In the figure, magnetic field points into the plane of paper and the conducting rod of length $l$ is moving in this field such that the lowest point has a velocity $v_{1}$ and the topmost point has the velocity $v_{2} (v_{2} > v_{1})$ . The emf induced is given by

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$B v_{1} l$

$B v_{2} l$

$\frac{1}{2} B (v_{1} + v_{2}) l$

$\frac{1}{2} B (v_{2} - v_{1}) l$

answer is C.

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Detailed Solution

$v = v_{1} + (\frac{v_{2} - v_{1}}{l}) x$

Small potential difference $= B v (d x)$

$∴$ Total potential difference $= \int_{0}^{l} B v d x = \frac{1}{2} B (v_{1} + v_{2}) l$

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