24. A wire of length 2 m is moving at a speed of 5ms–1 perpendicular to its length and a homogeneous magnetic field of 0.5 T. The ends of the wire are joined to a circuit of resistance 10 Ω. The rate at which work is being done to keep the wire moving at constant speed is

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512W

52W

53W

1 W

answer is B.

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Detailed Solution

Rate of work =Wt=P=Fv; also F=Bil=B BvlR l⇒P=B2v2l2R=(0.5)2×(5)2×(2)210=52W

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