884_Physics ProblemsTechnical Physics

# 884_Physics ProblemsTechnical Physics - 224 P31.29 Faradays...

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224 Faraday’s Law P31.29 (a) The force on the side of the coil entering the field (consisting of N wires) is F N ILB N IwB == af a f . The induced emf in the coil is ε = N d dt N dBwx dt NBwv B Φ . so the current is I R NBwv R counterclockwise. The force on the leading side of the coil is then: FN NBwv R wB NBwv R = F H G I K J = 22 2 to the left . (b) Once the coil is entirely inside the field, Φ B NBA constant , so = 0, I = 0 , and F = 0 . FIG. P31.29 (c) As the coil starts to leave the field, the flux decreases at the rate Bwv , so the magnitude of the current is the same as in part (a), but now the current is clockwise. Thus, the force exerted on the trailing side of the coil is: F R = to the left again . P31.30 Look in the direction of ba . The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current. Therefore, current
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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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