A rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R$ $\Omega $ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents ${I_1},{I_2}$ and $I$ are

Question

${I_1} = - {I_2} = {{Blv} \over {6R}},\,\,I = {{2Blv} \over {6R}}$

${I_1} = {I_2} = {{Blv} \over {3R}},\,\,I = {{2Blv} \over {3R}}$

${I_1} = {I_2} = I = {{Blv} \over R}$

${I_1} = {I_2} = {{Blv} \over {6R}},I = {{Blv} \over {3R}}$

Solution

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