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A conducting rod PQ of length 1m is moving with uniform velocity of 2m/s in a uniform magnetic field of 2T directed into the plane of paper. A capacitor of capacity $C = 10 μF$ is connected as shown. Then

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$q_{A} = - 40 μC, q_{B} = + 40 μC$

$q_{A} = q_{B} = 0$

$q_{A} = + 40 μC, q_{B} = - 40 μC$

$q_{A} = + 40 μC, q_{B} = + 40 μC$

answer is B.

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Detailed Solution

$e_{induced} = Blv = 4 V$
Q=C $e_{induced}$

$\Rightarrow Q = 40 μC$
Apply right hand rule to know direction of induced current

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