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YOUNG 11e - Chapter 24
32. 563 nJ/m³
Chapter 26
14. 25 Ohm - 7.00 A; 20 Ohm - 9.95 A
62. 8.60 V
84. (a) 7.10 J
(b) 3.62 kW
(c) 1.81 kW
WOLFSON - Chapter 26
3. 4.88 kJ
10. (a) 1.41 J
(b) 4.22 J
15. (a) proof
(b) dW = (2kq/a)dq
(c) (k/a)(Q²)
52. (a) circuit diagram
(b) 12.0 nF
(c) Q1 = 1.20 microC; Q2 = 0.40 microC; Q3 = 0.80 microC
(d) V1 = 60 V; V2 = V3 = 40 V
Halliday 2nd Ed Problems - Chapter 26
19. (a) k(lambda)ln((y+L)/y)
(b) k(lambda)L/(y+L)y
(c) zero, since V(x) must be a
max at P by symmetry
20. (a) kc(sqrt(L² + y²)-y)
(b) -kc((y/sqrt(L² + y²))-1)
(c) we don't know V(x) for any y
and symmetry cannot help
HRW PROBLEM SUPPLEMENT #1 - CHAPTER 26
53. 66.2 microJ
66. (a) 200 kV/m (b) 200 kV/m
(c) 1.77 microC/m² (d) 4.60 microC/m²
(e) -2.83 microC/m²
Chapter 27
76. (a) 0.0299 N⋅m in the +j direction
(b) 0.0173 N⋅m in the -j direction
(c) the answers for the two cases are the same;
you must carefully explain why
(d) 0.0173 J for the field in the +x dir;
0.00463 J for the field in the -z dir
78. (a) on PQ, F=0; on RP, F = 12 N into page;
on QR, F= 12 N out of page
(b) zero
(c) on PQ, tau=0; on RP, tau=0;
on QR, tau = 3.60 N⋅m to right
(d) 3.60 N⋅m to right; the results agree
(e) out of the plane
Halliday 2nd Ed Problems - Chapter 27
7. (a) 3.5 pF
(b) 3.5 pF
(c) 57.1 V
13. (a) (epsilon0)A/(d-b)
(b) Ui/Uf = d/(d-b)
(c) work done by the electric field is
Q²b/2(epsilon0)A; sucked in
18. (a) 7.18
(b) 0.766 microC
43. (a) 725 pF <== this number uses HRW's value of kappa
or for Pyrex, which is 4.7
863 pF <== this number uses Wolfson's value of kappa
for Pyrex, which is 5.6
(b) about 28 kV (more exactly 27.2 kV) <== these numbers
use 14 MV/m for the dielectric strength
of Pyrex (from HRW or Wolfson)
or
about 20 kV (more exactly 19.4 kV) <== these numbers
use 10 MV/m for the dielectric strength
of Pyrex (from Y&F)
45. (a) 13.4 pF
(b) 1.15 nC
(c) 11.3 kN/C
(d) 4.32 kN/C
WOLFSON - Chapter 27
4. 4.80 mA away from the screen
62. (a) 8700 A
(b) 15.1%
(c) 316 kV
72. (a) answer will be posted after this HW is due
(b) answer will be posted after this HW is due
(c) proof
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
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
27. (a) 3.48 mWb
(b) 15.5 microC
WOLFSON - Chapter 32
9. ((mu_0)l/2pi)ln((a+w)/a)
A Problems
A2. (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. this answer will appear after this question is due
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) 5.22 mA (b) zero
(c) 4.52 mA (d) taking energy (you must explain why)
2. (a) 39.1 mA (b) zero
(c) -33.9 mA (d) supplying energy (you must explain why)
4. 0.6 A for all frequencies
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
A3 and A4: proofs
A5: answer will appear only after this problem is due