230 Faraday’s LawP31.50The emf induced between the ends of the moving bar isε===B vA2 500 3508 007 00....TmmsVafafbg.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:+−=7 002 0001..VIΩafI13 50=.Aand+−=7 005 0003..VIΩafI31 40=.A.(b)The total power dissipated in the resistors of the circuit isPIIII=+=+=+=εεε13137 003 501 4034 3bg afaf....VAAW.(c)Method 1: The current in the sliding conductor is downward with valueI23 501 404 90=+=...AAA . The magnetic field exerts a force ofFI Bm===A4 900 3502 504 29....AmTNafafafdirected toward the right on thisconductor. An outside agent must then exert a force of
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