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

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

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I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

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I_{1} = I_{2} = I = \frac{B l v}{R}

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I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

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Solution

The correct option is B $I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}$
A moving conductor is equivalent to a battery of emf $= v B ℓ$ (motion emf) Equivalent circuit

$I = I_{1} + I_{2}$

Applying Kirchoff's law
$I_{1} R + I R - v B l = 0$ ............... (1)
$I_{2} R + I R - v B P = 0$ ............... (2)

Adding $(1)$ and $(2)$
$2 I R + I R = 2 v B l$
$I = \frac{2 v B l}{3 R}$

Solving, we get
$I_{1} = I_{2} = \frac{v B ℓ}{3 R}$

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