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896_Physics ProblemsTechnical Physics

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

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236 Faraday’s Law *P31.68 (a) ε θ θ θ ω = − = − °= − = − = − = − = N d dt BA d dt B a Ba d dt Ba cos cos . . . . 1 2 0 2 1 2 1 2 0 5 0 5 2 0 125 0 125 2 2 2 2 T m rad s V V clockwise a fa f The – sign indicates that the induced emf produces clockwise current, to make its own magnetic field into the page. (b) At this instant θ ω = = = t 2 0 25 0 5 rad s s rad . . a f . The arc PQ has length r θ == = 0 5 0 5 0 25 . . . rad m m a fa f . The length of the circuit is 0 5 0 5 0 25 1 25 . . . . m m m m + + = its resistance is 1 25 6 25 . . m 5 m b g = . The current is 0 125 0 020 0 . . 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 v B × as × = − away down b g = − = 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
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