896_Physics ProblemsTechnical Physics

896_Physics ProblemsTechnical Physics - 236 *P31.68...

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236 Faraday’s Law *P31.68 (a) εθ θ ω =− °=− = N d dt BA d dt B a Ba d dt Ba cos cos . . .. 1 2 0 2 1 2 1 2 05 2 0125 2 2 2 2 T m r a d s V V clockwise af a f The – sign indicates that the induced emf produces clockwise current, to make its own magnetic field into the page. (b) At this instant θω == = t 20 2 5 0 5 rad s s rad . The arc PQ has length r == = 025 ... rad m m a f . The length of the circuit is 0 5 0 5 0 25 1 25 . m m m m ++ = its resistance is 125 625 m5 m ΩΩ bg = . The current is 00200 . . V 6.25 A clockwise = . *P31.69 Suppose the field is vertically down. When an electron is moving away from you the force on it is in the direction given by q c vB × as −× = away down = left right . Therefore, the electrons circulate clockwise. FIG. P31.69 (a) As the downward field increases, an emf is induced to produce some current that in turn produces an upward field. This current is directed counterclockwise, carried by negative electrons moving clockwise. Therefore the original electron motion speeds up. (b) At the circumference, we have Fm a cc = : qvB mv r c sin90 2 °=
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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