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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=10μF is connected as shown in figure. Then




A
qA=+80μC and qB=80μC
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B
qA=80μC and qB=+80μC
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C
qA=0=qB
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D
Charge stored in the capacitor increases exponentially with time
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Solution

The correct option is A qA=+80μC and qB=80μC

Q=CV=C(Bvl)=10×106×4×2×1=80μC
According to Fleming's right hand rule induced current flows from Q to P. Hence P is at higher potential and Q is at lower potential. Therefore A is positively charged and B is negatively charged.



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