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A magnetic field vecB=B_0 sin (omegat)hatkcovers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d as in figure. The wires are in the xy-plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity ? |
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Answer» Solution :SUPPOSE wire AB is at x =x at .t. TIME, Induced EMF in wire AB, `epsilon=(dphi)/(dt)` `=d/(dt)(AB)` `=d/(dt)[(dx).(B_0 sin (omegat))]` `=dB_0 d/(dt)[x xx sin (omegat)]` `=dB_0 [x d/(dt) sin (omegat)+ sin(omegat)(dx)/(dt)]` `=dB_0[xomegacos (omegat)+VSIN(omegat)]` Now induced current , `I=epsilon/R` `I=(B_0d)/R (x omegacos (omegat)+v sin (omegat)]` Force required to move rod with constant velocity F=IlB `F=I(B_0d)/R [ xomegacos(omegat)+v sin (omegat)](d)[B_0 sin (omegat)]` `F=(B_0^2d^2)/R [ xomega cos (omegat)+ vsin (omegat)] xx sin omegat` |
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