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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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A conducting rod AB of length 1 m is moving with uniform speed of $2\;m/s$ in a uniform magnetic field of 4T which is directed perpendicular to the plane of motion of the rod AB. A capacitor of capacity $C= 10\;\mu F$ is connected. The charge on the capacitor is

$(a)\;zero\\ (b)\;is\; varying \\(c)\;is \;40\;\mu F \\(d)\;is 80\; \mu F $

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where $V$= is induced emf due to the motion of rod in magnetic field $V= Bvl$
$\therefore q= 10 \times 10^{-6} \times 4 \times \times 2 \times 1\;C$
$\qquad= 80 \;\mu C$
Hence d is the correct answer.
answered Mar 21, 2014 by meena.p

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