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Standard X

Physics

Faraday's Law

A conducting ...

Question

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

q A = + 40 μ C, q B = + 40 μ C

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q A = + 40 μ C, q B = - 40 μ C

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q A = - 40 μ C, q B = + 40 μ C

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q A = q B = 0

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Solution

The correct option is A $q A = + 40 μ C, q B = + 40 μ C$
Motional EMF, $v = B l V$
$= 2 \times 1 \times 2$
$= 4 V$
For a Capacitor,
$Q = C V$
$= ∠ O P F \times 4 V$
$Q = 40 μ C$
The induced any opposes the change that taking place.

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

The correct option is A qA=+40μC,qB=+40μCMotional EMF, v=BlV =2×1×2 =4VFor a Capacitor,Q=CV=∠OPF×4VQ=40μCThe induced any opposes the change that taking place.

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

The correct option is A $q A = + 40 μ C, q B = + 40 μ C$
Motional EMF, $v = B l V$
$= 2 \times 1 \times 2$
$= 4 V$
For a Capacitor,
$Q = C V$
$= ∠ O P F \times 4 V$
$Q = 40 μ C$
The induced any opposes the change that taking place.