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=+80μCandqB=–80μC
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B
qA=–80μCandqB=+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 AqA=+80μCandqB=–80μC
Q=CV=C(Bvl)=10×10–6×4×2×1=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.