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# Chapter-30 - Chapter 30 1(a The magnitude of the emf is = d...

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1175 Chapter 30 1. (a) The magnitude of the emf is ε = = + = + = + = d dt d dt t t t B Φ 6 0 7 0 12 7 0 12 2 0 7 0 31 2 . . . . . c h b g 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 . 2. (a) We use ε = – d Φ B / dt = – π r 2 dB / dt . For 0 < t < 2.0 s: ( 29 2 2 2 0.5T 0.12m 1.1 10 V. 2.0s dB r dt ε - = - = - = - × p p (b) For 2.0 s < t < 4.0 s: ε dB / dt = 0. (c) For 4.0 s < t < 6.0 s: ε = - = - - - F H G I K J = × - box4 box4 r dB dt 2 2 2 012 05 6 0 4 0 11 10 . . . . . . m T s s V b g 3. The amplitude of the induced emf in the loop is 6 2 0 0 4 (6.8 10 m )(4 T m A)(85400/ m)(1.28 A)(212 rad/s) 1.98 10 V. m A ni ε μ ϖ - - = = × × = × - 7 p 10 4. Using Faraday’s law, the induced emf is ( 29 ( ( 29( 29( 29 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 ε π Φ = - = - = - = - = - π = - π - = 5. The total induced emf is given by

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CHAPTER 30 1176 ( 29 2 0 0 0 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 ε μ μ μ π Φ = - = - = - = - = - = - × = - 7 p 10 p Ohm’s law then yields | |/ 0.016 V /5.3 0.030 A i R ε = = Ω = . 6. The resistance of the loop is ( 29 ( ( 29 8 3 2 m 1.69 10 m 1.1 10 . m / 4 L R A π ρ π - - -3 0.10 = = × Ω⋅ = × 2.5×10 We use i = | ε | / R = |d Φ B / dt| / R = ( π r 2 / R )| dB / dt |. Thus ( ( ( 29 3 2 2 10A 1.1 10 1.4 T s. m dB iR dt r π π - × = = = 0.05 7. 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. 8. From the datum at t = 0 in Fig. 30-41(b) we see 0.0015 A = V battery /R , which implies that the resistance is R = (6.00 μ V)/(0.0015 A) = 0.0040 . Now, the value of the current during 10 s < t < 20 s leads us to equate ( V battery + ε induced ) /R = 0.00050 A. This shows that the induced emf is ε induced = - 4.0 μ V. Now we use Faraday’s law: ε = - d Φ B dt = - A dB dt = - A a . Plugging in ε = - 4.0 × 10 - 6 V and A = 5.0 × 10 - 4 m 2 , we obtain a = 0.0080 T/s. 9. The flux Φ B BA = cos θ 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 zero.
1177 10. Fig. 30-43(b) demonstrates that / dB dt (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday’s law becomes ( ) B d d BA dB A dt dt dt ε Φ = - = - = - 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. 30-43(c) to be i = / dq dt = 0.002 A (the slope of that line). Therefore, the resistance of the loop is 4 2 | | | / | (8.0 10 m )(0.0030 T/s) 0.0012 0.0020 A A dB dt R i i ε - × = = = = . 11. (a) Let L be the length of a side of the square circuit. Then the magnetic flux through the circuit is Φ B L B = 2 2 / , and the induced emf is 2 . 2 B i d L dB dt dt ε Φ = - = - Now B = 0.042 – 0.870 t and dB / dt = –0.870 T/s. Thus, ε 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 ε + ε i = 20.0 V + 1.74 V = 21.7 V.

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