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Unformatted text preview: 1165 Chapter 30 1. The flux B BA cos does not change as the loop is rotated. Faradays law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is zero. 2. Using Faradays law, the induced emf is d B dt d BA dt B dA dt B d r 2 dt 2 rB dr dt 2 0.10m 0.800T 0.750m/s 0.377V. 3. The total induced emf is given by N d B dt NA dB dt NA d dt ( ni ) N nA di dt N n ( r 2 ) di dt (120)(4 T m A)(22000/m) 0.016m 2 1.5 A 0.025 s 0.16V. Ohms law then yields | | / 0.016 V / 5.3 0.030 A i R . 4. (a) We use = d B / dt = r 2 dB / dt . For 0 < t < 2.0 s: r 2 dB dt 0.12m 2 0.5T 2.0s 1.1 10 2 V. (b) For 2.0 s < t < 4.0 s: dB / dt = 0. (c) For 4.0 s < t < 6.0 s: r 2 dB dt 0.12m 2 0.5T 6.0s 4.0s 1.1 10 2 V. CHAPTER 30 1166 5. The field (due to the current in the straight wire) is out of the page in the upper half of the circle and is into the page in the lower half of the circle, producing zero net flux, at any time. There is no induced current in the circle. 6. From the datum at t = 0 in Fig. 30-35(b) we see 0.0015 A = V battery /R , which implies that the resistance is R = (6.00 V)/(0.0015 A) = 0.0040 . Now, the value of the current during 10 s < t < 20 s leads us to equate ( V battery + induced ) /R = 0.00050 A. This shows that the induced emf is induced = 4.0 V. Now we use Faradays law: = d B dt = A dB dt = A a . Plugging in = 4.0 10 6 V and A = 5.0 10 4 m 2 , we obtain a = 0.0080 T/s. 7. (a) The magnitude of the emf is d B dt d dt 6.0 t 2 7.0 t 12 t 7.0 12 2.0 7.0 31mV. (b) Appealing to Lenzs law (especially Fig. 30-5(a)) we see that the current flow in the loop is clockwise. Thus, the current is to the left through R . 8. The resistance of the loop is R L A 1.69 10 8 m m m 2 / 4 1.3 10 3 . We use i = | | / R = |d B / dt| / R = ( r 2 / R )| dB / dt |. Thus dB dt iR r 2 10A 1.3 10 3 m 2 1.7 T s . 9. The amplitude of the induced emf in the loop is 6 2 4 (6.8 10 m )(4 T m A)(85400 / m)(1.28 A)(212 rad/s) 1.98 10 V....
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- Spring '11