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Chapter_30_Solutions - 30.1 a 2 = M(di1/dt =(3.25 10-4...

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= M(di 1 /dt) = (3.25 × 10 4 H) (830 A/s) = 0.270 V, and is constant. 30.1: a) ε 2 b) If the second coil has the same changing current, then the induced voltage is the same and 1 = 0.270 V.
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30.2: For a toroidal solenoid, M = N 2 Φ B 2 /i 1 , and Φ B 2 = µ 0 N 1 i 1 A/2 π r. So, M = 0 AN 1 N 2 /2
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30.3: a) M = N 2 Φ B 2 /i 1 = (400) (0.0320 Wb)/(6.52 A) = 1.96 H. b) When i 2 = 2.54 A, Φ B 1 = i 2 M/N 1 = (2.54 A) (1.96 H)/(700) = 7.11 × 10 3 Wb.
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30.4: a) M = ε 2 /(di/dt) = 1.65 × 10 3 V/ ( 0.242 A/s) = 6.82 × 10 3 H. b) N 2 = 25, i 1 = 1.20 A, ⇒Φ B 2 = i 1 M/N 2 = (1.20 A) (6.82 × 10 3 H)/25 = 3.27 × 10 4 Wb. c) di 2 /dt = 0.360 A/s and 1 = Mdi 2 = (6.82 × 10 3 H) (0.360 A/s) = 2.45 mV.
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30.5: 1H = 1Wb/A = 1Tm 2 /A = 1Nm/A 2 = 1J/A 2 = 1 (J/AC)s = 1 (V/A)s = 1 s.
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30.6: For a toroidal solenoid, L = N Φ B /i = ε /(di/dt). So solving for we have: N N = i / Φ B ( di / dt ) = (12.6 × 10 3 V) (1.40 A) (0.00285 Wb) (0.0260 A/s) = 238 turns.
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30.7: a) ε = L(di 1 /dt) = (0.260 H) (0.0180 A/s) = 4.68 × 10 3 V. b) Terminal a is at a higher potential since the coil pushes current through from bto and if replaced by a battery it would have the a + terminal at . a
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30.8: a) L K m = K m µ 0 N 2 A/2 π r = (500 0 ) (1800) 2 (4.80 × 10 5 m 2 ) 2 (0.120 m) = 0.130 H. b) Without the material, L = 1 K m L K m = 1 500 (0.130 H) = 2.60 × 10 4 H.
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30.9: For a long, straight solenoid: L = N Φ B /i and Φ B = µ 0 NiA/l L = 0 N 2 A/l.
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30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according to Equation 30.8, di dt = V b V a L = 1. 0.260 04V H =− 4.00 A/s. Thus, the current is decreasing. b) From above we have that di = ( 4.00 A/s)dt. After integrating both sides of this expression with respect to we obtain , t i = ( 4.00 A/s) t i = (12.0 A) (4.00 A/s) (2.00 s) = 4.00 A.
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30.11: a) L = ε /(di/dt) = (0.0160 V)/(0.0640 A/s) = 0.250 H. b) Φ B = iL/N = (0.720 A) (0.250 H)/(400) = 4.50 × 10 4 Wb.
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30.12: a) U = 1 2 LI 2 = (12.0 H) (0.300 A) 2 /2 = 0.540 J. b) P = I 2 R = (0.300 A) 2 (180 ) = 16.2 W. c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, constant. = U
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30.13: U = 1 2 LI 2 = µ 0 N 2 AI 2 4 π r N = 4 rU 0 AI 2 = 4 (0.150 m) (0.390 J) 0 (5.00 × 10 4 m 2 ) (12.0 A) 2 = 2850 turns.
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30.14: a) U = Pt = (200 W) (24 h/day × 3600 s/h) = 1.73 × 10 7 J. b) U = 1 2 LI 2 L = 2U I 2 = 2(1.73 × 10 7 J) (80.0 A) 2 = 5406 H.
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30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability µ is used in place of 0 .
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30.16: a) free space: U = uV = B 2 2 µ 0 V = (0.560 T) 2 2 0 (0.0290 m 3 ) = 3619 J. b) material with K m = 450 U = uV = B 2 2K m 0 V = (0.560 T) 2 2(450) 0 (0.0290 m 3 ) = 8.04 J.
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30.17: a) u = U Vol = B 2 2 µ 0 Volume = 2 0 U B 2 = 2 0 (3.60 × 10 6 J) (0.600 T) 2 = 25.1 m 3 . b) B 2 = 2 0 U Vol = 2 0 (3.60 × 10 6 J) (0.400 m) 3 = 141.4 T 2 B = 11.9 T.
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30.18: a) B = µ 0 NI 2 π r = 0 (600) (2.50 A) 2 (0.0690 m) = 4.35 mT. b) From Eq. (30.10), u = B 2 2 0 = (4.35 × 10 3 T) 2 2 0 = 7.53 J/ m 3 . c) Volume V = 2 rA = 2 (0.0690 m) (3.50 × 10 6 m 2 ) = 1.52 × 10 6 m 3 . d) U = uV = (7.53 J/ m 3 )(1.52 × 10 6 m 3 ) = 1.14 × 10 5 J. e) L = 0 N 2 A 2 r = 0 (600) 2 (3.50 × 10 6 m 2 ) 2 (0.0690 m) = 3.65 × 10 6 H. U = 1 2 LI 2 = 1 2 (3.65 × 10 6 H)(2.50 A) 2 = 1.14 × 10 5 J same as (d).
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30.19: a) di dt = ε iR L . When i = 0 di dt = 6.00 V 2.50 H = 2.40 A /s. b) When i = 1.00 A di dt = 6.00 V (0.500 A) (8.00 ) 2.50 H = 0.800 A/s. c) At t = 0.200 s i = R (1 e (R/L)t ) = 6.00 V 8.00 e (8.00 / 2.50 H)(0.250 s) ) = 0.413 A. d) As t →∞⇒ i R = 6.00 V 8.00 = 0.750 A.
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30.20: (a) i max = 30 V 1000 = 0.030 A = 30 mA, long after closing the switch.
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Chapter_30_Solutions - 30.1 a 2 = M(di1/dt =(3.25 10-4...

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