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 == = Bv A 2 50 0 350 8 00 7 00 .. . . T m m s V a fa f bg . 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: +− = 700 200 0 1 V I af I 1 350 = . A and = 500 0 3 I I 3 140 = A . (b) The total power dissipated in the resistors of the circuit is PI I II =+= += + = εεε 13 1 3 343 a f ... . A A W . (c) Method 1 : The current in the sliding conductor is downward with value I 2 490 . A A A . The magnetic field exerts a force of FI B m = A 4 90 0 350 2 50 4 29 . . A m T N a f directed toward the right on this conductor. An outside agent must then exert a force of 429
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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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