Homework 10 Solutions

Homework 10 Solutions - Chapter 30 1 The flux B = BA cos...

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1165 Chapter 30 1. 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. 2. Using Faraday’s law, the induced emf is () ( ) ( ) ( ) 2 2 2 0.12m 0.800T 0.750m/s 0.452V. B dr dB A dd A d r BB r B dt dt dt dt dt π επ Φ =− = 3. The total induced emf is given by 2 00 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 NN AN A n i N n A N n r dt dt dt dt dt εμ μ Φ ⎛⎞ ⎜⎟ ⎝⎠ × = -7 p1 0 p Ohm’s law then yields | |/ 0.016 V / 5.3 0.030 A iR ε == Ω = . 4. (a) We use = – d Φ B / dt = – π r 2 dB / dt . For 0 < t < 2.0 s: 2 22 0.5T 0.12m 1.1 10 V. 2.0s dB r dt × pp (b) For 2.0 s < t < 4.0 s: dB / dt = 0. (c) For 4.0 s < t < 6.0 s: F H G I K J r dB dt 2 2 2 012 05 60 40 11 10 . . .. m T ss V bg 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.
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CHAPTER 30 1166 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 μ V)/(0.0015 A) = 0.0040 Ω . 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 induced = 4.0 μ V. Now we use Faraday’s law: = d Φ 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 dt d dt tt t B Φ 60 70 12 70 12 20 70 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 the left through R . 8. The resistance of the loop is () ( ) 83 2 m 1.69 10 m 1.1 10 . m/ 4 L R A π ρ −− −3 0.10 == × Ω = × Ω 2.5×10 We use i = | | / R = |d Φ 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 ×Ω = 0.05 9. The amplitude of the induced emf in the loop is 62 00 4 (6.8 10 m )(4 T m A)(85400/ m)(1.28 A)(212 rad/s) 1.98 10 V. m An i εμ ω × × -7 p1 0 10. (a) The magnetic flux Φ B through the loop is given by ( )( ) 2 22 c o s 4 5 B Br Φ= π °= 2 2 rB π .
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1167 Thus, ( ) 2 2 22 3 3 2 3.7 10 m 07 61 0T 4.5 10 s 2 5.1 10 V. B dd r B r B dt dt t π ππ ε × ⎛⎞ ΦΔ− × =− ⎜⎟ Δ× ⎝⎠ (a) The direction of the induced current is clockwise when viewed along the direction of G B . 11. (a) It should be emphasized that the result, given in terms of sin(2 ft ), could as easily be given in terms of cos(2 ft ) or even cos(2 ft + φ ) where is a phase constant as discussed in Chapter 15. The angular position θ of the rotating coil is measured from some reference line (or plane), and which line one chooses will affect whether the magnetic flux should be written as BA cos , BA sin or BA cos( + ). Here our choice is such that Φ B BA = cos . Since the coil is rotating steadily, increases linearly with time.
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Homework 10 Solutions - Chapter 30 1 The flux B = BA cos...

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