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Transcript
CHAPTER 29: Electromagnetic Induction and Faraday’s Law Solutions to Assigned Problems 1. The average induced emf is given by Eq. 29-2b. 38 Wb 58 Wb dB B e N N 2 460 V dt t 0.42s 5. Use Eq. 29-2a to calculate the emf. Setting the flux equal to the magnetic field multiplied by the area of the loop, A = p r 2 , and the emf equal to zero, we can solve for the rate of change in the coil radius. dB d dB dr e B r 2 r 2 2 Br 0 dt dt dt dt dr dB r 0.12 m 0.010 T/s 0.0012 m/s 1.2 mm/s dt dt 2 B 2 0.500 T 6. We choose up as the positive direction. The average induced emf is given by the “difference” version of Eq. 29-2a. e B t AB t 0.054 m 0.25T 0.68T 2 0.16s 5.3 102 V 21. The induced emf is given by Eq. 29-2a. Since the field is uniform and is perpendicular to the area, the flux is simply the field times the area of a circle. We calculate the initial radius from the initial area. To calculate the radius after one second we add the change in radius to the initial radius. d r2 dB dr A0 e (t ) B 2 Br A0 r02 r0 dt dt dt e (0) 2 0.28T 0.285 m 2 0.043m s 23mV 0.285 m 2 0.043m s 1.00s 0.043m/s 26 mV e (1.00s) 2 0.28T 31. The rod will descend at its terminal velocity when the magnitudes of the magnetic force (found in Example 29-8) and the gravitational force are equal. We set these two forces equal and solve for the terminal velocity. B 2 l 2 vt mg R 3 2 mgR 3.6 10 kg 9.80 m/s 0.0013 vt 2 2 0.39 m/s 2 2 B l 0.060 T 0.18 m © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 248 Electromagnetic Induction and Faraday’s Law Chapter 29 37. We find the number of turns from Eq. 29-4. The factor multiplying the sine term is the peak output voltage. epeak 24.0 V epeak NB A N 57.2 loops 2 B A 0.420 T 2 rad rev 60 rev s 0.0515 m 46. Because NS N P , this is a step-down transformer. Use Eq. 29-5 to find the voltage ratio, and Eq. 29-6 to find the current ratio. VS N S 85 turns I S N P 620 turns 0.14 7.3 VP N P 620 turns I P NS 85 turns 61. The charge on the capacitor can be written in terms of the voltage across the battery and the capacitance using Eq. 24-1. When fully charged the voltage across the capacitor will equal the emf of the loop, which we calculate using Eq. 29-2b. d B dB Q CV C CA 5.0 1012 F12 m2 10 T/s 0.60 nC dt dt 74. (a) Rfield I0 Rfield field Rarmature Rarmature I0 – + starting e (b) (c) – + I armature armature I field eback – + I0 full speed e I At startup there is no back emf. We therefore treat the circuit as two parallel resistors, each with the same voltage drop. The current through the battery is the sum of the currents through each resistor. e e 115 V 115 V I I0 I0 41.5 A Rfield Rarmature 36.0 3.00 field armature At full speed the back emf decreases the voltage drop across the armature resistor. e e eback 115V 115V 105V I I field I armature 6.53A Rfield Rarmature 36.0 3.00 © 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 249