Question
A rectangular loop has a sliding connector $PQ$ of length $2\ m$ and resistance $10\Omega $ and it is moving with a speed $5\ m/s$ as shown. The set-up is placed in a uniform magnetic field $3T$ going into the plane of the paper. The three currents $I_1$ , $I_2$ and $I$ are 

Answer

$I=\frac{B V \ell}{15}$

$I_{1}=I_{2}=\frac{I}{2}$

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