Consider parallel conducting rails separated by a distance l. There exists a uniform magnetic field B perpendicular to the plane of the rails as shown in figure. Two conducting wires each of length l are placed so as to slide on parallel conducting rails. One of the wires is given a velocity $v_{0}$ parallel to the rails. Till steady state is achieved, loss in kinetic energy of the system is

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zero

$\frac{3}{8} {mv}_{0}^{2}$

$\frac{3}{4} {mv}_{0}^{2}$

$\frac{1}{4} {mv}_{0}^{2}$

answer is C.

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Detailed Solution

In steady state, velocity of both wires will become same.

$2 mv = m v_{0} \Rightarrow v = v_{0} / 2$

Loss in KE $= \frac{1}{2} {mv}_{0}^{2} - 2 [\frac{1}{2} {mv}^{2}] = \frac{1}{4} {mv}_{0}^{2}$

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