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A circular loop of radius 1m is kept in a magnetic field of strength 2 T directed perpendicular to the plane of loop. Resistance of the loop wire is $2 / π Ω / m .$ A conductor of length 2 m is sliding with a speed 1 m/s as shown in the figure. Find the instantaneous force acting on the rod in Newtons [Assume that the rod has negligible resistance]

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answer is 16.

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Detailed Solution

$\begin{array}{l} B v (L) = I R \\ F = B I L \\ F = B^{2} v L^{2} / R \end{array}$

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