Problem%20Set%206%20Solutions

Problem%20Set%206%20Solutions - 10. Fig. 30-43(b)...

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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 dBA 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 42 | | | / | (8.0 10 m )(0.0030 T/s) 0.0012 0.0020 A AdBd t R ii × == = = .
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13. (a) It should be emphasized that the result, given in terms of sin(2 π ft ), could as easily be given in terms of cos(2 π ft ) or even cos(2 π 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. Thus, = ω t (equivalent to = 2 π ft ) if is understood to be in radians (and would be the angular velocity). Since the area of the rectangular coil is A=ab , Faraday’s law leads to () ( ) cos cos 2 2s i n 2 dB A d f t NN B A N B a b f f t dt dt ε π =− = π π which is the desired result, shown in the problem statement. The second way this is written ( 0 sin(2 π ft )) is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of 0 = 2 π f N abB . (b) We solve 0 = 150 V = 2 π f N abB when f = 60.0 rev/s and B = 0.500 T. The three unknowns are N, a , and b which occur in a product; thus, we obtain N ab = 0.796 m 2 .
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Faraday’s law, then, (with SI units and 3 significant figures understood) leads to () 00 2 0 0 ln ln 22 9 ln 10 91 0 ln . 2 B ib b dd a a d i dt dt b a b a dt b ad tt bad t bt a ba µµ ε µ ⎡⎤ Φ ⎛⎞ =− ⎜⎟ ⎢⎥ π− ⎝⎠ ⎣⎦ −− = With a = 0.120 m and b = 0.160 m, then, at t = 3.00 s, the magnitude of the emf induced in the rectangular loop is = ×− F H G I K J 4 10 016 9 3 10 2 012 016 012 598 10 7 7 π π ch bg b g . ln . .. V (b) We note that /0 di dt > at t = 3 s. The situation is roughly analogous to that shown in Fig. 30-5(c). From Lenz’s law, then, the induced emf (hence, the induced current) in the loop is counterclockwise. 24. (a) First, we observe that a large portion of the figure contributes flux which “cancels out.” The field (due to the current in the long straight wire) through the part of the rectangle above the wire is out of the page (by the right-hand rule) and below the wire it is into the page. Thus, since the height of the part above the wire is b – a , then a strip below the wire (where the strip borders the long wire, and extends a distance b – a away from it) has exactly the equal-but-opposite flux which cancels the contribution from the part above the wire. Thus, we obtain the non-zero contributions to the flux: ln .
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This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

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Problem%20Set%206%20Solutions - 10. Fig. 30-43(b)...

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