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JEE Main 2010 (AIEEE) · Motional EMF

Motional EMF question from JEE Main 2010 (AIEEE)

Q55JEE Main 2010 · AIEEEDifficult

Motional EMF

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

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Answer: (b)

PQ acts as a cell of emf $Blv$ and internal resistance $R$ .
The two side resistors $R$ are in parallel: $\dfrac{R}{2}$ .
$I = \dfrac{Blv}{R + R/2} = \dfrac{2Blv}{3R}$ , and it splits equally: $I_1 = I_2 = \dfrac{Blv}{3R}$
So the answer is (b).

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