Chapter 28
10. a, 2.00 microT; b, 0.705 microT;
c, 2.00 microT; d, zero;
e, 0.545 microT; also a figure
36. a, zero; b, -5.0 microT⋅m;
c, +2.5 microT⋅m; d, +5.0 microT⋅m
64. 79.7 microN towards the long straight wire
86. 0.5(mu_0)v(sigma)
WOLFSON - Chapter 28
12. 10.0 V
42. (a) 48.4 V
(b) 57.3 V
(c) 59.9 V
A6. (a) C>A=B>D=E
(b) 27.4 Ohms
(c) 267 mA
(d) 35.3 Ohms
(e) 312 mA
(f) C gets brighter, D and E turn off,
A and B get dimmer
A7. (a) -1010 km/s <== originally posted incorrectly
(b) -53.3 km/s <== originally posted incorrectly
corrected on 7/22 at 10 AM
(c) Since B is purely z, the z component of v
can have any possible value; the cross
product of two vectors having the same
direction is always zero since sin(0)=0.
(d) zero
A8: (a) graph; the energy axis should have
a range of +-1.23 J
(b) four drawings
(c) 2.45 J
A9: (a) side of loop value of int(Edotdl)
(-1,-1) to (-1,1) 600 V
(-1,1) to (1,1) 0 V
(1,1) to (1,-1) 600 V
(1,-1) to (-1,-1) 0 V
these values for CCW direction (CW direction values are NEG)
(b) No, all possible physical E fields
satisfy the loop rule; since int(Edotdl)
around the loop is not zero, this E field
is not physically possible.
Chapter 26
84. (a) 7.10 J
(b) 3.62 kW
(c) 1.81 kW
Fishbane Problems - Chapter 28
57. proof
Chapter 29
2. (a) before - 14.4 micro Wb; after - zero
(b) 0.360 mV
6. (a) 0.0302 V + (0.302 mV/s³)t³
(b) 0.113 mA
7. (d) the EMF direction is CCW
(e) 0.506 microV
25. (b) towards top of page (clockwise makes
no sense for the current in the rod)
26. (a) graph
(b) graph
28. (a) 170 microN/C
(b) 339 microN/C
(c) 530 microN/C
(d) counterclockwise
33. (a) 950 microV
(b) 109 microH
77. (a) from a to b
(b) (Rmg)tan(phi)/(L²B²cos(phi))
(c) (mg)tan(phi)/(LB)
(d) Rm²g²tan²(phi)/(L²B²)
(e) same answer as in (d) by a different method
Halliday 2nd Ed Problems - Chapter 29
22. (a) by integration, U_C = (1/2)C(EMF)² and
energy out from battery = C(EMF)²
(b) by integration, heat from resistor = (1/2)C(EMF)²
Chapter 30
46. (a) EMF(t)=(23.4 V)sin((240pi rad/s)t)
two graphs
(b) 23.4 V; zero
(c) 0.124 A; zero
48. (a) (mu_0)i/(2pir)
(b) Please ignore part (b) (but here is the
answer: [(mu_0)i/(2pir)]l(dr) )
(c) [(mu_0)i(l)/(2pi)]ln(b/a)
(d) trivial proof from part (c)
(e) [(mu_0)i²(l)/(4pi)]ln(b/a)
62. (a) V1 = 33.3 V; V2 = V4 = 16.7 V; V3 = 0 V
A1 = A3 = 0.333 A; A2 = 0 A
(b) V1 = 38.5 V; V2 = 0; V3 = V4 = 11.5 V
A1 = 0.385 A; A2 = 0.153 A; A3 = 0.230 A
68. (a) i1 = 1.50 A; i2 = 1.80 A
(b) 2.40 A
YOUNG 11e - Chapter 30
18. (a) 4.35 mT
(b) 7.53 J/m³
(c) 1.52 microm³
(d) 11.4 microJ
(e) 3.65 microH
(f) same as (d) by a different method
66. (a) V1 = 25 V; V2 = V4 = 50.0 V; V3 = 0 V
A1 = A3 = 0.500 A; A2 = 0 A
(b) V3 = 75.0 V; all other meters read zero
(c) 5.63 microC; long after S is closed
68. (a) 894 microC
(b) 20.0 mJ
(c) graph (cos function with amp 2.0 A)
(d) graph (sin function with amp 894 microC)
Halliday 2nd Ed Problems - Chapter 30
9. (a) 20 min
(b) 59.4 m(N⋅m)
22. (a) 3750 m/s
(b) drawing
29. (a) 2.60 x 10^6 m/s
(b) 0.109 microseconds
(c) 141 keV
(d) 70.3 kV
31. (-11.4 N/C)i - (6 N/C)j + (4.8 N/C)k
WOLFSON - Chapter 30
24. 7.14 microN West
42. (a) (mu0)J(|z|) (b) (mu0/2)Jh
48. (a) (mu0)I(sqrt(n²+(1/2piR)²))
(b) INVtan(1/2(pi)nR)
57. proof; out of the page
58. proof; out of the page
68. (a) (pi/3)R²J_0
(b) (mu0/6r)R²J_0
(c) (mu0/2)rJ_0(1-(2r/3R))
HRW PROBLEM SUPPLEMENT #1 - CHAPTER 30
73. (a) 4.8 mT
(b) 0.933 mT
(c) zero
Fishbane Problems - Chapter 30
32. (a) Bmax = 174 microT; Bmin = 139 microT <== these answers use
(b) 0.140 microWb (1 Wb = 1 T⋅m²) HRW's value of rho_CU
(c) heating rate is 1.57 W; doesn't get very hot (16.9 nOhms⋅m)
(a) Bmax = 175 microT; Bmin = 140 microT <== these answers use
(b) 0.141 microWb (1 Wb = 1 T⋅m²) Wolfson's value
(c) heating rate is 1.58 W; doesn't get very hot (16.8 nOhms⋅m)
(a) Bmax = 171 microT; Bmin = 137 microT <== these answers use
(b) 0.138 microWb (1 Wb = 1 T⋅m²) Y&F's value
(c) heating rate is 1.54 W; doesn't get very hot (17.2 nOhms⋅m)
Chapter 31
36. (a) 945 rad/s
(b) 70.6 ohms
(c) resistor - 120 V; capacitor - 450 V;
inductor - 450 V
(d) V4 = 0 V; V5 = 84.9 V
(e) V1 = 27.4 V; V2 = 70.1 V; V3 = 150.4 V;
V4 = 80.3 V; V5 = 84.9 V
(f) V1 = 36.6 V; V2 = 180.6 V; V3 = 104.0 V;
V4 = 76.6 V; V5 = 84.9 V
39. (a) use a transformer with half as many
secondary turns as in the primary
(b) 6.67 A
(c) 36 Ohms
WOLFSON - Chapter 31
33. (a) 0.1 V
(b) 25 mA
(c) 1.25 mN
(d) 2.5 mW from I²R
(e) 2.5 mW from F⋅v
YOUNG 11e - Chapter 31
60. C is 2.86 pF; R is 126 mOhms
Halliday 2nd Ed Problems - Chapter 31
5. proof
25. (a) 78.5 microT (b) 1.08 microN⋅m
Chapter 32
18. 15.9 microJ
20. (a) 240 microW
(b) 17.4 V/m
52. 61.4 kV/m and 205 microT
Halliday 2nd Ed Problems - Chapter 32
7. 0.452 V
20. (a) 1.51 Wb
(b) at 90 degrees, 1.20 Wb
at 180 degrees, 0.892 Wb
(c) at 0 degrees (posted after due)
(d) at 90 degrees and 270 degrees (posted after due)
(e) 155 Wb/s
(f) 155 V (posted after due)
27. (a) 3.48 mWb
(b) 15.5 microC
WOLFSON - Chapter 32
9. ((mu_0)l/2pi)ln((a+w)/a)
A Problems
A10 (a) 25 times per second
(b) 466 turns
(c) 20 Ohms
(d) 63.7 mWb
(e) 0.01 and 0.03 s (you must explain why)
HRW PROBLEM SUPPLEMENT #1 - CHAPTER 32
28. (a) 1.89 pT
(b) graph; at 10 cm, B = 0.565 pT
38. (a) (4/3) A
(b) at R/4 and at 4R
KNIGHT - Chapter 33
69. (a) drawing
(b) free-body diagram
(c) 8.64 mT
Chapter 33
28. (a) 63.4 degrees
(b) 71.6 degrees
34. 32.3 W/cm²
Halliday 2nd Ed Problems - Chapter 33
11. (a) 238.7 W (b) 154.8 W (c) 393.5 W
12. (a) 18.7 J (b) 5.10 J (c) 13.6 J
Fishbane Problems - Chapter 33
46. DiffEq is -I(t)R - L(dI/dt) = 0
HRW PROBLEM SUPPLEMENT #1 - CHAPTER 33
5. (a) n(5.00 microseconds) where n=1,2,3,4,...
(b) (2n+1)(2.50 microseconds) where n=0,1,2,3,4,...
(c) (2n+1)(1.25 microseconds) where n=0,1,2,3,4,...
12. (a) 3.6 mH (b) 1326 Hz (c) 189 microseconds
44. (a) 16.6 Ohms (b) 422 Ohms (c) 521 mA
(d) 33.2 Ohms (e) 408 Ohms (f) 539 mA
50. (posted after due) Since the AC voltmeter gives
the same reading for the inductor and capacitor
voltages, thus the circuit is at resonance.
At resonance, the EMF is in phase with the
resistor voltage, and has the same amplitude.
So, we may conclude that the reading is 100 V
RMS.
WOLFSON - Chapter 34
6. (a) 7.20 x 10^{11} (V/m)/s
(b) increasing
20. (a) 3.00 m (b) 10.0 cm
(c) 500 nm (d) 0.300 nm
HRW PROBLEM SUPPLEMENT #1 - CHAPTER 34
6. 4.74 m
40. (a) 15.9% (b) 84.1%
Knight 2nd Ed - Chapter 35
24. (a) 3.33 microT
(b) 1.67 microT in the negative x direction
(c) 50 m
Halliday 2nd Ed Problems - Chapter 36
1. (a) (EMF_max)sin((w_d)t)=L(dI/dt) (posted after due)
(b) I(t)=-(EMF_max/(Lw_d))cos((w_d)t) (posted after due)
(c) 5.22 mA (d) zero
(e) 4.52 mA (f) taking energy (you must explain why)
2. (a) (EMF_max)sin((w_d)t)=Q(t)/C; no (posted after due)
(b) I(t)=(EMF_max*Cw_d)cos((w_d)t) (posted after due)
(c) 39.1 mA (d) zero
(e) -33.9 mA (f) supplying energy (you must explain why)
4. (a) (EMF_max)sin((w_d)t)=I(t)R; no (posted after due)
(b) I(t)=(EMF_max/R)sin((w_d)t) (posted after due)
(c) 0.6 A
(d) 0.6 A
17. (a) 76.4 mH (b) yes; 17.8 Ohms; resistor would
consume energy, not store it
Knight 2nd Ed - Chapter 36
67. (a) 2.50 Ohms
(b) 1.93 microF
(c) 50900 rad/s
A11: (a) 0.405 J
(b) 31700 times per second
(c) 0.304 J
(d) 7.89 microseconds
(e) 1.06 A
A12: (a) 0.824 radians
(b) 0.680
(c) 356 Ohms
(d) 161 V
(e) 49.4 W
(f) 49.4 W (posted after due)
(g) zero (posted after due)
(h) 0.665 A
(i) 107 W
A13: (a) 0.260 pW/m²
(b) 8.66x10^-22 J/m³
(c) 14.0 microV/m
Satellite dishes can process amazingly small signals; of
course, in the first step of the process the signal is
concentrated at the receiver by the parabolic reflector
of the dish (the actual receiver is at the focal point
of the parabolic reflector).