MCQ
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 
  • A
    $I_1 = I_2 = 3A, I = 1A$
  • B
    $I_1 = I_2 = 5A, I = 2A$
  • $I_1 = I_2 = 1A, I = 2A$
  • D
    $I_1 = I_2 = I = 2A$

Answer

Correct option: C.
$I_1 = I_2 = 1A, I = 2A$
c
$I=\frac{B V \ell}{15}$

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

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