A rectangular loop has a sliding connector PQ of length l and resistance RΩ 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 I1, I2 and I are

\[I_{1}=I_{2}=I=\frac{Blv}{R}\]
\[I_{1}=I_{2}=\frac{Blv}{6R}=I=\frac{Blv}{3R}\]
\[I_{1}=-I_{2}=\frac{Blv}{R}=I=\frac{2Blv}{3R}\]
\[I_{1}=I_{2}=\frac{Blv}{3R}=I=\frac{2Blv}{3R}\]
No solution provided for this question.
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