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890_Physics ProblemsTechnical Physics

890_Physics ProblemsTechnical Physics - 230 P31.50 Faradays...

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230 Faraday’s Law P31.50 The emf induced between the ends of the moving bar is ε = = = B v A 2 50 0 350 8 00 7 00 . . . . T m m s V a fa fb g . The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I 1 represent the current flowing upward through the 2.00- resistor. The right-hand loop will carry counterclockwise current. Let I 3 be the upward current in the 5.00- resistor. (a) Kirchhoff’s loop rule then gives: + = 7 00 2 00 0 1 . . V I a f I 1 3 50 = . A and + = 7 00 5 00 0 3 . . V I a f I 3 1 40 = . A . (b) The total power dissipated in the resistors of the circuit is P I I I I = + = + = + = ε ε ε 1 3 1 3 7 00 3 50 1 40 34 3 b g a fa f . . . . V A A W . (c) Method 1 : The current in the sliding conductor is downward with value I 2 3 50 1 40 4 90 = + = . . . A A A . The magnetic field exerts a force of F I B m = = = A 4 90 0 350 2 50 4 29 . . . . A m T N a fa fa f directed toward the right on this conductor. An outside agent must then exert a force of
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