230 Faraday’s LawP31.50The emf induced between the ends of the moving bar isε===BvA2 500 3508 007 00....T mms Vafafbg.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 I1represent the current flowingupward through the 2.00-Ωresistor. The right-hand loop will carry counterclockwise current. Let I3be the upward current in the 5.00-Ωresistor.(a)Kirchhoff’s loop rule then gives:+−=70020001V IΩafI1350=. Aand=50003IΩI3140=A.(b)The total power dissipated in the resistors of the circuit isPI III=+=+=+=εεε1313343af....AAW.(c)Method 1: The current in the sliding conductor is downward with valueI2490.AAA . The magnetic field exerts a force ofFIBm=A4 900 3502 504 29..A mTNafdirected toward the right on thisconductor. 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 .