33 The change in the wavelength of a photon during Compton scattering is given by Eq. 45-17, h 0 = (1 cos ): mc We are not using the expression with the form because and E are not simply related. The wavelength is related to frequency by c = f , while the frequency is related to the energy by Eq. 45-1, E = hf . Then E = E E0 = hf hf0;

11 0 0 = hc : 0 Into this last expression we substitute the Compton formula. Then h2 (1 cos ) E = : m 0 Now E = hf = hc=, and we can divide this on both sides of the above equation. Also, 0 = c=f 0, and we can substitute this into the right hand side of the above equation. Both of these steps result in E hf 0 = (1 cos ): E mc2 Note that mc2 is the rest energy of the scattering particle (usually an electron), while hf 0 is the energy of the scattered photon.

E45-34 The wavelength is related to frequency by c = f , while the frequency is related to the energy by Eq. 45-1, E = hf . Then E = E E0 = hf hf0;

11 0 0 0 E E + But E=E = 3=4, so E45-35 The maximum shift occurs when = 180, so h (1240 eV nm) m = 2 = 2 = 2:641015m: mc 938 MeV) E45-36 Since E = hf frequency shifts are identical to energy shifts. Then we can use the results of Exercise 45-33 to get (0:9999)(6:2 keV) (511 keV) (b) (0:0001)(6:2 keV) = 0:62 eV:

(a) The change in wavelength is independent of the wavelength and is given by Eq. 45-17, hc (1240 eV nm) = (1 cos ) = 2 = 4:85103 nm: mc2 (0:511106 eV) (b) The change in energy is given by hc hc f i 11 i + i 11 = (1240 eV nm) = 42:1 keV (9:77 pm) + (4:85 pm) (9:77 pm) (c) This energy went to the electron, so the nal kinetic energy of the electron is 42:1 keV.

E45-38 For = 90 = h=mc. Then E hf 0 E hf + h=mc =: + h=mc (a) E=E = (2:43 pm)=(3:00 cm + 2:43 pm) = 8:11011: (b) E=E = (2:43 pm)=(500 nm + 2:43 pm) = 4:86106: (c) E=E = (2:43 pm)=(0:100 nm + 2:43 pm) = 0:0237: (d) E=E = (2:43 pm)=(1:30 pm + 2:43 pm) = 0:651:

E45-39 We can use the results of Exercise 45-33 to get (0:90)(215 keV) (511 keV) which has solution = 42=6.

E45-40 (a) A crude estimate is that the photons can’t arrive more frequently than once every (b) The power output would be (1240 eV nm) (550 nm) which is 3:61011 W!

E45-41 We can follow the example of Sample Problem 45-6, and apply = 0(1 v=c): (a) Solving for 0, (588:995 nm) 0 = = 588:9944 nm: (1 (300 m=s)(3108m=s) E45-37

(b) Applying Eq. 45-18, h (6:61034J s) v = = = 3102m=s: m (22)(1:71027kg)(590109m) (c) Emitting another photon will slow the sodium by about the same amount.

(b) If the argon averages a speed of 220 m=s, then it requires interactions at the rate of (2900)(220 m=s)=(1:0 m) = 6:4105=s if it is going to slow down in time.

The radiant intensity is given by Eq. 45-3, I = T 4. The power that is radiated through the opening is P = IA, where A is the area of the opening. But energy goes both ways through the opening; it is the di erence that will give the net power transfer. Then P45-1

4 4 P net = (I0 I1)A = A T T : 01 Put in the numbers, and Pnet=(5:67108W=m2K4)(5:20104m2)(488K)4(299K)4=1:44W:

P45-2 (a)I=T4andP=IA.ThenT4=P=A,or s (100 W) T = 4 = 3248 K: (5:67108W=m2 K4)(0:28103m)(1:8102m) (b) The rate that energy is radiated o is given by dQ=dt = mC dT =dt. The mass is found from m = V , where V is the volume. This can be combined with the power expression to yield AT4=VCdT=dt;

which can be integrated to yield V C t = (1=T 3 1=T 3): 3A 2 1 Putting in numbers, (19300kg=m3)(0:28103m)(132J=kgC) t = [1=(2748 K)3 1=(3248 K)3]; 3(5:67108W=m2 K4)(4) = 20 ms:

Light from the sun will \heat-up” the thin black screen. As the temperature of the screen increases it will begin to radiate energy. When the rate of energy radiation from the screen is equal to the rate at which the energy from the sun strikes the screen we will have equilibrium. We need rst to nd an expression for the rate at which energy from the sun strikes the screen. The temperature of the sun is TS. The radiant intensity is given by Eq. 45-3, IS = TS4. The total power radiated by the sun is the product of this radiant intensity and the surface area of the sun, so SSS P45-3

Assuming that the lens is on the surface of the Earth (a reasonable assumption), then we can nd the power incident on the lens if we know the intensity of sunlight at the distance of the Earth from the sun. That intensity is PS PS A 4RE2 where RE is the radius of the Earth’s orbit. Combining, 2 4 rS IE=TS RE

The total power incident on the lens is then 2 4 rS 2 RE

where rl is the radius of the lens. All of the energy that strikes the lens is focused on the image, so The screen radiates as the temperature increases. The radiant intensity is I = T 4, where T is the temperature of the screen. The power radiated is this intensity times the surface area, so P=IA=2r2T4: i

The factor of \2″ is because the screen has two sides, while ri is the radius of the image. Set this equal to P lens, 2 2 4 4 rS 2 RE or 2 1 rS rl T4 = T 4 : S 2 RE ri The radius of the image of the sun divided by the radius of the sun is the magni cation of the lens. But magni cation is also related to image distance divided by object distance, so ri i rS o Distances should be measured from the lens, but since the sun is so far from the Earth, we won’t be far o in stating o RE. Since the object is so far from the lens, the image will be very, very close to the focal point, so we can also state i f . Then ri f rS RE so the expression for the temperature of the thin black screen is considerably simpli ed to 2 1 rl T 4 = T S4 : 2f Now we can put in some of the numbers.

s 1 (1:9 cm) T = (5800 K) = 1300 K: 21=4 (26 cm)

P45-4 The derivative of R with respect to is c2 h 2 c3 h2 e( hkcT ) 10 + : 6 (e( hkcT ) 1) 7 (e( k T ) 1)2 k T hc

Ohh, that’s ugly. Setting it equal to zero allows considerable simpli cation, and we are left with (5 x)ex = 5;

where x = hc=kT . The solution is found numerically to be x = 4:965114232. Then (1240 eV nm) 2:898103m K = = : (4:965)(8:62105eV/K)T T P45-5 (a) If the planet has a temperature T , then the radiant intensity of the planet will be IT4, and the rate of energy radiation from the planet will be P = 4R2T 4;

A steady state planet temperature requires that the energy from the sun arrive at the same rate as the energy is radiated from the planet. The intensity of the energy from the sun a distance r from the sun is P sun=4r2;

and the total power incident on the planet is then 2 P sun P = R : 4r2 Equating, 2 4 2 P sun 4r2 4 P sun T= : 16r2 (b) Using the last equation and the numbers from Problem 3, s 1 (6:96108m) T = p (5800 K) = 279 K: 2 (1:51011m)

P45-6 (a) Change variables as suggested, then = hc=xkT and d = (hc=x2kT )dx. Integrate (note the swapping of the variables of integration picks up a minus sign): 2c2h (hc=x2kT )dx Z (hc=xkT )5 ex 1 4 4Z 3 2k T x dx h3c2 ex 1 25k4 = T 4: 15h3c2

P45-7 (a) P = E=t = nhf=t = (hc=)(n=t), where n=t is the rate of photon emission. Then (100 W)(589109m) n=t = = 2:961020=s: (6:631034J s)(3:00108m=s) (b) The ux at a distance r is the rate divided by the area of the sphere of radius r, so s (2:961020=s) r = = 4:8107m: 4(1104=m2 s) (c) The photon density is the ux divided by the speed of light; the distance is then s (2:961020=s) r = = 280 m: 4(1106=m3)(3108m=s) (d) The ux is given by (2:961020=s) = 5:891018=m2 s: 4(2:0 m)2 The photon density is (5:891018m2 s)=(3:00108m=s) = 1:961010=m3:

P45-8 Momentum conservation requires while energy conservation requires E + mc2 = Ee:

Square both sides of the energy expression and ee e p2 c2 + 2Emc2 = p2c2: e

But the momentum expression can be used here, and the result is 2Emc2 = 0:

P45-9 (a) Since qvB = mv2=r, v = (q=m)rB The kinetic energy of (non-relativistic) electrons will be 1 1 q2(rB)2 2 2m or 1 (1:61019C) K = (188106T m)2 = 3:1103 eV: 2 (9:11031kg) (b) Use the results of Exercise 45-1, (1240 eV nm) = 3:1103eV = 1:44104 eV: (71103nm)

P45-10 P45-11 (a) The maximum value of is 2h=mc. The maximum energy lost by the photon is then hc hc f i 11 i + i 2h=mc ( + 2h=mc) where in the last line we wrote for i. The energy given to the electron is the negative of this, so 2h2 Kmax = : m( + 2h=mc) Multiplying through by 12 = (E=hc)2 we get 2E2 Kmax = : mc2(1 + 2hc=mc2) or E2 Kmax = : mc2=2 + E (b) The answer is (17:5 keV)2 Kmax = = 1:12 keV: (511 eV)=2 + (17:5 keV)

E46-1 (a) Apply Eq. 46-1, = h=p. The momentum of the bullet is p = mv = (0:041 kg)(960 m=s) = 39kg m=s;

so the corresponding wavelength is = h=p = (6:631034J s)=(39kg m=s) = 1:71035 m: (b) This length is much too small to be signi cant. How much too small? If the radius of the galaxy were one meter, this distance would correspond to the diameter of a proton.

E46-2 (a) = h=p and p2=2m = K, then hc (1240 eV nm) 1:226 nm =p =p p = p : 2mc2K 2(511 keV) K K (b) K = eV , so r 1:226 nm 1:5 V = p = nm: eV V p E46-3 For non-relativistic particles = h=p and p2=2m = K, so = hc= 2mc2K. (a) For the electron, (1240 eV nm) = p = 0:0388 nm: 2(511 keV)(1:0 keV) (c) For the neutron, (1240 MeV fm) = p = 904 fm: 2(940 MeV)(0:001 MeV) (b) For ultra-relativistic particles K E pc, so hc (1240 eV nm) = = = 1:24 nm: E (1000 eV)

E46-4 For non-relativistic particles p = h= and p2=2m = K, so K = (hc)2=2mc22. Then (1240 eV nm)2 K = = 4:34106 eV: 2(5:11106eV)(589 nm)2

(a) Apply Eq. 46-1, p = h=. The proton speed would then be h hc (1240 MeV fm) v = = c = c = 0:0117c: m mc2 (938 MeV)(113 fm) This is good, because it means we were justi ed in using the non-relativistic equations. Then (b) The kinetic energy of this electron would be 11 K = mv2 = (938 MeV)(0:0117)2 = 64:2 keV: 22 The potential through which it would need to be accelerated is 64.2 kV.

p E46-6 (a) K = qV and p = 2mK. Then p p = 2(22)(932 MeV=c2)(325 eV) = 3:65106 eV=c: (b) = h=p, so hc (1240 eV nm) = = = 3:39104nm: pc (3:65106 eV=c)c

p E46-7 (a) For non-relativistic particles = h=p and p2=2m = K, so = hc= 2mc2K. For the alpha particle, (1240 MeV fm) = p = 5:2 fm: 2(4)(932 MeV)(7:5 MeV) (b) Since the wavelength of the alpha is considerably smaller than the distance to the nucleus we can ignore the wave nature of the alpha particle.

E46-8 (a) For non-relativistic particles p = h= and p2=2m = K, so K = (hc)2=2mc22. Then (1240 keV pm)2 K = = 15 keV: 2(511keV)(10 pm)2 (b) For ultra-relativistic particles K E pc, so hc (1240 keV pm) E = = = 124 keV: (10 pm)

The relativistic relationship between energy and momentum is E2 = p2c2 + m2c4;

and if the energy is very large (compared to mc2), then the contribution of the mass to the above expression is small, and E2 p2c2: Then from Eq. 46-1, h hc hc (1240 MeV f m) = = = = = 2:5102 fm: p pc E (50103 MeV) p E46-10 (a) K = 3kT=2, p = 2mK, and = h=p, so h hc 3mkT 3mc2kT (1240 MeV f m) = p = 74 pm: 3(4)(932MeV)(8:621011MeV/K)(291 K) (b) pV = N kT ; assuming that each particle occupies a cube of volume d3 = V0 then the inter- particle spacing is d, so E46-9

s p (1:381023J=K)(291 K) d = 3 V =N = 3 = 3:4 nm: (1:01105Pa)

E46-11 p = mv and p = h=, so m = h=v. Taking the ratio, me v = = (1:813104)(3) = 5:439104: m eve

The mass of the unknown particle is then (0:511 MeV=c2) m = = 939:5 MeV: (5:439104)

p E46-12 (a) For non-relativistic particles = h=p and p2=2m = K, so = hc= 2mc2K. For the electron, (1240 eV nm) = p = 1:0 nm: 2(5:11105 eV)(1:5 eV) For ultra-relativistic particles K E pc, so for the photon hc (1240 eV nm) = = = 830 nm: E (1:5 eV) (b) Electrons with energies that high are ultra-relativistic. Both the photon and the electron will then have the same wavelength;

hc (1240 MeV fm) = = = 0:83 fm: E (1:5 GeV) E46-13 (a) The classical expression for kinetic energy is p p = 2mK;

so h hc (1240 keV pm) = = p = p = 7:76 pm: p 2mc2K 2(511 keV)(25:0 keV) (a) The relativistic expression for momentum is p p pc = sqrtE2 m2c4 = (mc2 + K)2 m2c4 = K2 + 2mc2K: Then hc (1240 keV pm) = = p = 7:66 pm: pc (25:0 keV)2 + 2(511 keV)(25:0 keV)

E46-14 We want to match the wavelength of the gamma to that of the electron. For the gamma, = hc=E . For the electron, K = p2=2m = h2=2m2: Combining, h2 E2 K = E2 = : 2mh2c2 2mc2 With numbers, (136keV)2 K = = 18:1 keV: 2(511 keV) That would require an accelerating potential of 18:1 kV.

E46-15 First nd the wavelength of the neutrons. For non-relativistic particles = h=p and p p2=2m = K, so = hc= 2mc2K. Then (1240 keV pm) = p = 14 pm: 2(940103 keV)(4:2103 keV) Bragg re ection occurs when 2d sin = , so = arcsin(14 pm)=2(73:2 pm) = 5:5:

E46-16 This is merely a Bragg re ection problem. Then 2d sin = m, or = arcsin(3)(11 pm)=2(54:64 pm) = 17:6:

(a) Since sin 52 = 0:78, then 2(=d) = 1:57 > 1, so there is no di raction order other than the rst.

E46-17 (b) For an accelerating potential of 54 volts we have =d = 0:78. Increasing the potential will increase the kinetic energy, increase the momentum, and decrease the wavelength. d won’t change. pThe kinetic energy is increased by a factor of 60=54 = 1:11, the momentum increases by a factor of 1:11 = 1:05, so the wavelength changes by a factor of 1=1:05 = 0:952. The new angle is then = arcsin(0:952 0:78) = 48:

E46-18 First nd the wavelength of the electrons. For non-relativistic particles = h=p and p p2=2m = K, so = hc= 2mc2K. Then (1240 keV pm) = p = 62:9 pm: 2(511 keV)(0:380 keV) This is now a Bragg re ection problem. Then 2d sin = m, or = arcsin(9)(62:9 pm)=2(314 pm) = 64:3: But the odd orders vanish (see Chapter 43 for a discussion on this).

E46-19 Since f t 1=2, we have f = 1=2(0:23 s) = 0:69=s:

E46-20 Since f t 1=2, we have f = 1=2(0:10109s) = 1:61010=s: The bandwidth wouldn’t t in the frequency allocation! E46-21 Apply Eq. 46-9, h 4:141015 eV s) E = = 7:6105 eV: 2t 2(8:71012s) This is much smaller than the photon energy.

E46-22 Apply Heisenberg twice: 4:141015 eV s) E1 = = 5:49108 eV: 2(12109s) and 4:141015 eV s) E2 = = 2:86108 eV: 2(23109s) The sum is Etransition = 8:4108eV.

E46-23 Apply Heisenberg: 6:631034J s) p = = 8:81024kg m=s: 2(121012m) E46-24 p = (0:5 kg)(1:2 s) = 0:6 kg m=s. The position uncertainty would then be (0:6 J=s) x = = 0:16 m: 2(0:6 kg m=s)

We want v v, which means p p. Apply Eq. 46-8, and hh x : 2p 2p According to Eq. 46-1, the de Broglie wavelength is related to the momentum by = h=p;

E46-25 so x : 2 E46-26 (a) A particle con ned in a (one dimensional) box of size L will have a position uncertainty of no more than x L. The momentum uncertainty will then be no less than hh p : 2x 2L so (6:631034J s) p = 11024kg m=s: 2(1010 m)

(b) Assuming that p p, we have h 2L and then the electron will have a (minimum) kinetic energy of p2 h2 E : 2m 82mL2 or (hc)2 (1240 keV pm)2 E = = 0:004 keV: 82mc2L2 82(511 keV)(100 pm)2 E46-27 (a) A particle con ned in a (one dimensional) box of size L will have a position uncer- tainty of no more than x L. The momentum uncertainty will then be no less than hh p : 2x 2L so (6:631034J s) p = 11020kg m=s: 2(1014 m) (b) Assuming that p p, we have h 2L and then the electron will have a (minimum) kinetic energy of p2 h2 E : 2m 82mL2 or (hc)2 (1240 MeV fm)2 E = = 381 MeV: 82mc2L2 82(0:511 MeV)(10 fm)2 This is so large compared to the mass energy of the electron that we must consider relativistic e ects. It will be very relativistic (381 0:5!), so we can use E = pc as was derived in Exercise 9. Then hc (1240 MeV fm) E = = = 19:7 MeV: 2L 2(10 fm) This is the total energy; so we subtract 0.511 MeV to get K = 19 MeV.

E46-28 We want to nd L when T = 0:01. This means solving

EE U0 U0 (5:0 eV) (5:0 eV) 0 (6:0 eV) (6:0 eV) 5:31010m = L:

E46-29 The wave number, k, is given by 2 p k = 2mc2(U0 E): hc (a) For the proton mc2 = 938 MeV, so 2 p 1 k = 2(938 MeV)(10 MeV 3:0 MeV) = 0:581 fm : (1240 MeV fm) The transmission coecient is then

(3:0 MeV) 2(0:581 fm )(10 fm) 5 (3:0 MeV) 1 T = 16 1 e = 3:010 : (10 MeV) (10 MeV) (b) For the deuteron mc2 = 2 938 MeV, so 2 p 1 k = 2(2)(938 MeV)(10 MeV 3:0 MeV) = 0:821 fm : (1240 MeV fm) The transmission coecient is then

(3:0 MeV) 2(0:821 fm )(10 fm) 7 (3:0 MeV) 1 T = 16 1 e = 2:510 : (10 MeV) (10 MeV) E46-30 The wave number, k, is given by 2 p k = 2mc2(U0 E): hc (a) For the proton mc2 = 938 MeV, so 2 p k = 2(938 MeV)(6:0 eV 5:0 eV) = 0:219 pm1: (1240 keV pm) We want to nd T . This means solving

EE U0 U0 (5:0 eV) (5:0 eV) 12 9 = 16 1 e2(0:21910 )(0:710 ); (6:0 eV) (6:0 eV) = 1:610133: A current of 1 kA corresponds to N = (1103C=s)=(1:61019C) = 6:31021=s protons per seconds. The time required for one proton to pass is then t = 1=(6:31021=s)(1:610133) = 9:910110s: That’s 10104 years!

We will interpret low energy to mean non-relativistic. Then hh = =p : p 2mnK The di raction pattern is then given by p d sin = m = mh= 2mnK;

where m is di raction order while mn is the neutron mass. We want to investigate the spread by taking the derivative of with respect to K, P46-1

mh d cos d = p dK: 2 2mnK3 Divide this by the original equation, and then cos dK d = : sin 2K Rearrange, change the di erential to a di erence, and then K = tan : 2K We dropped the negative sign out of laziness; but the angles are in radians, so we need to multiply by 180= to convert to degrees.

P46-2 P46-3 We want to solve EE U0 U0 for E. Unfortunately, E is contained in k since 2 p k = 2mc2(U0 E): hc We can do this by iteration. The maximum possible value for

EE 1 U0 U0 is 1=4; using this value we can get an estimate for k: 5:92= nm = k: Now solve for E: 0

Put this value for E back into the transmission equation to nd a new k:

EE U0 U0 (4:7 eV) (4:7 eV) (6:0 eV) (6:0 eV) 5:65= nm = k: Now solve for E using this new, improved, value for k: E = U0 (hc)2k28mc22;

(1240 eV nm)2(5:65=nm)2 82(5:11105eV) = 4:78 eV: P46-4 (a) A one percent increase in the barrier height means U0 = 6:06 eV. For the electron mc2 = 5:11105 eV, so 2 p k = 2(5:11105 eV)(6:06 eV 5:0 eV) = 5:27 nm1: (1240 eV nm) We want to nd T . This means solving

EE U0 U0 (5:0 eV) (5:0 eV) (6:06 eV) (6:06 eV) = 1:44103:

(b) A one percent increase in the barrier thickness means L = 0:707 nm. For the electron mc2 = 5:11105 eV, so 2 k = p2(5:11105 eV)(6:0 eV 5:0 eV) = 5:12 nm1: (1240 eV nm) We want to nd T . This means solving

EE U0 U0 (5:0 eV) (5:0 eV) (6:0 eV) (6:0 eV) = 1:59103: For the electron mc2 = 5:11105 eV, so 2 k = p2(5:11105 eV)(6:0 eV 5:05 eV) = 4:99 nm1: (1240 eV nm)

We want to nd T . This means solving EE U0 U0 (5:05 eV) (5:05 eV) (6:0 eV) (6:0 eV) = 1:97103: That’s a 14% increase.

P46-5 First, the rule for exponents ei(a+b) = eia eib: Then apply Eq. 46-12, ei = cos + i sin , cos(a + b) + i sin(a + b) = (cos a + i sin a)(sin b + i sin b): Expand the right hand side, remembering that i2 = 1, cos(a + b) + i sin(a + b) = cos a cos b + i cos a sin b + i sin a cos b sin a sin b: Since the real part of the left hand side must equal the real part of the right and the imaginary part of the left hand side must equal the imaginary part of the right, we actually have two equations. They are cos(a + b) = cos a cos b sin a sin b and sin(a + b) = cos a sin b + sin a cos b:

(a) The ground state energy level will be given by h2 (6:63 1034J s)2 E1 = = = 3:1 1010 J: 8mL2 8(9:11 1031kg)(1:4 1014 m)2 The answer is correct, but the units make it almost useless. We can divide by the electron charge to express this in electron volts, and then E = 1900 MeV. Note that this is an extremely relativistic (b) We can repeat what we did above, or we can apply a \trick” that is often used in solving these problems. Multiplying the top and the bottom of the energy expression by c2 we get (hc)2 E1 = 8(mc2)L2 Then (1240 MeV fm)2 E1 = = 1:0 MeV: 8(940 MeV)(14 fm)2 (c) Finding an neutron inside the nucleus seems reasonable; but nding the electron would not. The energy of such an electron is considerably larger than binding energies of the particles in the nucleus.

E47-1 E47-2 Solve n2(hc)2 En = 8(mc2)L2 for L, then nhc 8mc2En (3)(1240 eV nm) 8(5:11105eV)(4:7 eV) = 0:85 nm:

E47-3 Solve for E4 E1: 42(hc)2 12(hc)2 8(mc2)L2 8(mc2)L2 (16 1)(1240 eV nm)2 8(5:11105)(0:253 nm)2 = 88:1 eV:

E47-4 Since E / 1=L2, doubling the width of the well will lower the ground state energy to (1=2)2 = 1=4, or 0.65 eV.

E47-5 (a) Solve for E2 E1: 22h2 12h2 8mL2 8mL2 (3)(6:631034J s)2 8(40)(1:671027kg)(0:2 m)2 = 6:21041J: (b) K = 3kT =2 = 3(1:381023J=K)(300 K)=2 = 6:211021: The ratio is 11020. (c) T = 2(6:21041J)=3(1:381023J=K) = 3:01018K:

E47-6 (a) The fractional di erence is (En+1 En)=En, or E h2 h2 h2

En 8mL2 8mL2 8mL2 (n + 1)2 n2 n2 2n + 1 =: n2 (b) As n ! 1 the fractional di erence goes to zero; the system behaves as if it is continuous.

E47-7 (a) We will take advantage of the \trick” that was developed in part (b) of Exercise 47-1. Then (hc)2 (1240 eV nm)2 En = n2 = (15)2 = 8:72 keV: 8mc2 L 8(0:511 106 eV)(0:0985 nm)2 (b) The magnitude of the momentum is exactly known because E = p2=2m. This momentum is given by p p pc = 2mc2E = 2(511 keV)(8:72 keV) = 94:4 keV: What we don’t know is in which direction the particle is moving. It is bouncing back and forth between the walls of the box, so the momentum could be directed toward the right or toward the left. The uncertainty in the momentum is then p = p which can be expressed in terms of the box size L by p r n2h2 nh p = p = 2mE = = : 4L2 2L (c) The uncertainty in the position is 98.5 pm; the electron could be anywhere inside the well.

E47-8 The probability distribution function is 2 2x P2 = sin2 : LL We want to integrate over the central third, or L=6 2 Z 2x L=6 L L 1 =3 Z =3 = 0:196:

E47-9 (a) Maximum probability occurs when the argument of the cosine (sine) function is k ([k + 1=2]). This occurs when x = N L=2n (b) Minimum probability occurs when the argument of the cosine (sine) function is [k + 1=2] (k). This occurs when x = N L=2n for even N .

E47-10 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2: (a) The Lyman series is the series which ends on E1. The least energetic state starts on E2. The transition energy is E E = (13:6 eV)(1=12 1=22) = 10:2 eV: 21 The wavelength is hc (1240 eV nm) = = = 121:6 nm: E (10:2 eV) (b) The series limit is 0 E1 = (13:6 eV)(1=12) = 13:6 eV: The wavelength is hc (1240 eV nm) = = = 91:2 nm: E (13:6 eV) E47-11 The ground state of hydrogen, as given by Eq. 47-21, is me4 (9:109 1031 kg)(1:602 1019 C)4 E1 = = = 2:179 1018 J: 82h2 8(8:854 1012 F=m)2(6:626 1034 J s)2 0

In terms of electron volts the ground state energy is E1 = (2:179 1018 J)=(1:602 1019 C) = 13:60 eV: E47-12 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2: (c) The transition energy is E = E E = (13:6 eV)(1=12 1=32) = 12:1 eV: 31

(a) The wavelength is hc (1240 eV nm) = = = 102:5 nm: E (12:1 eV) (b) The momentum is p = E=c = 12:1 eV=c:

E47-13 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2: (a) The Balmer series is the series which ends on E2. The least energetic state starts on E3. The transition energy is E E = (13:6 eV)(1=22 1=32) = 1:89 eV: 32 The wavelength is hc (1240 eV nm) = = = 656 nm: E (1:89 eV)

(b) The next energetic state starts on E4. The transition energy is E4 E2 = (13:6 eV)(1=22 1=42) = 2:55 eV: The wavelength is hc (1240 eV nm) = = = 486 nm: E (2:55 eV) (c) The next energetic state starts on E5. The transition energy is E5 E2 = (13:6 eV)(1=22 1=52) = 2:86 eV: The wavelength is hc (1240 eV nm) = = = 434 nm: E (2:86 eV) (d) The next energetic state starts on E6. The transition energy is E6 E2 = (13:6 eV)(1=22 1=62) = 3:02 eV: The wavelength is hc (1240 eV nm) = = = 411 nm: E (3:02 eV) (e) The next energetic state starts on E7. The transition energy is E7 E2 = (13:6 eV)(1=22 1=72) = 3:12 eV: The wavelength is hc (1240 eV nm) = = = 397 nm: E (3:12 eV) E47-14 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2: The transition energy is hc (1240 eV nm) E = = = 10:20 eV: (121:6 nm) This must be part of the Lyman series, so the higher state must be En = (10:20 eV) (13:6 eV) = 3:4 eV: That would correspond to n = 2.

E47-15 The binding energy is the energy required to remove the electron. If the energy of the electron is negative, then that negative energy is a measure of the energy required to set the electron The rst excited state is when n = 2 in Eq. 47-21. It is not necessary to re-evaluate the constants in this equation every time, instead, we start from E1 En = where E1 = 13:60 eV: n2 Then the rst excited state has energy (13:6 eV) E2 = = 3:4 eV: (2)2 The binding energy is then 3.4 eV.

E47-16 rn = a0n2, so p n = (847 pm)=(52:9 pm) = 4: E47-17 (a) The energy of this photon is hc (1240 eV nm) E = = = 0:96739 eV: (1281:8 nm) The nal state of the hydrogen must have an energy of no more than 0:96739, so the largest possible n of the nal state is p n < 13:60 eV=0:96739 eV = 3:75;

so the nal n is 1, 2, or 3. The initial state is only slightly higher than the nal state. The jump from n = 2 to n = 1 is too large (see Exercise 15), any other initial state would have a larger energy So what level might be above n = 2? We’ll try p n = 13:6 eV=(3:4 eV 0:97 eV) = 2:36;

which is so far from being an integer that we don’t need to look farther. The n = 3 state has energy 13:6 eV=9 = 1:51 eV. Then the initial state could be p n = 13:6 eV=(1:51 eV 0:97 eV) = 5:01;

which is close enough to 5 that we can assume the transition was n = 5 to n = 3. (b) This belongs to the Paschen series.

E47-18 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2:

(a) The transition energy is E = E4 E1 = (13:6 eV)(1=12 1=42) = 12:8 eV:

(b) All transitions n ! m are allowed for n 4 and m < n. The transition energy will be of the form En Em = (13:6 eV)(1=m2 1=n2): The six possible results are 12.8 eV, 12.1 eV, 10.2 eV, 2.55 eV, 1.89 eV, and 0.66 eV.

E47-19 E = h=2t, so E = (4:141015 eV s)=2(1108s) = 6:6108eV:

E47-20 (a) According to electrostatics and uniform circular motion, mv2=r = e2=40r2;

or s s e2 e4 e2 v = == = : 40mr 42h2n2 20hn 0

Putting in the numbers, (1:61019C)2 2:18106m=s v= = : 2(8:851012F=m)(6:631034J s)n n

(b) = h=mv, = (6:631034J s)=(9:111031kg)(2:18106m=s) = 3:341010m: (c) =a0 = (3:341010m)=(5:291011) = 6:31 2: Actually, it is exactly 2.

In order to have an inelastic collision with the 6.0 eV neutron there must exist a transition with an energy di erence of less than 6.0 eV. For a hydrogen atom in the ground state E1 = 13:6 eV the nearest state is E2 = (13:6 eV)=(2)2 = 3:4 eV: Since the di erence is 10.2 eV, it will not be possible for the 6.0 eV neutron to have an inelastic collision with a ground state hydrogen atom.

E47-21 E47-22 (a) The atom is originally in the state n given by p n = (13:6 eV)=(0:85 eV) = 4: The state with an excitation energy of 10:2 eV, is p n = (13:6 eV)=(13:6 eV 10:2 eV) = 2: The transition energy is then E = (13:6 eV)(1=22 1=42) = 2:55 eV:

E47-23 According to electrostatics and uniform circular motion, 0

or s s e2 e4 e2 v = == = : 40mr 42h2n2 20hn 0

The de Broglie wavelength is then h 20hn = = : mv me2 The ratio of =r is 20hn r me2a0n2 where k is a constant. As n ! 1 the ratio goes to zero.

E47-24 In Exercise 47-21 we show that the hydrogen levels can be written as En = (13:6 eV)=n2: The transition energy is E = E4 E1 = (13:6 eV)(1=12 1=42) = 12:8 eV: The momentum of the emitted photon is p = E=c = (12:8 eV)=c: This is the momentum of the recoiling hydrogen atom, which then has velocity p pc (12:8 eV) v = = c = (3:00108m=s) = 4:1 m=s: m mc2 (932 MeV)

The rst Lyman line is the n = 1 to n = 2 transition. The second Lyman line is the n = 1 to n = 3 transition. The rst Balmer line is the n = 2 to n = 3 transition. Since the photon frequency is proportional to the photon energy (E = hf ) and the photon energy is the energy di erence between the two levels, we have E47-25

Em En fn!m = h where the En is the hydrogen atom energy level. Then E3 E1 h E3 E2 + E2 E1 E3 E2 E2 E1 hhh = f2!3 + f1!2:

E47-26 Use En = Z2(13:6 eV)=n : 2 (a) The ionization energy of the ground state of He+ is E = (2)2(13:6 eV)=(1)2 = 54:4 eV: n

(b) The ionization energy of the n = 3 state of Li2+ is E = (3)2(13:6 eV)=(3)2 = 13:6 eV: n

(a) The energy levels in the He+ spectrum are given by where Z = 2, as is discussed in Sample Problem 47-6. The photon wavelengths for the n = 4 series are then hc hc=E4 En E4 1 En=E4 E47-27

which can also be written as 16hc=(54:4 eV) 1 16=n2 16hcn2=(54:4 eV) n2 16 Cn2 n2 16 (b) The wavelength of the rst line is the transition from n = 5, (365 nm)(5)2 = = 1014 nm: (5)2 (4)2 The series limit is the transition from n = 1, so = 365 nm: (c) The series starts in the infrared (1014 nm), and ends in the ultraviolet (365 nm). So it must also include some visible lines.

E47-28 We answer these questions out of order! (f) According to electrostatics and uniform circular motion, 0

or s s e2 e4 e2 v = == = : 40mr 42h2n2 20hn 0

Putting in the numbers, (1:61019C)2 v = = 2:18106m=s: 2(8:851012F=m)(6:631034J s)(1) (d) p = (9:111031kg)(2:18106m=s) = 1:991024kg m=s: (e) ! = v=r = (2:18106m=s)=(5:291011m) = 4:121016rad/s: (c) l = pr = (1:991024kg m=s)(5:291011m) = 1:051034J s: (g) F = mv2=r, so F = (9:111031kg)(2:18106m=s)2=(5:291011m) = 8:18108N: (h) a = (8:18108N)=(9:111031kg) = 8:981022m=s2: (i) K = mv2=r, or (9:111031kg)(2:18106m=s)2 K = = 2:161018J = 13:6 eV: 2

p E47-30 (a) Using the results of Exercise 45-1, (1240 eV nm) E1 = = 1:24105 eV: (0:010 nm) (b) Using the results of Problem 45-11, E2 (1:24105 eV)2 Kmax = = = 40:5104 eV: mc2=2 + E (5:11105 eV)=2 + (1:24105 eV) (c) This would likely knock the electron way out of the atom.

E47-31 The energy of the photon in the series limit is given by Elimit = (13:6 eV)=n2;

where n = 1 for Lyman, n = 2 for Balmer, and n = 3 for Paschen. The wavelength of the photon is (1240 eV nm) limit = n2 = (91:17 nm)n : 2 (13:6 eV) The energy of the longest wavelength comes from the transition from the nearest level, or (13:6 eV) (13:6 eV) 2n + 1 Elong = = (13:6 eV) : (n + 1)2 n2 [n(n + 1)]2 The wavelength of the photon is (1240 eV nm)[n(n + 1)]2 [n(n + 1)]2 long = = (91:17 nm) : (13:6 eV)n2 2n + 1 (a) The wavelength interval long limit, or n2(n + 1)2 n2(2n + 1) n4 = (91:17 nm) = (91:17 nm) : 2n + 1 2n + 1 For n = 1, = 30:4 nm. For n = 2, = 292 nm. For n = 3, = 1055 nm: (b) The frequency interval is found from Elimit Elong (13:6 eV) 1 (3:291015=s) f = = = : h (4:141015eV s) (n + 1)2 (n + 1)2 For n = 1, f = 8:231014Hz: For n = 2, f = 3:661014Hz: For n = 3, f = 2:051014Hz:

E47-32 E47-33 (a) We’ll use Eqs. 47-25 and 47-26. At r = 0 11 a3 a3 00 while P (0) = 4(0)2 2(0) = 0: (b) At r = a0 we have e2 0 a3 0

and P (a0) = 4(a0)2 2(a0) = 10:2 nm1: E47-34 Assume that (a0) is a reasonable estimate for (r) everywhere inside the small sphere. Then e2 0:1353 2= = : a3 a3 00 The probability of nding it in a sphere of radius 0:05a0 is 0:05a0 (0:1353)4r2 dr 4 Z = (0:1353)(0:05)3 = 2:26105: 0 a3 3 0

E47-35 Using Eq. 47-26 the ratio of the probabilities is P (a ) (a )2e2(a0)=a0 e2 00 = = = 1:85: P (2a ) (2a )2e2(2a0)=a0 4e4 00

E47-36 The probability is 1:016a0 2e2r=a Z 4r 0 a3 a0 0 Z 2:032 1 22 = 0:00866:

(b) From Eq. 47-31, = arccos(ml=pl(l + 1)), so can equal 90, 73:2, 54:7, or 30:0. (c) The magnitude of L~ is given by Eq. 47-28, p hp L = l(l + 1) = 3h=: 2 E47-38 The maximum possible value of ml is 5. Apply Eq. 47-31: (5) = arccos p = 24:1: (5)(5 + 1) E47-37

h 2 rh r 2 x h r 2 h L = : 2 E47-40 Note that there is a typo in the formula; P (r) must have dimensions of one over length. The probability is Z 1 4er=a r0 0 24a5 0 Z1 1 24 0 = 1:00 What does it mean? It means that if we look for the electron, we will nd it somewhere.

dP r(2a r)(4a2 6ra + r2) = er = 0: dr 8a6 0

The solutions are r = 0, r = 2a, and 4a2 6ra + r2 = 0. The rst two correspond to minima (see Fig. 47-14). The other two are the solutions to the quadratic, or r = 0:764a0 and r = 5:236a0. (b) Substitute these two values into Eq. 47-36. The results are P (0:764a0) = 0:981 nm1:

and P (5:236a0) = 3:61 nm1: E47-42 The probability is 5:01a0 2(2 r=a0)2er=a Z r0 8a3 0 5:00a0 = 0:01896:

E47-43 n = 4 and l = 3, while ml can be any of E47-44 n must be greater than l, so n 4. jmlj must be less than or equal to l, so jmlj 3. ms is 1=2 or 1=2.

E47-47 Each is in the n = 1 shell, the l = 0 angular momentum state, and the ml = 0 state. But one is in the state ms = +1=2 while the other is in the state ms = 1=2.

E47-48 Apply Eq. 47-31: (+1=2) = arccos p = 54:7 (1=2)(1=2 + 1) and (1=2) = arccos p = 125:3: (1=2)(1=2 + 1)

E47-50 There are n possible values for l (start at 0!). For each value of l there are 2l + 1 possible values for ml. If n = 1, the sum is 1. If n = 2, the sum is 1+3 = 4. If n = 3, the sum is 1+3+5 = 9. The pattern is clear, the sum is n2. But there are two spin states, so the number of states is 2n2.

P47-1 We can simplify the energy expression as h2 E = E0 n2x + n2 + n2 y z where E0 = : 8mL2 To nd the lowest energy levels we need to focus on the values of nx, ny, and nz. It doesn’t take much imagination to realize that the set (1; 1; 1) will result in the smallest value for n2 + n2 + n2. The next choice is to set one of the values equal to 2, and try the set (2; 1; 1). xyz Then it starts to get harder, as the next lowest might be either (2; 2; 1) or (3; 1; 1). The only way to nd out is to try. I’ll tabulate the results for you: nx ny nz n2 + n2 + n2 xyz Mult. nx ny nz n2 + n2 + n2 111 211 221 311 222 3 6 9 11 12 1 3 3 3 1 321 322 411 331 421 14 17 18 19 21 6 3 3 3 6

We are now in a position to state the ve lowest energy levels. The fundamental quantity is (hc)2 (1240 eV nm)2 E0 = = 8mc2L2 8(0:511106 The ve lowest levels are found by multiplying table above.

probability distribution function P (x) = 2 L = 6:02106 eV: eV)(250 nm)2 this fundamental quantity by the numbers in the

we nd the probability of nding the particle in the region 0 x L=3 is Z L=3 2 nx 0LL 1 sin(2n=3) = 1 : 3 2n=3 (e) Classically the probability distribution function is uniform, so there is a 1/3 chance of nding it in the region 0 to L/3.

P47-3 The region of interest is small compared to the variation in P (x); as such we can approxi- mate the probability with the expression P = P (x)x: (b) Evaluating, 2 4x LL 2 4(L=8) LL = 0:0006: (b) Evaluating, 2 4x LL 2 4(3L=16) LL = 0:0003:

P47-4 (a) P = , or P = A2e2m!x2=h: 0 (b) Integrating, Z1 0 1 Z1 r 0 2m! 1 r hp 0 2m! r 4 2m! = A0: h

Doing the math one derivative at a time, 2 d dd dx2 dx dx d (2m!x=h)em!x2=h

dx = (2m!=h) (2m!x=h)2 0: = (2m!=h)

In the last line we factored out 0. This will make our lives easier later on. Now we want to go to Schrodinger’s equation, and make some substitutions.

h2 d2 82m dx2 h2 82m h2 82m where in the last line we divided through by 0. Now for some algebra,

h2 U = E + (2m!x=h)2 (2m!=h) 82m m!2×2 h! = E+ : 2 4 But we are given that E = h!=4, so this simpli es to m!2×2 U= 2 which looks like a harmonic oscillator type potential.

P47-6 Assume the electron is originally in the state n. The classical frequency of the electron is f0, where f0 = v=2r: According to electrostatics and uniform circular motion, mv2=r = e2=40r2;

or s s e2 e4 e2 v = == = : 40mr 42h2n2 20hn 0

Then e2 1 me2 me4 2E1 f0 = = = 20hn 2 0h2n2 42h3n3 hn3 0

Photon frequency is related to energy according to f = Enm=h, where Enm is the energy of transition from state n down to state m. Then

where E1 = 13:6 eV. Combining the fractions and letting m = n , where is an integer, E1 m2 n2 h m2n2 E1 (n m)(m + n) h m2n2 E1 (2n + ) h (n + )2n2 E1 (2n) h (n)2n2 2E1 = = f0: hn3 P47-7 We need to use the reduced mass of the muon since the muon and proton masses are so close together. Then (207)(1836) m = me = 186me: (207) + (1836) (a) Apply Eq. 47-20 1/2: a = a0=(186) = (52:9 pm)=(186) = 0:284 pm: (b) Apply Eq. 47-21: E = E1(186) = (13:6 eV)(186) = 2:53 keV: (c) = (1240 keV pm)=(2:53 pm) = 490 pm:

P47-8 (a) The reduced mass of the electron is (1)(1) m = me = 0:5me: (1) + (1) The spectrum is similar, except for this additional factor of 1/2; hence pos = 2H: (b) apos = a0=(186) = (52:9 pm)=(1=2) = 105:8 pm. This is the distance between the particles, but they are both revolving about the center of mass. The radius is then half this quantity, or 52:9 pm.

This problem isn’t really that much of a problem. Start with the magnitude of a vector in terms of the components, xyz P47-9

and then rearrange, L2 + L2 = L2 L2: xy z According to Eq. 47-28 L2 = l(l + 1)h2=42, while according to Eq. 47-30 Lz = mlh=2. Substitute that into the equation, and h2 L2 + L2 = l(l + 1)h2=42 m2h2=42 = l(l + 1) m2 : x y l l 42 Take the square root of both sides of this expression, and we are done.

The maximum value for ml is l, while the minimum value is 0. Consequently, q q p xy l

and qp q L2 + L2 = l(l + 1) m2 h=2 l h=2: xy l

P47-10 Assume that (0) is a reasonable estimate for (r) everywhere inside the small sphere. Then e0 1 2= = : a3 a3 00 The probability of nding it in a sphere of radius 1:11015m is Z 1:11015m 4r2 dr 4 (1:11015m)3 = 1:21014: = 0 a3 3 (5:291011m)3 0

P47-11 Assume that (0) is a reasonable estimate for (r) everywhere inside the small sphere. Then (2)2e0 1 2= = : 32a3 8a3 00 The probability of nding it in a sphere of radius 1:11015m is Z 1:11015m 2 (1:11015 3 4r dr 1 m) = = 1:51015: 0 8a3 6 (5:291011m)3 0

P47-12 (a) The wave function squared is e2r=a0 2= a3 0 The probability of nding it in a sphere of radius r = xa0 is Z xa0 4r2e2r=a dr 0

0 a3 0 Zx 0 = 1 e2x(1 + 2x + 2×2): (b) Let x = 1, then P = 1 e2(5) = 0:323: P47-13 We want to evaluate the di erence between the values of P at x = 2 and x = 2. Then P (2) P (1) = 1 e4(1 + 2(2) + 2(2)2) 1 e2(1 + 2(1) + 2(1)2) ;

P47-14 Using the results of Problem 47-12, or e2x = 1 + 2x + 2×2:

P47-15 The probability of nding it in a sphere of radius r = xa0 is xa0 r2(2 r=a )2er=a0 dr Z 0 P= 0 8a3 0 Zx 1 = x2(2 x)2ex dx 80 = 1 ex(y4=8 + y2=2 + y + 1): The minimum occurs at x = 2, so P = 1 e2(2 + 2 + 2 + 1) = 0:0527:

The highest energy x-ray photon will have an energy equal to the bombarding electrons, as is shown in Eq. 48-1, hc min = eV E48-1

Insert the appropriate values into the above expression, (4:14 1015 eV s)(3:00 108 m=s) 1240 109 eV m min = = : eV eV The expression is then 1240 109 V m 1240 kV pm min = = : VV So long as we are certain that the \V ” will be measured in units of kilovolts, we can write this as min = 1240 pm=V: E48-2 f = c= = (3:00108m=s)=(31:11012m) = 9:6461018=s. Planck’s constant is then E (40:0 keV) h = = = 4:141015 eV s: f (9:6461018=s) E48-3 Applying the results of Exercise 48-1, (1240kV pm) V = = 9:84 kV: (126 pm)

E48-4 (a) Applying the results of Exercise 48-1, (1240kV pm) min = = 35:4 pm: (35:0 kV) (b) Applying the results of Exercise 45-1, (1240keV pm) K = = 49:6 pm: (25:51 keV) (0:53 keV) (c) Applying the results of Exercise 45-1, (1240keV pm) K = = 56:5 pm: (25:51 keV) (3:56 keV) E48-5 (a) Changing the accelerating potential of the x-ray tube will decrease min. The new value will be (using the results of Exercise 48-1) min = 1240 pm=(50:0) = 24:8 pm: (b) K doesn’t change. It is a property of the atom, not a property of the accelerating potential of the x-ray tube. The only way in which the accelerating potential might make a di erence is if (c) K doesn’t change. See part (b).

E48-6 (a) Applying the results of Exercise 45-1, (1240keV pm) E = = 64:2 keV: (19:3 pm) (b) This is the transition n = 2 to n = 1, so E = (13:6 eV)(1=12 1=22) = 10:2 eV:

E48-7 Applying the results of Exercise 45-1, (1240keV pm) E = = 19:8 keV: (62:5 pm) and (1240keV pm) E = = 17:6 keV: (70:5 pm) The di erence is E = (19:8 keV) (17:6 keV) = 2:2 eV:

E48-8 Since E = hf = hc=, and = h=mc = hc=mc2, then E = hc= = mc2:

or V = E=e = mc2=e = 511 kV: E48-9 The 50.0 keV electron makes a collision and loses half of its energy to a photon, then the photon has an energy of 25.0 keV. The electron is now a 25.0 keV electron, and on the next collision again loses loses half of its energy to a photon, then this photon has an energy of 12.5 keV. On the third collision the electron loses the remaining energy, so this photon has an energy of 12.5 keV. The wavelengths of these photons will be given by (1240 keV pm) E

E48-10 (a) The x-ray will need to knock free a K shell electron, so it must have an energy of at (b) Applying the results of Exercise 48-1, (1240kV pm) min = = 17:8 pm: (69:5 kV) (c) Applying the results of Exercise 45-1, (1240keV pm) K = = 18:5 pm: (69:5 keV) (2:3 keV) Applying the results of Exercise 45-1, (1240keV pm) K = = 21:3 pm: (69:5 keV) (11:3 keV)

E48-11 (a) Applying the results of Exercise 45-1, (1240keV pm) EK = = 19:7 keV: (63 pm) Again applying the results of Exercise 45-1, (1240keV pm) EK = = 17:5 keV: (71 pm) (b) Zr or Nb; the others will not signi cantly absorb either line.

E48-12 Applying the results of Exercise 45-1, (1240keV pm) K = = 154:5 pm: (8:979 keV) (0:951 keV) Applying the Bragg re ection relationship, (154:5 pm) d = = = 282 pm: 2 sin 2 sin(15:9)

p E48-13 Plot the data. The plot should look just like Fig 48-4. Note that the vertical axis is f , pp which is related to the wavelength according to f = c=:

E48-14 Remember that the m in Eq. 48-4 refers to the electron, not the nucleus. This means pp that the constant C in Eq. 48-5 is the same for all elements. Since f = c=, we have 2 1 Z2 1 =: 2 Z1 1 For Ga and Nb the wavelength ratio is then 2 Nb (31) 1 = = 0:5625: Ga (41) 1

(a) The ground state question is fairly easy. The n = 1 shell is completely occupied by the rst two electrons. So the third electron will be in the n = 2 state. The lowest energy angular momentum state in any shell is the s sub-shell, corresponding to l = 0. There is only one choice for ml in this case: ml = 0. There is no way at this level of coverage to distinguish between the energy of either the spin up or spin down con guration, so we’ll arbitrarily pick spin up. (b) Determining the con guration for the rst excited state will require some thought. We could assume that one of the K shell electrons (n = 1) is promoted to the L shell (n = 2). Or we could assume that the L shell electron is promoted to the M shell. Or we could assume that the L shell electron remains in the L shell, but that the angular momentum value is changed to l = 1. The question that we would need to answer is which of these possibilities has the lowest energy. The answer is the last choice: increasing the l value results in a small increase in the energy of multi-electron atoms.

E48-15 E48-16 Refer to Sample Problem 47-6: a0(1)2 (5:291011m) r1 = = = 5:751013 m: Z (92)

E48-17 We will assume that the ordering of the energy of the shells and sub-shells is the same. That ordering is 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p < 8s: If there is no spin the s sub-shell would hold 1 electron, the p sub-shell would hold 3, the d sub-shell 5, and the f sub-shell 7. inert gases occur when a p sub-shell has lled, so the rst three inert gases would be element 1 (Hydrogen), element 1 + 1 + 3 = 5 (Boron), and element 1 + 1 + 3 + 1 + 3 = 9 Is there a pattern? Yes. The new inert gases have half of the atomic number of the original inert gases. The factor of one-half comes about because there are no longer two spin states for each set We can save time and simply divide the atomic numbers of the remaining inert gases in half: element 18 (Argon), element 27 (Cobalt), element 43 (Technetium), element 59 (Praseodymium).

E48-18 The pattern is or 2(12 + 22 + 22 + 33 + 33 + 42 + 42 + x2) The unknown is probably x = 5, the next noble element is probably 118 + 2 52 = 168:

E48-19 (a) Apply Eq. 47-23, which can be written as (13:6 eV)Z2 En = : n2 For the valence electron of sodium n = 3, s (5:14 eV)(3)2 (13:6 eV) while for the valence electron of potassium n = 4, s (4:34 eV)(4)2 (13:6 eV) (b) The ratios with the actual values of Z are 0.167 and 0.119, respectively.

E48-20 (a) There are three ml states allowed, and two ms states. The rst electron can be in any one of these six combinations of M1 and m2. The second electron, given no exclusion principle, could also be in any one of these six states. The total is 36. Unfortunately, this is wrong, because we can’t distinguish electrons. Of this total of 36, six involve the electrons being in the same state, while 30 involve the electron being in di erent states. But if the electrons are in di erent states, then they could be swapped, and we won’t know, so we must divide this number by two. The total number of distinguishable states is then (30=2) + 6 = 21: (b) Six. See the above discussion.

E48-21 (a) The Bohr orbits are circular orbits of radius rn = a0n2 (Eq. 47-20). The electron is orbiting where the force is e2 40r2 n

and this force is equal to the centripetal force, so mv2 e2 =: rn 40r2 n

where v is the velocity of the electron. Rearranging, s e2 v= : 40mrn

The time it takes for the electron to make one orbit can be used to calculate the current, s q e e e2 i= = = : t 2rn=v 2rn 40mrn

The magnetic moment of a current loop is the current times the area of the loop, so s e e2 2rn 40mrn n

which can be simpli ed to s e e2 = rn: 2 40mrn

But rn = a0n2, so s e a0e2 =n : 2 40m This might not look right, but a0 = 0h2=me2, so the expression can simplify to r 2 e h eh = n = n = nB: 2 42m2 4m (b) In reality the magnetic moments depend on the angular momentum quantum number, not the principle quantum number. Although the Bohr theory correctly predicts the magnitudes, it does not correctly predict when these values would occur.

E48-22 (a) Apply Eq. 48-14: dBz 24 3T=m) = 1:51025 Fz = z = (9:2710 J=T)(1610 N: dz (b) a = F=m, z = at2=2, and t = y=vy. Then F y2 (1:51025 N)(0:82 m)2 z = = = 3:2105m: 2mv2 2(1:671027kg)(970 m=s)2 y

E48-23 a = (9:271024J=T)(1:4103T=m)=(1:71025kg) = 7:6104m=s2:

E48-24 (a) U = 2B, or U = 2(5:79105eV=T)(0:520 T) = 6:02105eV: (b) f = E=h = (6:02105eV)(4:141015eV s) = 1:451010Hz: (c) = c=f = (3108m=s)=(1:451010Hz) = 2:07102m:

The energy change can be derived from Eq. 48-13; we multiply by a factor of 2 because the spin is completely ipped. Then E48-25

E = 2zBz = 2(9:271024 J=T)(0:190 T) = 3:521024J: The corresponding wavelength is hc (6:631034J s)(3:00108m=s) = = = 5:65102m: E (3:521024J) This is somewhere near the microwave range.

E48-26 The photon has an energy E = hc=. This energy is related to the magnetic eld in the vicinity of the electron according to E = 2B;

so hc (1240 eV nm) B = = = 0:051 T: 2 2(5:79105J=T)(21107nm) E48-27 Applying the results of Exercise 45-1, (1240 eV nm) E = = 1:55 eV: (800nm) The production rate is then (5:0103W) R = = 2:01016=s: (1:55 eV)(1:61019J=eV)

E48-28 (a) x = (3108m=s)(121012s) = 3:6103m: (b) Applying the results of Exercise 45-1, (1240 eV nm) E = = 1:786 eV: (694:4nm) The number of photons in the pulse is then N = (0:150J)=(1:786 eV)(1:61019J=eV) = 5:251017: E48-29 We need to nd out how many 10 MHz wide signals can t between the two wavelengths. The lower frequency is c (3:00 108 m=s) f1 = = = 4:29 1014 Hz: 1 700 109 m)

The higher frequency is c (3:00 108 m=s) f = = = 7:50 1014 Hz: 1 1 400 109 m) The number of signals that can be sent in this range is f2 f1 (7:50 1014 Hz) (4:29 10 Hz) 14 = = 3:21 107: (10 MHz) (10 106 Hz) That’s quite a number of television channels.

E48-30 Applying the results of Exercise 45-1, (1240 eV nm) E = = 1:960 eV: (632:8nm) The number of photons emitted in one minute is then (2:3103W)(60 s) N = = 4:41017: (1:960 eV)(1:61019J=eV) E48-31 Apply Eq. 48-19. E13 E11 = 2(1:2 eV):. The ratio is then n13 (2:4 eV)=(8:62105eV/K)(2000 K) = e = 9107: n11 E48-32 (a) Population inversion means that the higher energy state is more populated; this can only happen if the ratio in Eq. 48-19 is greater than one, which can only happen if the argument of (b) If n2 = 1:1n1 then the ratio is 1:1, so (2:26 eV T = = 2:75105K: (8:62105eV/K) ln(1:1) E48-33 (a) At thermal equilibrium the population ratio is given by N eE2=kT 2 E=kT = =e : N eE1=kT 1

But E can be written in terms of the transition photon wavelength, so this expression becomes N2 = N1ehc=kT : Putting in the numbers,

N2 = (4:01020)e(1240 eVnm)=( eV=K)(300 K)) = 6:621016 582 nm)(8:62105 :

(b) If the population of the upper state were 7:01020, then in a single laser pulse hc (6:631034J s)(3:00108m=s) E = N = (7:01020) = 240 J: (582109m)

E48-34 The allowed wavelength in a standing wave chamber are n = 2L=n. For large n we can write 2L 2L 2L n+1 = : n + 1 n n2 The wavelength di erence is then 2L 2 n2 2L which in this case is (533109m)2 = = 1:71012m: 2(8:3102m) E48-35 (a) The central disk will have an angle as measured from the center given by d sin = (1:22);

and since the parallel rays of the laser are focused on the screen in a distance f , we also have R=f = sin . Combining, and rearranging, 1:22f R= : d (c) I = P=A = (5:21 W)=(1:5 mm)2 = 7:37105W=m2: (d) I = P=A = (0:84)(5:21 W)=(7:2 m)2 = 2:71010W=m2:

E48-36 P48-1 Let 1 be the wavelength of the rst photon. Then 2 = 1 + 130 pm. The total energy transfered to the two photons is then hc hc E1 + E2 = + = 20:0 keV: 1 2 We can solve this for 1, 20:0 keV 1 1 hc 1 1 + 130 pm 21 + 130 pm 1(1 + 130 pm)

which can also be written as 2 + (6 pm)1 (8060 pm2) = 0: 1

This equation has solutions 1 = 86:8 pm and 92:8 pm: Only the positive answer has physical meaning. The energy of this rst photon is then (1240 keV pm) E1 = = 14:3 keV: (86:8 pm) (a) After this rst photon is emitted the electron still has a kinetic energy of 20:0 keV 14:3 keV = 5:7 keV: (b) We found the energy and wavelength of the rst photon above. The energy of the second photon must be 5.7 keV, with wavelength 2 = (86:8 pm) + 130 pm = 217 pm:

P48-2 Originally, 1 = p = 2:412: 1 (2:73108m=s)2=(3108m=s)2 The energy of the electron is E0 = mc2 = (2:412)(511 keV) = 1232 keV: Upon emitting the photon the new energy is E = (1232 keV) (43:8 keV) = 1189 keV;

so the new gamma factor is and the new speed is p v = c 1 1=(2:326)2 = (0:903)c:

P48-3 Switch to a reference frame where the electron is originally at rest. Momentum conservation requires while energy conservation requires mc2 = E + Ee: Rearrange to Ee = mc2 E: Square both sides of this energy expression and ee e p2 c2 2E mc2 = p2c2: e But the momentum expression can be used here, and the result is 2Emc2 = 0: Not likely.

P48-4 (a) In the Bohr theory we can assume that the K shell electrons \see” a nucleus with charge Z. The L shell electrons, however, are shielded by the one electron in the K shell and so they \see” a nucleus with charge Z 1. Finally, the M shell electrons are shielded by the one electron in the K shell and the eight electrons in the K shell, so they \see” a nucleus with charge Z 9. The transition wavelengths are then 1 E E0(Z 1)2 1 1

hc hc 22 12 E0(Z 1)2 3 =: hc 4 and 1 E E0 1 1 hc hc 32 12 E0(Z 9)2 8 =: hc 9

The ratio of these two wavelengths is 27 (Z 1)2 =: 32 (Z 9)2 Note that the formula in the text has the square in the wrong place!

P48-5 (a) E = hc=; the energy di erence is then 11 1 2

2 1 = hc : 21 hc = : 21 Since 1 and 2 are so close together we can treat the product 12 as being either 2 or . Then 2 12 (1240 eV nm) E = (0:597 nm) = 2:1103eV: (589 nm)2 (b) The same energy di erence exists in the 4s ! 3p doublet, so (1139 nm)2 = (2:1103eV) = 2:2 nm: (1240 eV nm) P48-6 (a) We can assume that the K shell electron \sees” a nucleus of charge Z 1, since the other electron in the shell screens it. Then, according to the derivation leading to Eq. 47-22, rK = a0=(Z 1):

(b) The outermost electron \sees” a nucleus screened by all of the other electrons; as such Z = 1, and the radius is r = a0

We assume in this crude model that one electron moves in a circular orbit attracted to the helium nucleus but repelled from the other electron. Look back to Sample Problem 47-6; we need to use some of the results from that Sample Problem to solve this problem. The factor of e2 in Eq. 47-20 (the expression for the Bohr radius) and the factor of (e2)2 in Eq. 47-21 (the expression for the Bohr energy levels) was from the Coulomb force between the single electron and the single proton in the nucleus. This force is e2 F= : 40r2 In our approximation the force of attraction between the one electron and the helium nucleus is 2e2 F1 = : 40r2 The factor of two is because there are two protons in the helium nucleus. There is also a repulsive force between the one electron and the other electron, e2 40(2r)2 P48-7

where the factor of 2r is because the two electrons are on opposite side of the nucleus. The net force on the rst electron in our approximation is then 2e2 e2 40r2 40(2r)2 which can be rearranged to yield e2 e2 17 F net = 2 = : 40r2 4 40r2 4

It is apparent that we need to substitute 7e2=4 for every occurrence of e2. (a) The ground state radius of the helium atom will then be given by Eq. 47-20 with the appropriate substitution, 0h2 4 r = = a0: m(7e2=4) 7 (b) The energy of one electron in this ground state is given by Eq. 47-21 with the substitution of 7e2=4 for every occurrence of e2, then m(7e2=4)2 49 me4 E= = : 84h2 16 84h2 00

One last thing. There are two electrons, so we need to double the above expression. The ground state energy of a helium atom in this approximation is 49 E0 = 2 (13:6 eV) = 83:3eV: 16 (c) Removing one electron will allow the remaining electron to move closer to the nucleus. The energy of the remaining electron is given by the Bohr theory for He+, and is EHe+ = (4)(13:60 eV) = 54:4 eV;

P48-8 Applying Eq. 48-19: (3:2 eV) T = = 1:0104K: (8:62105eV=K) ln(6:11013=2:51015)

P48-9 sin r=R, where r is the radius of the beam on the moon and R is the distance to the moon. Then 1:22(600109m)(3:82108m) r = = 2360 m: (0:118 m) The beam diameter is twice this, or 4740 m.

P48-10 (a) N = 2L=n, or 2(6102m)(1:75) N = = 3:03105: (694109)

(b) N = 2nLf =c, so c (3108m=s) f = = = 1:43109=s: 2nL 2(1:75)(6102m) Note that the travel time to and fro is t = 2nL=c! (c) f =f is then f (694109) = = = 3:3106: f 2nL 2(1:75)(6102m)

E49-1 (a) Equation 49-2 is pp 8 2m3=2 8 2(mc2)3=2 n(E) = E1=2 = E1=2: h3 (hc)3 We can evaluate this by substituting in all known quantities, p 8 2(0:511 106 eV)3=2 n(E) = E1=2 = (6:81 1027 m3 eV3=2)E1=2: (1240 109 eV m)3 Once again, we simpli ed the expression by writing hc wherever we could, and then using hc = (b) Then, if E = 5:00 eV, n(E) = (6:81 1027 m3 eV3=2)(5:00 eV)1=2 = 1:52 1028 m3 eV : 1

E49-2 Apply the results of Ex. 49-1: n(E) = (6:81 1027 m3 eV3=2)(8:00 eV)1=2 = 1:93 1028 m3 eV1:

E49-3 Monovalent means only one electron is available as a conducting electron. Hence we need only calculate the density of atoms: N N (19:3103kg=m3)(6:021023mol1) = A = = 5:901028=m3: V Ar (0:197 kg=mol) E49-4 Use the ideal gas law: pV = N kT . Then N p = kT = (8:491028m3)(1:381023J= K)(297 K) = 3:48108Pa: V E49-5 (a) The approximate volume of a single sodium atom is (0:023 kg=mol) V1 = = 3:931029m3: (6:021023part/mol)(971 kg=m3) The volume of the sodium ion sphere is 4 V2 = (981012 m)3 = 3:941030 m3: 3 The fractional volume available for conduction electrons is V V (3:931029m3) (3:941030 m3) 12 = = 90%: V (3:931029m3) 1

(b) The approximate volume of a single copper atom is (0:0635 kg=mol) V1 = = 1:181029m3: (6:021023part/mol)(8960 kg=m3) The volume of the copper ion sphere is 4 V = (961012 m)3 = 3:711030 m3: 2 3 The fractional volume available for conduction electrons is V V (1:181029m3) (3:711030 m3) 12 = = 69%: V (1:181029m3) 1

E49-6 (a) Apply Eq. 49-6: hi (0:0730 eV)=(8:62105eV=K)(0 K) p = 1= e + 1 = 0: (b) Apply Eq. 49-6: hi (0:0730 eV)=(8:62105eV=K)(320 K) 2 p = 1= e + 1 = 6:6210 :

E49-7 Apply Eq. 49-6, remembering to use the energy di erence: hi (1:1) eV)=(8:62105eV=K)(273 K) p = 1= e + 1 = 1:00;

hi (0:1) eV)=(8:62105eV=K)(273 K) hi (0:0) eV)=(8:62105eV=K)(273 K) p = 1= e + 1 = 0:5;

hi (0:1) eV)=(8:62105eV=K)(273 K) hi (1:1) eV)=(8:62105eV=K)(273 K) p = 1= e + 1 = 0:0: (b) Inverting the equation, E k ln(1=p 1) so (0:1 eV) T = = 700 K (8:62105eV=K) ln(1=(0:16) 1) E49-8 The energy di erences are equal, except for the sign. Then 11 e+E=kt + 1 eE=kt + 1 eE=2kt e+E=2kt e+E=2kt + eE=2kt eE=2kt + e+E=2kt eE=2kt + e+E=2kt = 1: eE=2kt + e+E=2kt E49-9 The Fermi energy is given by Eq. 49-5, h2 3n 2=3 8m where n is the density of conduction electrons. For gold we have (19:3 g/cm3)(6:021023part/mol) = 5:901022 3 3 n = elect./cm = 59 elect./nm (197 g/mol) The Fermi energy is then !2=3 (1240 eV nm)2 3(59 electrons/nm3) EF = = 5:53 eV: 8(0:511106 eV)

E49-10 Combine the results of Ex. 49-1 and Eq. 49-6: p CE no = : eE=kt + 1 Then for each of the energies we have (6:811027 m3 eV3=2)p(4 eV) no = = 1:361028=m3 eV; e(3:06 eV)=(8:62105eV=K)(1000 K) + 1 (6:811027 m3 eV3=2)p(6:75 eV) no = = 1:721028=m3 eV; e(0:31 eV)=(8:62105eV=K)(1000 K) + 1 (6:811027 m3 eV3=2)p(7 eV) no = = 9:021027=m3 eV; e(0:06 eV)=(8:62105eV=K)(1000 K) + 1 (6:811027 m3 eV3=2)p(7:25 eV) no = = 1:821027=m3 eV; e(0:19 eV)=(8:62105eV=K)(1000 K) + 1 (6:811027 m3 eV3=2)p(9 eV) no = = 3:431018=m3 eV: e(1:94 eV)=(8:62105eV=K)(1000 K) + 1

E49-11 Solve n2(hc)2 En = 8(mc2)L2 for n = 50, since there are two electrons in each level. Then (50)2(1240 eV nm)2 Ef = = 6:53104eV: 8(5:11105eV)(0:12 nm)2

E49-12 We need to be much higher than T E49-13 Equation 49-5 is EF =

and if we collect the constants, = (7:06 eV)=(8:62105eV=K) = 8:2104K:

h2 2=3 3n 8m h2 3 2=3 8m where, if we multiply the top and bottom by c2 (hc)2 3 2=3 (1240 109 eV m)2 2=3 3 A = = = 3:65 1019 m2 eV: 8mc2 8(0:511 106 eV) E49-14 (a) Inverting Eq. 49-6, so E = (8:62105eV=K)(1050 K) ln(1=(0:91) 1) = 0:209 eV: Then E = (0:209 eV) + (7:06 eV) = 6:85 eV: (b) Apply the results of Ex. 49-1: n(E) = (6:81 1027 m3 eV3=2)(6:85 eV)1=2 = 1:78 1028 m3 eV 1 :

(c) n = np = (1:78 1028 m3 eV1)(0:910) = 1:62 1028 m3 eV 1 o:

E49-15 Equation 49-5 is 2 2=3 h 3n 8m and if we rearrange, 3h3 16 2m3=2 Equation 49-2 is then p 8 2m3=2 3 n(E) = E1=2 = nE 3=2E1=2: F h3 2

E49-16 ph = 1 p, so 1 eE=kT + 1 eE=kT + 1 1 =: 1 + eE=kT

The steps to solve this exercise are equivalent to the steps for Exercise 49-9, except now the iron atoms each contribute 26 electrons and we have to nd the density. First, the density is m (1:991030kg) = = = 1:84109kg=m3 4r3=3 4(6:37106m)3=3 Then (26)(1:84106 g/cm3)(6:021023part/mol) n = = 5:11029 elect./cm3; (56 g/mol) = 5:1108 elect./nm3 The Fermi energy is then E49-17

!2=3 (1240 eV nm)2 3(5:1108 elect./nm3) EF = = 230 keV: 8(0:511106 eV)

E49-18 First, the density is m 2(1:991030kg) = = = 9:51017kg=m3 4r3=3 4(10103m)3=3 Then n = (9:51017kg=m3)=(1:671027kg) = 5:691044=m3:

The Fermi energy is then !2=3 (1240 MeV fm)2 3(5:69101/fm3) EF = = 137 MeV: 8(940 MeV)

E49-19 E49-20 (a) EF = 7:06 eV, so 3(8:62105eV K)(0 K) 2(7:06 eV) (b) f = 3(8:62105eV K)(300 K)=2(7:06 eV) = 0:0055: (c) f = 3(8:62105eV K)(1000 K)=2(7:06 eV) = 0:0183: E49-21 Using the results of Exercise 19, 2fEF 2(0:0130)(4:71 eV) T = = = 474 K: 3k 3(8:62105eV K)

E49-22 f = 3(8:62105eV K)(1235 K)=2(5:5 eV) = 0:029: E49-23 (a) Monovalent means only one electron is available as a conducting electron. Hence we need only calculate the density of atoms: N NA (10:5103kg=m3)(6:021023mol1) = = = 5:901028=m3: V Ar (0:107 kg=mol) (b) Using the results of Ex. 49-13, E = (3:65 1019 m2 eV)(5:901028=m3)2=3 = 5:5 eV: F p (c) v = 2K=m, or p v = 2(5:5 eV)(5:11105eV=c2) = 1:4108m=s: (d) = h=p, or (6:631034J s) = = 5:21012m: (9:111031kg)(1:4108m=s)

E49-24 (a) Bivalent means two electrons are available as a conducting electron. Hence we need to double the calculation of the density of atoms: N N 2(7:13103kg=m3)(6:021023mol1) = A = = 1:321029=m3: V Ar (0:065 kg=mol) (b) Using the results of Ex. 49-13, EF = (3:65 1019 m2 eV)(1:321029=m3)2=3 = 9:4 eV: p (c) v = 2K=m, or p v = 2(9:4 eV)(5:11105eV=c2) = 1:8108m=s: (d) = h=p, or (6:631034J s) = = 4:01012m: (9:111031kg)(1:8108m=s)

(a) Refer to Sample Problem 49-5 where we learn that the mean free path can be written in terms of Fermi speed vF and mean time between collisions as E49-25

= vF : The Fermi speed is p p 3 vF = c 2EF =mc2 = c 2(5:51 eV)=(5:11105 eV) = 4:6410 c:

The time between collisions is m (9:111031kg) = = = 3:741014s: ne2 (5:861028m3)(1:601019C)2(1:62108 m) We found n by looking up the answers from Exercise 49-23 in the back of the book. The mean free path is then = (4:64103)(3:00108m=s)(3:741014s) = 52 nm: (b) The spacing between the ion cores is approximated by the cube root of volume per atom. This atomic volume for silver is (108 g/mol) V = = 1:711023cm3: (6:021023part/mol)(10:5 g/cm3)

The distance between the ions is then p 3 l = V = 0:257 nm: The ratio is =l = 190:

E49-26 (a) For T = 1000 K we can use the approximation, so for diamond p = e(5:5 eV)=2(8:62105eV=K)(1000 K) = 1:41014 ;

while for silicon, p = e(1:1 eV)=2(8:6210 eV=K)(1000 K) = 1:710 5 3 ;

(b) For T = 4 K we can use the same approximation, but now E kT and the exponential function goes to zero.

E49-27 (a) E EF 0:67 eV=2 = 0:34 eV:. The probability the state is occupied is then hi (0:34) eV)=(8:62105eV=K)(290 K) 6 p = 1= e + 1 = 1:210 :

(b) E EF 0:67 eV=2 = 0:34 eV:. The probability the state is unoccupied is then 1 p, or hi (0:34) eV)=(8:62105eV=K)(290 K) 6 p = 1 1= e + 1 = 1:210 :

E49-28 (a) E EF 0:67 eV=2 = 0:34 eV:. The probability the state is occupied is then hi (0:34) eV)=(8:62105eV=K)(289 K) 6 p = 1= e + 1 = 1:210 :

E49-29 (a) The number of silicon atoms per unit volume is (6:021023part/mol)(2:33 g/cm3) n = = 4:991022 part./cm3: (28:1 g/mol) If one out of 1:0eex7 are replaced then there will be an additional charge carrier density of 4:991022 3=1:0107 = 4:991015 3 = 4:991021 3 part./cm part./cm m : (b) The ratio is (4:991021m3)=(2 1:51016m3) = 1:7105: The extra factor of two is because all of the charge carriers in silicon (holes and electrons) are charge carriers.

E49-30 Since one out of every 5106 silicon atoms needs to be replaced, then the mass of phos- phorus would be 1 30 m = = 2:1107 g: 5106 28

E49-31 l = p3 1=1022=m3 = 4:6108m: E49-32 The atom density of germanium is N N (5:32103kg=m3)(6:021023mol1) = A = = 1:631028=m3: V Ar (0:197 kg=mol) The atom density of the impurity is (1:631028=m3)=(1:3109) = 1:251019: The average spacing is l = p3 1=1:251019=m3 = 4:3107m: E49-33 The rst one is an insulator because the lower band is lled and band gap is so large; The second one is an extrinsic n-type semiconductor: it is a semiconductor because the lower band is lled and the band gap is small; it is extrinsic because there is an impurity; since the impurity The third sample is an intrinsic semiconductor: it is a semiconductor because the lower band is The fourth sample is a conductor; although the band gap is large, the lower band is not completely The fth sample is a conductor: the Fermi level is above the bottom of the upper band. The sixth one is an extrinsic p-type semiconductor: it is a semiconductor because the lower band is lled and the band gap is small; it is extrinsic because there is an impurity; since the impurity level is close to the bottom of the band gap the impurity is an acceptor.

E49-36 (a) A region with some potential di erence exists that has a gap between the charged (b) C = Q=V . Using the results in Sample Problem 49-9 for q and V , n0eAd=2 C = = 20A=d: n0ed2=40

E49-37 (a) Apply that ever so useful formula hc (1240 eV nm) = = = 225 nm: E (5:5 eV) Why is this a maximum? Because longer wavelengths would have lower energy, and so not enough (b) Ultraviolet.

E49-38 Apply that ever so useful formula hc (1240 eV nm) E = = = 4:20 eV: (295 nm) E49-39 The photon energy is hc (1240 eV nm) E = = = 8:86 eV: (140 nm) which is enough to excite the electrons through the band gap. As such, the photon will be absorbed, which means the crystal is opaque to this wavelength.

E49-40 P49-1 We can calculate the electron density from Eq. 49-5,

8mc2 3=2 EF 3 (hc)2 8(0:511106 3=2 eV)(11:66 eV) 3 (1240 eV nm)2 3 = 181 electrons/nm : From this we calculate the number of electrons per particle,

3 (181 electrons/nm )(27:0 g/mol) (2:70 g/cm3)(6:021023 particles/mol) which we can reasonably approximate as 3.

P49-2 At absolute zero all states below EF are lled, an none above. Using the results of Ex. 49-15, ZE 1F n0 3E Z F

20 32 25 3 = EF: 5 P49-3 (a) The total number of conduction electron is (0:0031 kg)(6:021023mol1) n = = 2:941022: (0:0635 kg=mol) The total energy is

3 E = (7:06 eV)(2:941022) = 1:241023eV = 2104J: 5 (b) This will light a 100 W bulb for t = (2104 J)=(100 W) = 200 s:

P49-4 (a) First do the easy part: nc = Ncp(Ec), so Nc : e(EcEF)=kT + 1

Then use the results of Ex. 49-16, and write Nv nv = Nv[1 p(Ev)] = : e(EvEF)=kT + 1

Since each electron in the conduction band must have left a hole in the valence band, then these two (b) If the exponentials dominate then we can drop the +1 in each denominator, and Nc Nv e(EcEF)=kT e(EvEF)=kT N c e(E 2E +E )=kT Nv 1 EF = (Ec+Ev+kTln(Nc=Nv)): 2

(a) We want to use Eq. 49-6; although we don’t know the Fermi energy, we do know the di erences between the energies in question. In the un-doped silicon E EF = 0:55 eV for the bottom of the conduction band. The quantity P49-5

Then 1 p = = 2:81010: e(0:55 eV)=(0:025 eV) + 1 In the doped silicon E EF = 0:084 eV for the bottom of the conduction band. Then

1 p = = 3:4102: e(0:084 eV)=(0:025 eV) + 1

(b) For the donor state E EF = 0:066 eV, so 1 p = = 0:93: e(0:066 eV)=(0:025 eV) + 1

P49-6 (a) Inverting Eq. 49-6, so E = (1:1 eV 0:11 eV) (8:62105 eV=K)(290 K) ln(1=(4:8105) 1) = 0:74 eV F

(b) E EF = (1:1 eV) (0:74 eV) = 0:36 eV, so 1 p = = 5:6107: e(0:36 eV)=(0:025 eV) + 1 P49-7 (a) Plot the graph with a spreadsheet. It should look like Fig. 49-12. (b) kT = 0:025 eV when T = 290 K. The ratio is then if e(0:5 eV)=(0:025 eV) + 1 = = 4:9108: ir e(0:5 eV)=(0:025 eV) + 1

E50-1 We want to follow the example set in Sample Problem 50-1. The distance of closest approach is given by qQ 40K (2)(29)(1:601019C)2 4(8:851012C2=Nm2)(5:30MeV)(1:60 1013J=MeV) = 1:571014 m:

E50-2 (a) The gold atom can be treated as a point particle: q1q2 40r2 (2)(79)(1:601019C)2 4(8:851012C2=Nm2)(0:16109m)2 = 1:4106N:

(b) W = F d, so (5:3106eV)(1:61019J=eV) d = = 6:06107m: (1:4106N) That’s 1900 gold atom diameters.

E50-3 Take an approach similar to Sample Problem 50-1: qQ 40d (2)(79)(1:601019C)2 4(8:851012C2=Nm2)(8:781015m)(1:60 1019J=eV) = 2:6107eV:

E50-5 We can make an estimate of the mass number A from Eq. 50-1,

where R0 = 1:2 fm. If the measurements indicate a radius of 3.6 fm we would have

A = (R=R0)3 = ((3:6 fm)=(1:2 fm))3 = 27: E50-6 E50-7 The mass number of the sun is A = (1:991030kg)=(1:671027kg) = 1:21057:

The radius would be p R = (1:21015 3 1:21057 = 1:3104 m) m:

E50-8 239Pu is composed of 94 protons and 239 94 = 145 neutrons. The combined mass of the free particles is M = Zmp + Nmn = (94)(1:007825 u) + (145)(1:008665 u) = 240:991975 u: The binding energy is the di erence EB = (240:991975 u 239:052156 u)(931:5 MeV/u) = 1806:9 MeV;

and the binding energy per nucleon is then E50-9 (1806:9 MeV)=(239) = 7:56 MeV:

62Ni is composed of 28 protons and 62 28 = 34 neutrons. The combined mass of the free particles is M = Zmp + Nmn = (28)(1:007825 u) + (34)(1:008665 u) = 62:513710 u: The binding energy is the di erence EB = (62:513710 u 61:928349 u)(931:5 MeV/u) = 545:3 MeV;

and the binding energy per nucleon is then (545:3 MeV)=(62) = 8:795 MeV:

E50-10 (a) Multiply each by 1=1:007825, so and 238U = 236:202500: m

E50-11 (a) Since the binding energy per nucleon is fairly constant, the energy must be proportional (b) Coulomb repulsion acts between pairs of protons; there are Z protons that can be chosen as rst in the pair, and Z 1 protons remaining that can make up the partner in the pair. That makes for Z(Z 1) pairs. The electrostatic energy must be proportional to this. (c) Z2 grows faster than A, which is roughly proportional to Z.

E50-12 Solve (0:7899)(23:985042) + x(24:985837) + (0:2101 x)(25:982593) = 24:305 for x. The result is x = 0:1001, and then the amount 26Mg is 0:1100.

The neutron con ned in a nucleus of radius R will have a position uncertainty on the order of x R. The momentum uncertainty will then be no less than hh p : 2x 2R Assuming that p p, we have h 2R E50-13

and then the neutron will have a (minimum) kinetic energy of p2 h2 E : 2m 82mR2 But R = R0A1=3, so (hc)2 E : 82mc2R2A2=3 0

For an atom with A = 100 we get (1240 MeV fm)2 E = 0:668 MeV: 82(940 MeV)(1:2 fm)2(100)2=3 This is about a factor of 5 or 10 less than the binding energy per nucleon.

E50-14 (a) To remove a proton, E = [(1:007825) + (3:016049) (4:002603)] (931:5 MeV) = 19:81 MeV: To remove a neutron, E = [(1:008665) + (2:014102) (3:016049)] (931:5 MeV) = 6:258 MeV: To remove a proton, E = [(1:007825) + (1:008665) (2:014102)] (931:5 MeV) = 2:224 MeV: (b) E = (19:81 + 6:258 + 2:224)MeV = 28:30 MeV: (c) (28:30 MeV)=4 = 7:07 MeV:

E50-15 (a) = [(1:007825) (1)](931:5 MeV) = 7:289 MeV: (b) = [(1:008665) (1)](931:5 MeV) = 8:071 MeV: (c) = [(119:902197) (120)](931:5 MeV) = 91:10 MeV: E50-16(a)EB=(ZmH+NmNm)c2.Substitutethedenitionformassexcess,mc2=Ac2+, and BHN = ZH + NN : (b) For 197Au, EB = (79)(7:289 MeV) + (197 79)(8:071 MeV) (31:157 MeV) = 1559 MeV;

and the binding energy per nucleon is then (1559 MeV)=(197) = 7:92 MeV:

E50-17 The binding energy of 63Cu is given by M = Zmp + Nmn = (29)(1:007825 u) + (34)(1:008665 u) = 63:521535 u: The binding energy is the di erence EB = (63:521535 u 62:929601 u)(931:5 MeV/u) = 551:4 MeV: The number of atoms in the sample is (0:003 kg)(6:021023mol1) n = = 2:871022: (0:0629 kg=mol) The total energy is then (2:871022)(551:4 MeV)(1:61019J=eV) = 2:531012J:

E50-18 (a) For ultra-relativistic particles E = pc, so (1240 MeV fm) = = 2:59 fm: (480 MeV) (b) Yes, since the wavelength is smaller than nuclear radii.

We will do this one the easy way because we can. This method won’t work except when there is an integer number of half-lives. The activity of the sample will fall to one-half of the initial decay rate after one half-life; it will fall to one-half of one-half (one-fourth) after two half-lives. So two half-lives have elapsed, for a total of (2)(140 d) = 280 d.

E50-20 N = N0(1=2)t=t1=2, so N = (481019)(0:5)(26)=(6:5) = 3:01019:

E50-21 (a) t1=2 = ln 2=(0:0108/h) = 64:2 h: (b) N = N0(1=2)t=t , so 1=2

N=N0 = (0:5)(3) = 0:125: (c) N = N (1=2)t=t1=2 , so 0 N=N0 = (0:5)(240)=(64:2) = 0:0749: E50-22 (a) = (dN=dt)=N , or = (12=s)=(2:51018) = 4:81018=s: (b) t1=2 = ln 2=, so 1=2 which is 4.5 billion years.

E50-23 (a) The decay constant for 67Ga can be derived from Eq. 50-8, ln 2 ln 2 = = = 2:461106s1: t1=2 (2:817105 s) The activity is given by R = N , so we want to know how many atoms are present. That can be found from 1 u 1 atom 3:42 g = 3:0771022 atoms: 1:66051024 g 66:93 u So the activity is R = (2:461106=s1)(3:0771022 atoms) = 7:5721016 decays/s: (b) After 1:728105 s the activity would have decreased to R = R0et = (7:5721016 decays/s)e(2:461106=s )(1:72810 s) = 4:9491016 decays/s: 1 5

E50-24 N = N0et, but = ln 2=t1=2, so t=t1=2 1 N = N e ln 2t=t1=2 = N (2)t=t1=2 = N : 000 2

E50-25 The remaining 223 is N = (4:71021)(0:5)(28)=(11:43) = 8:61020: The number of decays, each of which produced an alpha particle, is (4:71021) (8:61020) = 3:841021:

E50-26 The amount remaining after 14 hours is m = (5:50 g)(0:5)(14)=(12:7) = 2:562 g: The amount remaining after 16 hours is m = (5:50 g)(0:5)(16)=(12:7) = 2:297 g: The di erence is the amount which decayed during the two hour interval: (2:562 g) (2:297 g) = 0:265 g: E50-27 (a) Apply Eq. 50-7, R = R0et: We rst need to know the decay constant from Eq. 50-8, ln 2 ln 2 = = = 5:618107s1: t1=2 (1:234106 s) And the the time is found from 1R R0 1 (170 counts/s) (5:618107s1) (3050 counts/s) = 5:139106 s 59:5 days:

Note that counts/s is not the same as decays/s. Not all decay events will be picked up by a detector and recorded as a count; we are assuming that whatever scaling factor which connects the initial count rate to the initial decay rate is valid at later times as well. Such an assumption is a reasonable (b) The purpose of such an experiment would be to measure the amount of phosphorus that is taken up in a leaf. But the activity of the tracer decays with time, and so without a correction factor we would record the wrong amount of phosphorus in the leaf. That correction factor is R0=R; we need to multiply the measured counts by this factor to correct for the decay. In this case R 7 1 5 = et = e(5:61810 s )(3:00710 s) = 1:184: R0

E50-28 The number of particles of 147Sm is (0:001 kg)(6:021023mol1) n = (0:15) = 6:1431020: (0:147 kg=mol) The decay constant is = (120=s)=(6:1431020) = 1:951019=s: The half-life is t1=2 = ln 2=(1:951019=s) = 3:551018s;

E50-29 The number of particles of 239Pu is (0:012 kg)(6:021023mol1) n0 = = 3:0231022: (0:239 kg=mol) The number which decay is hi n0 n = (3:0251022) 1 (0:5)(20000)=(24100) = 1:321022:

The mass of helium produced is then (0:004 kg=mol)(1:321022) m=

E50-30 Let R33=(R33 + R32) time x = 0:9. Rearranging, so = 8:78105kg: (6:021023mol1)

= x, where x0 = 0:1 originally, and we want to nd out at what (R33 + R32)=R33 = 1=x;

R32=R33 = 1=x 1: Since R = R0(0:5)t=t1=2 we can write a ratio

11 1 = 1 (0:5)t=t32t=t33 : x x0 Put in some of the numbers, and

E50-31 The Q values are Q5 = (235:043923 230:033127 5:012228)(931:5 MeV) = 1:33 MeV: E50-33

E50-34 (a) For the 14C decay, Q = (223:018497 208:981075 14:003242)(931:5 MeV) = 31:84 MeV: For the 4He decay, Q = (223:018497 219:009475 4:002603)(931:5 MeV) = 5:979 MeV: E50-35 Q = (136:907084 136:905821)(931:5 MeV) = 1:17 MeV: E50-36 Q = (1:008665 1:007825)(931:5 MeV) = 0:782 MeV:

(a) The kinetic energy of this electron is signi cant compared to the rest mass energy, so we must use relativity to nd the momentum. The total energy of the electron is E = K + mc2, the momentum will be given by pp p = (1:00 MeV)2 + 2(1:00 MeV)(0:511 MeV) = 1:42 MeV: The de Broglie wavelength is then hc (1240 MeV fm) = = = 873 fm: pc (1:42 MeV) (b) The radius of the emitting nucleus is R = R0A1=3 = (1:2 fm)(150) = 6:4 fm: 1=3

(c) The longest wavelength standing wave on a string xed at each end is twice the length of the string. Although the rules for standing waves in a box are slightly more complicated, it is a fair assumption that the electron could not exist as a standing a wave in the nucleus. (d) See part (c).

E50-38 The electron is relativistic, so p p = 2:16 MeV: This is also the magnitude of the momentum of the recoiling 32S. Non-relativistic relations are K = p2=2m, so (2:16 MeV)2 K = = 78:4 eV: 2(31:97)(931:5 MeV)

E50-39 N = mNA=Mr will give the number of atoms of 198Au; R = N will give the activity; = ln 2=t1=2 will give the decay constant. Combining, N Mr Rt1=2Mr m= = : NA ln 2NA

Then for the sample in question (250)(3:71010=s)(2:693)(86400 s)(198 g/mol) m = = 1:02103g: ln 2(6:021023=mol) E50-40 R = (8722=60 s)=(3:71010=s) = 3:93109Ci:

The radiation absorbed dose (rad) is related to the roentgen equivalent man (rem) by E50-41 the quality factor, so for the chest x-ray (25 mrem) = 29 mrad: (0:85) Each rad corresponds to the delivery of 105 J/g, so the energy absorbed by the patient is

1 (0:029)(105 J/g) (88 kg) = 1:28102 J: 2 E50-42 (a) (75 kg)(102J=kg)(0:024 rad) = 1:8102J: (b) (0:024 rad)(12) = 0:29 rem:

E50-43 R = R0(0:5)t=t1=2, so R0 = (3:94 Ci)(2)(6:048105s)=(1:82105s) = 39:4 Ci:

E50-44 (a) N = mNA=MR, so (2103g)(6:021023/mol) N = = 5:081018: (239 g/mol) (b) R = N = ln 2N=t1=2, so R = ln 2(5:081018)=(2:411104y)(3:15107s/y) = 4:64106=s: (c) R = (4:64106=s)=(3:71010 decays/s Ci) = 1:25104Ci:

E50-45 The hospital uses a 6000 Ci source, and that is all the information we need to nd the number of disintegrations per second: (6000 Ci)(3:71010 decays/s Ci) = 2:221014 decays/s: We are told the half life, but to nd the number of radioactive nuclei present we want to know the decay constant. Then ln 2 ln 2 = = = 4:17109 s1: t1=2 (1:66108 s) The number of 60Co nuclei is then R (2:221014 decays/s) N = = = 5:321022: (4:17109 s1)

E50-46 The annual equivalent does is (12104rem/h)(20 h/week)(52 week/y) = 1:25 rem:

E50-47 (a) N = mNA=MR and MR = (226) + 2(35) = 296, so (1101g)(6:021023/mol) N = = 2:031020: (296 g/mol) (b) R = N = ln 2N=t1=2, so R = ln 2(2:031020)=(1600 y)(3:15107s/y) = 2:8109Bq: (c) (2:8109)=(3:71010) = 76 mCi:

E50-48 R = N = ln 2N=t1=2, so (4:6106)(3:71010=s)(1:28109y)(3:15107s/y) N = = 9:91021; ln 2 N = mNA=MR, so (40 g/mol)(9:91021) m = = 0:658 g: (6:021023/mol) E50-49 We can apply Eq. 50-18 to nd the age of the rock, t1=2 NF ln 2 NI (4:47109y) (2:00103g)=(206 g/mol)

= ln 1 + ; ln 2 (4:20103g)=(238 g/mol) = 2:83109y:

E50-50 The number of atoms of 238U originally present is (3:71103g)(6:021023/mol) N = = 9:381018: (238 g/mol) The number remaining after 260 million years is N = (9:381018)(0:5)(260 My)=(4470 My) = 9:011018:

The di erence decays into lead (eventually), so the mass of lead present should be (206 g/mol)(0:371018) m = = 1:27104 g: (6:021023/mol) E50-51 We can apply Eq. 50-18 to nd the age of the rock, t1=2 NF ln 2 NI (4:47109y) 6 (15010 g)=(206 g/mol) = ln 1 + ; ln 2 (860106g)=(238 g/mol) = 1:18109y: Inverting Eq. 50-18 to nd the mass of 40K originally present, NF NI

so (since they have the same atomic mass) the mass of 40K is (1:6103 g) m = = 1:78103g: 2(1:18)=(1:28) 1

E50-52 (a) There is an excess proton on the left and an excess neutron, so the unknown must be (b) We’ve added two protons but only one (net) neutron, so the element is Ti and the mass (c) The mass number doesn’t change but we swapped one proton for a neutron, so 7Li.

E50-53 Do the math: Q = (58:933200 + 1:007825 58:934352 1:008665)(931:5 MeV) = 1:86 MeV:

We will write these reactions in the same way as Eq. 50-20 represents the reaction of Eq. 50-19. It is helpful to work backwards before proceeding by asking the following question: what The goal is 60Co, which has 27 protons and 60 27 = 33 neutrons.

1. Removing a proton will leave 26 protons and 33 neutrons, which is 59Fe; but that nuclide is unstable.

E50-55 2. Removing a neutron will leave 27 protons and 32 neutrons, which is 59Co; and that nuclide is stable.

3. Removing a deuteron will leave 26 protons and 32 neutrons, which is 58Fe; and that nuclide is stable.

It looks as if only 59Co(n)60Co and 58Fe(d)60Co are possible. If, however, we allow for the possibility of other daughter particles we should also consider some of the following reactions.

E50-56 (a) The possible results are 64Zn, 66Zn, 64Cu, 66Cu, 61Ni, 63Ni, 65Zn, and 67Zn. (b) The stable results are 64Zn, 66Zn, 61Ni, and 67Zn.

E50-57 E50-59 E50-60 Shells occur at numbers 2, 8, 20, 28, 50, 82. The shells occur separately for protons and neutrons. To answer the question you need to know both Z and N = A Z of the isotope. (c) One vacancy in a shell are 13C, 40K, 49Ti, 205Tl, and 207Pb.

(a) The binding energy of this neutron can be found by considering the Q value of the reaction 90Zr(n)91Zr which is (89:904704 + 1:008665 90:905645)(931:5 MeV) = 7:19 MeV: E50-61

(b) The binding energy of this neutron can be found by considering the Q value of the reaction 89Zr(n)90Zr which is (88:908889 + 1:008665 89:904704)(931:5 MeV) = 12:0 MeV: (c) The binding energy per nucleon is found by dividing the binding energy by the number of nucleons: (401:007825 + 511:008665 90:905645)(931:5 MeV) = 8:69 MeV: 91 The neutron in the outside shell of 91Zr is less tightly bound than the average nucleon in 91Zr.

Before doing anything we need to know whether or not the motion is relativistic. The rest mass energy of an particle is P50-1

and since this is much greater than the kinetic energy we can assume the motion is non-relativistic, and we can apply non-relativistic momentum and energy conservation principles. The initial velocity of the particle is then p p p 2 v = 2K=m = c 2K=mc2 = c 2(5:00 MeV)=(3:73 GeV) = 5:1810 c: For an elastic collision where the second particle is at originally at rest we have the nal velocity of the rst particle as m m (4:00u) (197u) v1;f = v1;i 2 1 = (5:18102 2 m2 + m1 (4:00u) + (197u) while the nal velocity of the second particle is 2m1 2 2(4:00u) 3 v2;f = v1;i = (5:1810 c) = 2:0610 c: m2 + m1 (4:00u) + (197u) (a) The kinetic energy of the recoiling nucleus is 11 K = mv2 = m(2:06103c)2 = (2:12106)mc2 22 = (2:12106)(197)(931:5 MeV) = 0:389 MeV: (b) Energy conservation is the fastest way to answer this question, since it is an elastic collision. Then (5:00 MeV) (0:389 MeV) = 4:61 MeV:

P50-2 The gamma ray carries away a mass equivalent energy of m = (2:2233 MeV)=(931:5 MeV/u) = 0:002387 u: The neutron mass would then be mN = (2:014102 1:007825 + 0:002387)u = 1:008664 u:

P50-3 (a) There are four substates: mj can be +3/2, +1/2, -1/2, and -3/2. (b) E = (2=3)(3:26)(3:15108eV=T)(2:16 T) = 1:48107eV: (c) = (1240 eV nm)=(1:48107eV) = 8:38 m: (d) This is in the radio region.

P50-4 (a) The charge density is = 3Q=4R3. The charge on the shell of radius r is dq = 4r2 dr. The potential at the surface of a solid sphere of radius r is q r2 V= = : 40r 30 The energy required to add a layer of charge dq is 42r4 30

which can be integrated to yield 42R5 3Q2 U= = : 30 200R (b) For 239Pu, 3(94)2(1:61019C) U = = 1024106eV: 20(8:851012F=m)(7:451015m) (c) The electrostatic energy is 10:9 MeV per proton.

P50-5 The decay rate is given by R = N , where N is the number of radioactive nuclei present. If R exceeds P then nuclei will decay faster than they are produced; but this will cause N to decrease, which means R will decrease until it is equal to P . If R is less than P then nuclei will be produced faster than they are decaying; but this will cause N to increase, which means R will increase until it is equal to P . In either case equilibrium occurs when R = P , and it is a stable equilibrium because it is approached no matter which side is larger. Then P = R = N at equilibrium, so N = P=.

(b) (8:881010=s)(1e0:269t), where t is in hours. The factor 0:269 comes from ln(2)=(2:58) = . (c) N = P= = (8:881010=s)(3600 s/h)=(0:269/h) = 1:191015: (d) m = NMr=NA, or

(1:191015)(55:94 g/mol) m = = 1:10107g: (6:021023/mol)

P50-7 (a) A = N , so ln 2mNA ln 2(1103g)(6:021023/mol) A = = = 3:66107=s: t1=2Mr (1600)(3:15107s)(226 g/mol)

(c)N=P==t1=2P=ln2,so (3:82)(86400 s)(3:66107=s)(222 g/mol) m = = 6:43109g: (6:021023/mol) ln 2 P50-8 The number of water molecules in the body is N = (6:021023/mol)(70103g)=(18 g/mol) = 2:341027:

There are ten protons in each water molecule. The activity is then A = (2:341027) ln 2=(11032y) = 1:62105/y: The time between decays is then 1=A = 6200 y:

Assuming the 238U nucleus is originally at rest the total initial momentum is zero, which means the magnitudes of the nal momenta of the particle and the 234Th nucleus are equal. The particle has a nal velocity of v = p2K=m = cp2K=mc2 = cp2(4:196 MeV)=(4:0026931:5 MeV) = 4:744102c: Since the magnitudes of the nal momenta are the same, the 234Th nucleus has a nal velocity of (4:0026 u) (4:744102c) = 8:113104c: (234:04 u) The kinetic energy of the 234Th nucleus is 11 K = mv2 = m(8:113104c)2 = (3:291107)mc2 22 = (3:291107)(234:04)(931:5 MeV) = 71:75 keV: P50-9

The Q value for the reaction is then P50-10 (a) The Q value is Q = (238:050783 4:002603 234:043596)(931:5 MeV) = 4:27 MeV: (b) The Q values for each step are Q = (238:050783 237:048724 1:008665)(931:5 MeV) = 6:153 MeV;

Q = (235:045432 234:043596 1:007825)(931:5 MeV) = 5:579 MeV: (c) The total Q for part (b) is 24:026 MeV. The di erence between (a) and (b) is 28:296 MeV. The binding energy for the alpha particle is E = [2(1:007825) + 2(1:008665) 4:002603](931:5 MeV) = 28:296 MeV:

P50-11 (a) The emitted positron leaves the atom, so the mass must be subtracted. But the daughter particle now has an extra electron, so that must also be subtracted. Hence the factor (b) The Q value is Q = [11:011434 11:009305 2(0:0005486)](931:5 MeV) = 0:961 MeV:

P50-12 (a) Capturing an electron is equivalent to negative beta decay in that the total number of electrons is accounted for on both the left and right sides of the equation. The loss of the K shell electron, however, must be taken into account as this energy may be signi cant. (b) The Q value is Q = (48:948517 48:947871)(931:5 MeV) (0:00547 MeV) = 0:596 MeV:

P50-13 The decay constant for 90Sr is ln 2 ln 2 = = = 7:581010s1: t1=2 (9:15108 s) The number of nuclei present in 400 g of 90Sr is (6:021023=mol) (89:9 g/mol) so the overall activity of the 400 g of 90Sr is R = N = (7:581010s1)(2:681024)=(3:71010=Ci s) = 5:49104 Ci: This is spread out over a 2000 km2 area, so the \activity surface density” is (5:49104 Ci) = 2:74105 Ci=m2: (20006 m2) If the allowable limit is 0.002 mCi, then the area of land that would contain this activity is (0:002103 Ci) = 7:30102m2: (2:74105 Ci=m2)

P50-14 (a) N = mNA=Mr, so N = (2:5103g)(6:021023/mol)=(239 g/mol) = 6:31018: (b) A = ln 2N=t1=2, so the number that decay in 12 hours is ln 2(6:31018)(12)(3600 s) = 2:51011: (24100)(3:15107s) (c) The energy absorbed by the body is E = (2:51011)(5:2 MeV)(1:61019J=eV) = 0:20 J: (d) The dose in rad is (0:20 J)=(87 kg) = 0:23 rad: (e) The biological dose in rem is (0:23)(13) = 3 rem:

P50-15 (a) The amount of 238U per kilogram of granite is (4106kg)(6:021023/mol) N = = 1:011019: (0:238 kg/mol) The activity is then ln 2(1:011019) A = = 49:7=s: (4:47109y)(3:15107s/y) The energy released in one second is E = (49:7=s)(51:7 MeV) = 4:11010J: The amount of 232Th per kilogram of granite is (13106kg)(6:021023/mol) N = = 3:371019: (0:232 kg/mol)

The activity is then ln 2(3:371019) A = = 52:6=s: (1:411010y)(3:15107s/y) The energy released in one second is E = (52:6=s)(42:7 MeV) = 3:61010J: The amount of 40K per kilogram of granite is (4106kg)(6:021023/mol) N = = 6:021019: (0:040 kg/mol) The activity is then ln 2(6:021019) A = = 1030=s: (1:28109y)(3:15107s/y) The energy released in one second is E = (1030=s)(1:32 MeV) = 2:21010J: (b) The total for the Earth is 2:71013 W.

P50-16 (a) Since only a is moving originally then the velocity of the center of mass is mava + mX (0) ma V = = va : mX + ma ma + mX (b) Moving to the center of mass frame gives the velocity of X as V , and the velocity of a as va V . The kinetic energy is now 1 K = m V 2 + m (v V )2 ;

cm X a a 2 v2 m2 m2

Xa 2 (m + m )2 (m + m )2 aX aX m v2 m m + m2 aaaXX 2 (ma + mX )2 mX = Klab : ma + mX p (c) va = 2K=m, so p va = 2(15:9 MeV)=(1876 MeV)c = 0:130c: The center of mass velocity is (2) V = (0:130c) = 2:83103c: (2) + (90) Finally, (90) Kcm = (15:9 MeV) = 15:6 MeV: (2) + (90)

P50-18 (a) Removing a proton from 209Bi: E = (207:976636 + 1:007825 208:980383)(931:5 MeV) = 3:80 MeV: Removing a proton from 208Pb: E = (206:977408 + 1:007825 207:976636)(931:5 MeV) = 8:01 MeV: (b) Removing a neutron from 209Pb: E = (207:976636 + 1:008665 208:981075)(931:5 MeV) = 3:94 MeV: Removing a neutron from 208Pb: E = (206:975881 + 1:008665 207:976636)(931:5 MeV) = 7:37 MeV:

E51-1 (a) For the coal, m = (1109J)=(2:9107J=kg) = 34 kg: (b) For the uranium, m = (1109J)=(8:21013J=kg) = 1:2105kg:

E51-2 (a) The energy from the coal is E = (100 kg)(2:9107J=kg) = 2:9109J: (b) The energy from the uranium in the ash is E = (3106)(100 kg)(8:21013J) = 2:51010J: E51-3 (a) There are (1:00 kg)(6:021023mol1) = 2:561024 (235g/mol) (b) If each atom releases 200 MeV, then (200106 eV)(1:61019J= eV)(2:561024) = 8:191013 J (c) This amount of energy would keep a 100-W lamp lit for (8:191013 J) t = = 8:191011s 26; 000 y! (100 W)

E51-4 2 W = 1:251019eV=s. This requires (1:251019eV=s)=(200106eV) = 6:251010=s as the ssion rate.

E51-5 There are (1:00 kg)(6:021023mol1) = 2:561024 (235g/mol) atoms in 1.00 kg of 235U. If each atom releases 200 MeV, then (200106 eV)(1:61019J= eV)(2:561024) = 8:191013 J of energy could be released from 1.00 kg of 235U. This amount of energy would keep a 100-W lamp lit for (8:191013 J) t = = 8:191011s 30; 000 y! (100 W) E51-6 There are (1:00 kg)(6:021023mol1) = 2:521024 (239g/mol) atoms in 1.00 kg of 239Pu. If each atom releases 180 MeV, then (180106 eV)(1:61019J= eV)(2:521024) = 7:251013 J of energy could be released from 1.00 kg of 239Pu.

When the 233U nucleus absorbs a neutron we are given a total of 92 protons and 142 neutrons. Gallium has 31 protons and around 39 neutrons; chromium has 24 protons and around 28 neutrons. There are then 37 protons and around 75 neutrons left over. This would be rubidium, but the number of neutrons is very wrong. Although the elemental identi cation is correct, because we must conserve proton number, the isotopes are wrong in our above choices for neutron numbers.

E51-7 E51-8 Beta decay is the emission of an electron from the nucleus; one of the neutrons changes into a proton. The atom now needs one more electron in the electron shells; by using atomic masses (as opposed to nuclear masses) then the beta electron is accounted for. This is only true for negative beta decay, not for positive beta decay.

E51-9 (a) There are (1:0 g)(6:021023mol1) = 2:561021 (235g/mol) atoms in 1.00 g of 235U. The ssion rate is A = ln 2N=t1=2 = ln 2(2:561021)=(3:51017y)(365d/y) = 13:9=d: (b) The ratio is the inverse ratio of the half-lives: (3:51017y)=(7:04108y) = 4:97108:

E51-10 (a) The atomic number of Y must be 92 54 = 38, so the element is Sr. The mass number (b) The atomic number of Y must be 92 53 = 39, so the element is Y. The mass number is (c) The atomic number of X must be 92 40 = 52, so the element is Te. The mass number is (d) The mass number di erence is 235 + 1 141 92 = 3, so b = 3.

E51-11 The Q value is Q = [51:94012 2(25:982593)](931:5 MeV) = 23 MeV: The negative value implies that this ssion reaction is not possible.

E51-12 The Q value is Q = [97:905408 2(48:950024)](931:5 MeV) = 4:99 MeV: The two fragments would have a very large Coulomb barrier to overcome.

E51-13 The energy released is (235:043923 140:920044 91:919726 21:008665)(931:5 MeV) = 174 MeV:

E51-15 (a) The uranium starts with 92 protons. The two end products have a total of 58 + 44 = (b) The Q value for this process is Q = (238:050783 + 1:008665 139:905434 98:905939)(931:5 MeV) = 231 MeV:

E51-16 (a) The other fragment has 92 32 = 60 protons and 235 + 1 83 = 153 neutrons. That (b) Since K = p2=2m and momentum is conserved, then 2m1K1 = 2m2K2. This means that K2 = (m1=m2)K1. But K1 + K2 = Q, so m2 + m1 m2 or m2 m1 + m2 with a similar expression for K2. Then for 83Ge (153) (83 + 153) while for 153Nd (83) (83 + 153) (c) For 83Ge, s r 2K 2(110 MeV) m (83)(931 MeV) while for 153Nd s r 2K 2(60 MeV) v = = c = 0:029c: m (153)(931 MeV)

Since 239Pu is one nucleon heavier than 238U only one neutron capture is required. The atomic number of Pu is two more than U, so two beta decays will be required. The reaction series is then 239Np ! 239Pu + + : E51-17

E51-18 Each ssion releases 200 MeV. The total energy released over the three years is (190106W)(3)(3:15107s) = 1:81016J: That’s (1:81016J)=(1:61019J=eV)(200106eV) = 5:61026 ssion events. That requires m = (5:61026)(0:235 kg/mol)=(6:021023/mol) = 218 kg: But this is only half the original amount, or 437 kg.

E51-19 According to Sample Problem 51-3 the rate at which non- ssion thermal neutron capture occurs is one quarter that of ssion. Hence the mass which undergoes non- ssion thermal neutron capture is one quarter the answer of Ex. 51-18. The total is then (437 kg)(1 + 0:25) = 546 kg:

E51-20 (a) Qe = E=N, where E is the total energy released and N is the number of decays. This can also be written as P P t1=2 P t1=2Mr A ln 2N ln 2NAm where A is the activity and P the power output from the sample. Solving, (2:3 W)(29 y)(3:15107s)(90 g/mol) Qe = = 4:531013J = 2:8 MeV: ln 2(6:021023/mol)(1 g) (b) P = (0:05)m(2300 W=kg), so (150 W) m = = 1:3 kg: (0:05)(2300 W=kg)

Let the energy released by one ssion be E1. If the average time to the next ssion event is tgen, then the \average” power output from the one ssion is P1 = E1=tgen. If every ssion event results in the release of k neutrons, each of which cause a later ssion event, then after every time period tgen the number of ssion events, and hence the average power output from all of the ssion For long enough times we can write P (t) = P kt=tgen : 0 E51-21

E51-22 Invert the expression derived in Exercise 51-21: tgen=t (1:3103s)=(2:6 s) P (350) k = = = 0:99938: P0 (1200) E51-23 Each ssion releases 200 MeV. Then the ssion rate is (500106W)=(200106eV)(1:61019J=eV) = 1:61019=s The number of neutrons in \transit” is then (1:61019=s)(1:0103s) = 1:61016:

E51-24 Using the results of Exercise 51-21: P = (400 MW)(1:0003)(300 s)=(0:03 s) = 8030 MW: E51-25 The time constant for this decay is ln 2 = = 2:501010s1: (2:77109 s) The number of nuclei present in 1.00 kg is (1:00 kg)(6:021023 mol1) N = = 2:531024: (238 g/mol) The decay rate is then R = N = (2:501010s1)(2:531024) = 6:331014s1: The power generated is the decay rate times the energy released per decay, P = (6:331014s1)(5:59106 eV)(1:61019 J/eV) = 566 W:

E51-26 The detector detects only a fraction of the emitted neutrons. This fraction is A (2:5 m2) = = 1:62104: 4R2 4(35 m)2 The total ux out of the warhead is then (4:0=s)=(1:62104) = 2:47104=s: The number of 239Pu atoms is A (2:47104=s)(1:341011y)(3:15107s/y) N = = = 6:021022: ln 2(2:5) That’s one tenth of a mole, so the mass is (239)=10 = 24 g.

E51-27 Using the results of Sample Problem 51-4, ln[R(0)=R(t)] 5 8

so ln[(0:03)=(0:0072)] t = = 1:72109y: (0:984 0:155)(1109/y)

E51-28 (a) (15109W y)(2105y) = 7:5104W: (b) The number of ssions required is (15109W y)(3:15107s/y) N = = 1:51028: (200 MeV)(1:61019J=eV) The mass of 235U consumed is m = (1:51028)(0:235kg/mol)=(6:021023/mol) = 5:8103kg:

If 238U absorbs a neutron it becomes 239U, which will decay by beta decay to rst 239Np and then 239Pu; we looked at this in Exercise 51-17. This can decay by alpha emission according to 239Pu !235 U + : E51-29

E51-30 The number of atoms present in the sample is N = (6:021023/mol)(1000 kg)=(2:014g/mol) = 2:991026: It takes two to make a fusion, so the energy released is (3:27 MeV)(2:991026)=2 = 4:891026MeV: That’s 7:81013J, which is enough to burn the lamp for t = (7:81013J)=(100 W) = 7:81011s = 24800 y:

E51-31 The potential energy at closest approach is (1:61019C)2 U = = 9105eV: 4(8:851012F=m)(1:61015m)

E51-32 The ratio can be written as r n(K ) K n(K2) K2

so the ratio is s (5000 eV) 5 7 e(3100 eV)=(8:6210 eV=K)(1:510 K) = 0:15: (1900 eV)

E51-35 The energy released is (34:002603 12:0000000)(931:5 MeV) = 7:27 MeV:

E51-36 (a) The number of particle of hydrogen in 1 m3 is N = (0:35)(1:5105kg)(6:021023/mol)=(0:001 kg/mol) = 3:161031 (b) The density of particles is N=V = p=kT ; the ratio is (3:161031)(1:381023J=K)(298 K) = 1:2106: (1:01105Pa)

E51-37 (a) There are (1:00 kg)(6:021023mol1) = 6:021026 (1g/mol) atoms in 1.00 kg of 1H. If four atoms fuse to releases 26.7 MeV, then (26:7 MeV)(6:021026)=4 = 4:01027MeV (b) There are (1:00 kg)(6:021023mol1) = 2:561024 (235g/mol) atoms in 1.00 kg of 235U. If each atom releases 200 MeV, then (200 MeV)(2:561024) = 5:11026MeV of energy could be released from 1.00 kg of 235U.

E51-38 (a) E = mc2, so (3:91026J=s) m = = 4:3109kg=s: (3:0108m=s)2 (b) The fraction of the Sun’s mass \lost” is (4:3109kg=s)(3:15107s/y)(4:5109y) = 0:03 %: (2:01030kg)

E51-39 The rate of consumption is 6:21011kg=s, the core has 1/8 the mass but only 35% is hydrogen, so the time remaining is t = (0:35)(1=8)(2:01030kg)=(6:21011kg=s) = 1:41017s;

E51-40 For the rst two reactions into one: Q = [2(1:007825) (2:014102)](931:5 MeV) = 1:44 MeV: For the second, Q = [(1:007825) + (2:014102) (3:016029)](931:5 MeV) = 5:49 MeV: For the last, Q = [2(3:016029) (4:002603) 2(1:007825)](931:5 MeV) = 12:86 MeV:

E51-41 (a) Use mNA=Mr = N , so (0:012 kg/mol) 1 (3:3107J=kg) = 4:1 eV: (6:021023/mol) (1:61019J=eV) (b) For every 12 grams of carbon we require 32 grams of oxygen, the total is 44 grams. The total mass required is then 40=12 that of carbon alone. The energy production is then (3:3107J=kg)(12=44) = 9106J=kg: (c) The sun would burn for (21030kg)(9106J=kg) = 4:61010s: (3:91026W) That’s only 1500 years!

E51-42 The rate of fusion events is (5:31030W) = 4:561042=s: (7:27106eV)(1:61019J=eV) The carbon is then produced at a rate (4:561042=s)(0:012 kg/mol)=(6:021023/mol) = 9:081016kg=s: The process will be complete in (4:61032kg) = 1:6108y: (9:081016kg=s)(3:15107s/y) E51-43 (a) For the reaction d-d, Q = [2(2:014102) (3:016029) (1:008665)](931:5 MeV) = 3:27 MeV: (b) For the reaction d-d, Q = [2(2:014102) (3:016029) (1:007825)](931:5 MeV) = 4:03 MeV: (c) For the reaction d-t, Q = [(2:014102) + (3:016049) (4:002603) (1:008665)](931:5 MeV) = 17:59 MeV:

E51-44 One liter of water has a mass of one kilogram. The number of atoms of 2H is (6:021023/mol) (0:00015 kg) = 4:51022: (0:002 kg/mol) The energy available is (3:27106eV)(1:61019J=eV)(4:51022)=2 = 1:181010J: The power output is then (1:181010J) = 1:4105W (86400 s) E51-45 Assume momentum conservation, then p = pn or vn=v = m =mn:

The ratio of the kinetic energies is then Kn m v2 m = n n = 4: K m v2 mn Then Kn = 4Q=5 = 14:07 MeV while K = Q=5 = 3:52 MeV.

E51-46 The Q value is Q = (6:015122 + 1:008665 3:016049 4:002603)(931:5 MeV) = 4:78 MeV: Combine the two reactions to get a net Q = 22:37 MeV. The amount of 6Li required is N = (2:61028MeV)=(22:37 MeV) = 1:161027: The mass of LiD required is (1:161027)(0:008 kg/mol) m = = 15:4 kg: (6:021023/mol) P51-1 (a) Equation 50-1 is R = R0A1=3;

where R0 = 1:2 fm. The distance between the two nuclei will be the sum of the radii, or

R (140)1=3 + (94)1=3 : 0 The potential energy will be 1 q1q2 40 r e2 (54)(38) 40R0 (140)1=3 + (94)1=3 (1:601019C)2 4(8:851012C2=Nm2)(1:2 fm) = 253 MeV: (b) The energy will eventually appear as thermal energy.

p P51-2 (a) Since R = R 3 A, the surface area a is proportional to A2=3. The fractional change in 0 surface area is (a1 + a2) a0 (140)2=3 + (96)2=3 (236)2=3 = = 25 %: a0 (236)2=3 p (c) Since U / Q2=R, U / Q2= 3 A. The fractional change in the electrostatic potential energy is

U1 + U2 U0 (54)2(140)1=3 + (38)2(96)1=3 (92)2(236)1=3 = = 36 %: U0 (92)2(236)1=3

P51-3 (a) There are (2:5 kg)(6:021023mol1) = 6:291024 (239g/mol) atoms in 2.5 kg of 239Pu. If each atom releases 180 MeV, then (180 MeV)(6:291024)=(2:61028MeV/megaton) = 44 kiloton is the bomb yield.

P51-4 (a) In an elastic collision the nucleus moves forward with a speed of 2mn mn + m

so the kinetic energy when it moves forward is m 4m2 mnm 2 0 (m + mn)2 (mn + m)2

where we can write K because in an elastic collision whatever energy kinetic energy the nucleus (b) For hydrogen, K 4(1)(1) = = 1:00: K (1 + 1)2 For deuterium, K 4(1)(2) = = 0:89: K (1 + 2)2 For carbon, K 4(1)(12) = = 0:28: K (1 + 12)2 For lead, K 4(1)(206) = = 0:019: K (1 + 206)2 (c) If each collision reduces the energy by a factor of 1 0:89 = 0:11, then the number of collisions required is the solution to (0:025 eV) = (1106eV)(0:11)N ;

P51-5 The radii of the nuclei are p 3 R = (1:2 fm) 7 = 2:3 fm: The using the derivation of Sample Problem 51-5, (3)2(1:61019C)2 K = = 1:4106eV: 16(8:851012F=m)(2:31015m)

P51-6 (a) Add up the six equations to get 12C +1H +13 N +13 C +1 H +14 N +1 H +15 O +15 N +1 H ! 13N + +13 C + e+ + +14 N + +15 O + +15 N + e+ + +12 C +4 He: Cancel out things that occur on both sides and get 1H +1 H +1 H +1 H ! + e+ + + + + e+ + +4 He: (b) Add up the Q values, and then add on 4(0:511 MeV for the annihilation of the two positrons.

(a) Demonstrating the consistency of this expression is considerably easier than deriving it from rst principles. From Problem 50-4 we have that a uniform sphere of charge Q and radius R has potential energy 3Q2 U= : 200R P51-7

This expression was derived from the fundamental expression 1 q dq dU = : 40 r For gravity the fundamental expression is Gm dm r

so we replace 1=40 with G and Q with M . But like charges repel while all masses attract, so we (b) The initial energy would be zero if R = 1, so the energy released is 3GM2 3(6:71011Nm2=kg2)(2:01030kg)2 U = = = 2:31041J: 5R 5(7:0108m) At the current rate (see Sample Problem 51-6), the sun would be (2:31041J) (3:91026W) or 187 million years old.

P51-8 (a) The rate of fusion events is (3:91026W) = 9:31037=s: (26:2106eV)(1:61019J=eV) Each event produces two neutrinos, so the rate is 1:861038=s: (b) The rate these neutrinos impinge on the Earth is proportional to the solid angle subtended by the Earth as seen from the Sun: r2 (6:37106m)2 4R2 4(1:501011m)2 so the rate of neutrinos impinging on the Earth is (1:861038=s)(4:51010) = 8:41028=s:

P51-9 (a) Reaction A releases, for each d Reaction B releases, for each d (1=3)(17:59 MeV) + (1=3)(4:03 MeV) = 7:21 MeV: Reaction B is better, and releases (7:21 MeV) (2=02 MeV) = 5:19 MeV more for each N .

P51-10 (a) The mass of the pellet is 4 m = (2:0105m)3(200 kg=m3) = 6:71012kg: 3

The number of d-t pairs is (6:71012kg)(6:021023/mol) N = = 8:061014; (0:005 kg/mol) and if 10% fuse then the energy release is (17:59 MeV)(0:1)(8:061014)(1:61019J=eV) = 230 J: (b) That’s (230 J)=(4:6106J=kg) = 0:05 kg (c) The power released would be (230 J)(100=s) = 2:3104W.

(a) The gravitational force is given by Gm2=r2, while the electrostatic force is given by q2=40r2. The ratio is 40Gm2 4(8:851012C2=Nm2)(6:671011Nm2=kg2)(9:111031kg)2 q2 (1:601019C)2 = 2:41043: Gravitational e ects would be swamped by electrostatic e ects at any separation. (b) The ratio is 40Gm2 4(8:851012C2=Nm2)(6:671011Nm2=kg2)(1:671027kg)2 q2 (1:601019C)2 = 8:11037: E52-1

E52-3 The gravitational force from the lead sphere is GmeM 4GmeR =: R2 3 Setting this equal to the electrostatic force from the proton and solving for R, 3e2 162 Gm a2 0 e0 or 3(1:61019C)2 162(8:851012F=m)(6:671011Nm2=kg2)(11350 kg=m3)(9:111031kg)(5:291011m)2 which means R = 2:851028m:

E52-4 Each takes half the energy of the pion, so (1240 MeV fm) = = 18:4 fm: (135 MeV)=2 E52-5 The energy of one of the pions will be pp E = (pc)2 + (mc2)2 = (358:3 MeV)2 + (140 MeV)2 = 385 MeV: There are two of these pions, so the rest mass energy of the 0 is 770 MeV.

E52-6 E = mc2, so = (1:5106eV)=(20 eV) = 7:5104: The speed is given by p v = c 1 1= 2 c c=2 2;

where the approximation is true for large . Then v = c=2(7:5104)2 = 2:7102m=s:

E52-7 d = ct = hc=2E. Then (1240 MeV fm) d = = 2:16103fm: 2(91200 MeV)

(a) Baryon number is conserved by having two \p” on one side and a \p” and a 0 on the other. Charge will only be conserved if the particle x is positive. Strangeness will only be conserved if x is strange. Since it can’t be a baryon it must be a meson. Then x is K+. (b) Baryon number on the left is 0, so x must be an anti-baryon. Charge on the left is zero, so x must be neutral because \n” is neutral. Strangeness is everywhere zero, so the particle must be n. (c) There is one baryon on the left and one on the right, so x has baryon number 0. The charge on the left adds to zero, so x is neutral. The strangeness of x must also be 0, so it must be a 0.

E52-9 E52-10 There are two positive on the left, and two on the right. The anti-neutron must then be neutral. The baryon number on the right is one, that on the left would be two, unless the anti- neutron has a baryon number of minus one. There is no strangeness on the right or left, except possible the anti-neutron, so it must also have strangeness zero.

E52-11 (a) Annihilation reactions are electromagnetic, and this involves ss. (d) Strangeness is conserved, so this is neither weak nor electromagnetic, so it must be strong.

E52-12 (a) K0 ! e+ + e, (b) K0 ! + + 0, (c) K0 ! + + + + , (d) K0 ! + + 0 + 0,

E52-14 A strangeness of +1 corresponds to the existence of an s anti-quark, which has a charge of +1/3. The only quarks that can combine with this anti-quark to form a meson will have charges of -1/3 or +2/3. It is only possible to have a net charge of 0 or +1. The reverse is true for strangeness -1.

quarks Q S C particle uc 0 0 -1 D0 dc -1 0 -1 D sc -1 -1 -1 D E52-19 We’ll construct a table: s cc 0 0 0 c cu 0 0 1 D0 cd 1 0 1 D+ cs 1 1 1 D+ s

E52-20 (a) Write the quark content out then cancel out the parts which are the same on both sides: dds ! udd + du;

so the fundamental process is s ! u + d + u: (b) Write the quark content out then cancel out the parts which are the same on both sides: ds ! ud + du;

so the fundamental process is s ! u + d + u: (c) Write the quark content out then cancel out the parts which are the same on both sides: ud + uud ! uus + us;

so the fundamental process is d + d ! s + s: (d) Write the quark content out then cancel out the parts which are the same on both sides: + udd ! uud + du;

so the fundamental process is ! u + u: E52-21 The slope is (7000 km/s) = 70 km/s Mpc: (100 Mpc) E52-22 c = Hd, so d = (3105 km/s)=(72 km/s Mpc) = 4300 Mpc:

E52-23 The question should read \What is the…” The speed of the galaxy is v = Hd = (72 km/s Mpc)(240 Mpc) = 1:72107m=s: The red shift of this would then be p 1 (1:72107m=s)2=(3108m=s)2 = (656:3nm) = 695 nm: 1 (1:72107m=s)=(3108m=s)

E52-24 We can approximate the red shift as so 0 (590 nm) u = c 1 = c 1 = 0:02c: (602 nm) The distance is d = v=H = (0:02)(3108m=s)=(72 km/s Mpc) = 83 Mpc:

The minimum energy required to produce the pairs is through the collision of two 140 MeV photons. This corresponds to a temperature of E52-25

T = (140 MeV)=(8:62105eV=K) = 1:621012K: This temperature existed at a time (1:51010s1=2K)2 t = = 86 s: (1:621012K)2 (b) f = (3108m=s)=(0:002 m) = 1:51011Hz: (c) E = (1240 eV nm)=(2106nm) = 6:2104eV:

E52-27 (a) Use Eq. 52-3: p (1:51010 sK)2 t = = 91012s: (5000 K)2 (b) kT = (8:62105eV=K)(5000 K) = 0:43 eV: (c) The ratio is (109)(0:43 eV) = 0:457: (940106eV) P52-1 The total energy of the pion is 135 + 80 = 215 MeV. The gamma factor of relativity is = E=mc2 = (215 MeV)=(135 MeV) = 1:59;

so the velocity parameter is p = 1 1= 2 = 0:777: The lifetime of the pion as measured in the laboratory is t = (8:41017 s)(1:59) = 1:341016s;

so the distance traveled is d = vt = (0:777)(3:00108m=s)(1:341016s) = 31 nm:

p P52-2 (a) E = K + mc2 and pc = E2 (mc2)2, so p pc = (2200 MeV + 1777 MeV)2 (1777 MeV)2 = 3558 MeV: That’s the same as (3558106eV) p = (1:61019J=eV) = 1:901018kg m=s (3108m=s) .

(b) qvB = mv2=r, so p=qB = r. Then (1:901018kg m=s) r = = 9:9 m: (1:61019C)(1:2 T)

P52-3 (a) Apply the results of Exercise 45-1: (1240 MeV fm) E = = (4:281010MeV=K)T: (2898 m K)T (b) T = 2(0:511 MeV)=(4:281010MeV=K) = 2:39109K:

P52-4 (a) Since p 1 2 1 we have p 1 2 0 1 or p 1 2+ 1 z= : 1 Now invert, p z(1 ) + 1 = 1 2;

(z2 + 2z + 2) 2 2(z2 + 2z + 1) + (z2 + 2z) = 0: Solve this quadratic for , and z2 + 2z =: z2 + 2z + 2 (b) Using the result, (4:43)2 + 2(4:43) = = 0:934: (4:43)2 + 2(4:43) + 2 (c) The distance is d = v=H = (0:934)(3108m=s)=(72 km/s Mpc) = 3893 Mpc:

P52-5 (a) Using Eq. 48-19, n1 E = kT ln : n2 Here n1 = 0:23 while n2 = 1 0:23, then

E = (8:62105eV=K)(2:7 K) ln(0:23=0:77) = 2:8104eV: (b) Apply the results of Exercise 45-1: (1240 eV nm) = = 4:4 mm: (2:8104eV)

P52-6 (a) Unlimited expansion means that v Hr, so we are interested in v = Hr. Then p 3H2=8G = : (b) Evaluating, 3[72103m=s (3:0841022m)]2 (6:021023/mol) = 5:9=m3: 8(6:671011N m2=kg2) (0:001 kg/mol) P52-7 (a) The force on a particle in a spherical distribution of matter depends only on the matter contained in a sphere of radius smaller than the distance to the center of the spherical distribution. And then we can treat all that relevant matter as being concentrated at the center. If M is the total mass, then r3 R3 is the fraction of matter contained in the sphere of radius r < R. The force on a star of mass m a distance r from the center is F = GmM0=r2 = GmMr=R3: This force is the source of the centripetal force, so the velocity is ppp v = ar = F r=m = r GM=R3: The time required to make a revolution is then 2r p T = = 2 R3=GM : v

Note that this means that the system rotates as if it were a solid body! (b) If, instead, all of the mass were concentrated at the center, then the centripetal force would be F = GmM=r2;

so ppp and the period would be 2r p T = = 2 r3=GM : v

P52-8 We will need to integrate Eq. 45-6 from 0 to min, divide this by I(T), and set it equal to z = 0:2109. Unfortunately, we need to know T to perform the integration. Writing what we do know and then letting x = hc=kT , 15c2 3 m 2c2 Z h h d 25k4T 4 5 ehc=kT 1 0 15c2 3 4 4 xm x3 Z h 2k T dx 2 k T h c2 1 ex 1 544 3 Z 1 x3 15 dx =: 4 ex 1 xm

The result is a small number, so we expect that xm is fairly large. We can then ignore the 1 in the denominator and then write Z1 z4=15 = x3exdx xm

which easily integrates to z4=15 xm3exm: The solution is so (2:2106eV) T = = 8:5108K: (8:62105eV=K)(30)