Phys 181b Problem Set 7 Solution

Phys 181b Problem Set 7 Solution - 11(a It should be...

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11. (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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24. (a) We assume the flux is entirely due to the field generated by the long straight wire (which is given by Eq. 29-17). We integrate according to Eq. 30-1, not worrying about the possibility of an overall minus sign since we are asked to find the absolute value of the flux. /2 00 2 2 || () l n . 22 b rb B b r ii a adr rr µµ + §· + Φ= = ¨¸ ππ ©¹ ³ When 1.5 = , we have 8 (4 T m A)(4.7A)(0.022m) | | ln(2.0) 1.4 10 Wb. 2 B π ×⋅ = × −7 π1 0 (b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and recognizing that dr dt v = . The magnitude of the induced emf divided by the loop resistance then gives the induced current: 2 loop 2 3 42 5 ln [ ( / 2 ) ] (4 T m A)(4.7A)(0.022m)(0.0080m)(3.2 10 m/s) 2 (4.0 10 )[2(0.0080m) ] 1.0 10 A. b b r ia iabv d i RR d t r R r b ε + == = π− π × = ×Ω −7 0
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26. Noting that | B | = B , we find the thermal energy is 2 2 22 2 thermal 42 2 6 2 63 10 11 ( 2 . 0
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

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Phys 181b Problem Set 7 Solution - 11(a It should be...

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