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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 mangetic field going into the plane of the paper. The three currents l₁, I₂ and I are

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

$I_{1} = I_{2} = \frac{B / V}{3 R}, I = \frac{2 B / v}{3 R}$

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

$I_{1} = I_{2} = \frac{B / V}{6 R}, I = \frac{B / V}{3 R}$

answer is B.

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Detailed Solution

A moving conductor is equivalent to a battery of
emf = vBl (motion emf )
Equivalent circuit ; l = l₁+ l₂
Applying Kirchhoff’s law

$l_{1} R + I R - v B I = 0 - - - - - (i) l_{2} R + I R - v B I = 0 - - - - - (i i)$

Adding Eqs. (i) and (ii)

$\begin{array}{l} 2 IR + IR = 2 vBI; I = \frac{2 vBI}{3 R} \end{array}$

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