A rod PQ of mass ‘m’ and resistance ‘r’ is moving on two fixed, resistance less, smooth conducting rails (closed on both sides by resistance R1 and R2). Then the current in the rod at the instant when its velocity V is (A uniform magnetic field B→ exists in the region)

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i=BlVR1−R2

i=BlVr+R1+R2

i=BlVR1+R2

i=BlVR1+R2R1R2+rR1+R2

answer is D.

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Detailed Solution

Given mass= mEquivalent Resistance of R1 and R2 in parallel =R1R2R1+R2Velocity= VWe know current i=er+ReqAlso, e=BlVThen , i=BlVr+R1R2R1+R2 Therefore, the correct answer is (D).

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