$A$ 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:

  • A
    $I_1 = I_2 = \frac{Bvl}{6R}, I = \frac{Blv}{R}$
  • B
    $I_1 = -I_2 = \frac{Bvl}{R}, I = 2\frac{Blv}{R}$
  • C
    $I_1 = I_2 = \frac{Bvl}{3R}, I = \frac{2Blv}{3R}$
  • D
    $I_1 = I_2 = I = \frac{Blv}{R}$

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