# CH-30 - Chapter 30 1 The amplitude of the induced emf in...

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Unformatted text preview: Chapter 30 1. 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. m A ni e m w--= = ﾴ ﾴ ﾴ = ﾴ ± 7 p 10 2. (a) = = = = = d dt d dt t t t B 6 0 7 0 12 7 0 12 2 0 7 0 31 2 . . . . . c h bg mV. (b) Appealing to Lenz’s law (especially Fig. 30-5(a)) we see that the current flow in the loop is clockwise. Thus, the current is to left through R . 3. (a) We use = – d B / dt = – r 2 dB / dt . For 0 < t < 2.0 s: 2 2 2 0.5T 0.12m 1.1 10 V. 2.0s dB r dt-ﾴ ﾴ =-=-=- ﾴ ﾴ ﾴ ﾴ p p (b) 2.0 s < t < 4.0 s: dB / dt = 0. (c) 4.0 s < t < 6.0 s: =-=---F H G I K J = -ﾴ ﾴ r dB dt 2 2 2 012 05 6 0 4 0 11 10 . . . . . . m T s s V b g 4. The resistance of the loop is 8 3 2 m 1.69 10 m 1.1 10 . / 4 L R A r--= = W = W ﾴ ± 3 . 010 . 25 10 We use i = | | / R = |d B / dt| / R = ( r 2 / R )| dB / dt |. Thus 3 2 2 10A 1.1 10 1.4 T s. m dB iR dt r-ﾴ W = = = . 005 5. The total induced emf is given by 1201 CHAPTER 30 2 2 ( ) ( ) 1.5 A (120)(4 T m A)(22000/m) 0.016m 0.025 s 0.16V. B d dB d di di N NA NA ni N nA N n r dt dt dt dt dt e m =-=-=-=-=-=-ﾴ ﾴ = ± 7 p 10 p Ohm’s law then yields | |/ 0.016 V / 5.3 0.030 A i R = = W= . 6. Using Faraday’s law, the induced emf is 2 2 2 0.12m 0.800T 0.750m/s 0.452V. B d r d BA d dA dr B B rB dt dt dt dt dt =-=-=-=-=- =- -= 7. The flux B BA = cos does not change as the loop is rotated. Faraday’s law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is 0. 8. 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. 9. (a) Let L be the length of a side of the square circuit. Then the magnetic flux through the circuit is B L B = 2 2 / , and the induced emf is 2 . 2 B i d L dB dt dt =-=-Now B = 0.042 – 0.870 t and dB / dt = –0.870 T/s. Thus, i = ( . ( . / 2 00 2 0870 m) T s) = 1.74 V. 2 The magnetic field is out of the page and decreasing so the induced emf is counterclockwise around the circuit, in the same direction as the emf of the battery. The total emf is + i = 20.0 V + 1.74 V = 21.7 V. (b) The current is in the sense of the total emf (counterclockwise). 178 10. Fig. 30-41(b) demonstrates that (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday’s law becomes e = -= -A where A = 8 ×10-4 m 2 . We related the induced emf to resistance and current using Ohm’s law. The current is estimated from Fig. 30-41(c) to be i = = 0.002 A (the slope of that line). Therefore, the resistance of the loop is R = | | / i = = 0.0012 W ....
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## This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.

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CH-30 - Chapter 30 1 The amplitude of the induced emf in...

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