A rectangular loop has a sliding connector PQ of length $ℓ$ and resistance $R Ω$ 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

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A moving conductor is equivalent to a battery of emf $= vB ℓ$ (motion emf) Equivalent circuit $I = l_{1} + l_{2}$ applying Kirchoff's law $I_{1} R + I R - v B ℓ = 0 I_{2} R + I R - v B ℓ = 0$ adding (1) \& (2) $2 IR + IR = 2 vB ℓ I = \frac{2 vB ℓ}{3 R} I_{1} = I_{2} = \frac{vB ℓ}{3 R}$

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