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A conducting rod of mass $m$ and length $\ell$ is placed over a smooth table. A uniform magnetic field is acting perpendicular to the rod. Charge $q$ is suddenly passed through the rod and it acquires a velocity $v$ on the surface, then $q$ is equal to

$\begin {array} {1 1} (a)\;\large\frac{2mv}{B\ell} & \quad (b)\;\large\frac{mv}{B\ell} \\ (c)\;\large\frac{B\ell}{mv} & \quad (d)\;\large\frac{B\ell}{2mv} \end {array}$


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Impulse = Change in linear momentum
$\int F\: dt = \int \: (i\ell B)\: dt = mv$
$\int \: i\: dt = q$
$mv = B\ell q$
Ans : (b)
answered Feb 22, 2014 by thanvigandhi_1

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