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Question

A conducting rod PQ of length l=1.0 m is moving with a uniform speed v=2 m/s in a uniform magnetic field B=4.0 T directed into the paper. A capacitor of capacity C=10 μF is connected as shown in the figure. Then charge on plate A(in μC) is

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Solution

According to Flemings right-hand rule induced current flows from P to Q. Hence P is at higher potential and Q is at lower potential. Therefore A is positively charged and B is negatively charged.


Due to change in magnetic flux, potential difference generated in the loop is V=Bvl
Charge on capacitor is
Q=CV=C(Bvl)=10×106×4×2×1=80 μC

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