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Q.

Consider smooth parallel conducting fixed rails separated by a distance 'l'. There exists uniform magnetic field ‘B’ perpendicular to the plane of the rails as shown in figure. Two conducting wires each of length 'l'  and mass  m are placed perpendicular to rails, so as to slide on parallel conducting rails. One of the wires is given a velocity v0 parallel to the rails. Till steady state is achieved, loss in kinetic energy of the system is  mv02n. Then find 'n'?

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answer is 4.

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Detailed Solution

ΔKE=12μv02=12.m2.v02=mv024

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Consider smooth parallel conducting fixed rails separated by a distance 'l'. There exists uniform magnetic field ‘B’ perpendicular to the plane of the rails as shown in figure. Two conducting wires each of length 'l'  and mass  m are placed perpendicular to rails, so as to slide on parallel conducting rails. One of the wires is given a velocity v0 parallel to the rails. Till steady state is achieved, loss in kinetic energy of the system is  mv02n. Then find 'n'?