A square loop of side $l$ and resistance $R$ is placed in a uniform magnetic field B acting normally to the plane of the loop. If we attempt to pull it out of the field with a constant velocity $v$ , the power needed is

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$\frac{B^{2} l^{2} v^{2}}{R}$

$\frac{B l^{2} v^{2}}{R}$

$\frac{B v l}{R}$

$B R l v$

answer is B.

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Detailed Solution

$\begin{array}{rcl} P & = & F \cdot v = F m \cdot v \\ = & i l B \cdot v = (\frac{B v l}{R}) \cdot l B \cdot v \\ = & \frac{B^{2} l^{2} v^{2}}{R} \end{array}$

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