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(D) The sliding wire $PQ$ acts as a motional electromotive force (emf) source with $\varepsilon = B l v$. This emf source is in series with the resist

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$\frac{2 B l v}{3 R}$

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(D) The sliding wire $PQ$ acts as a motional electromotive force (emf) source with $\varepsilon = B l v$. This emf source is in series with the resistance $R$ of the wire $PQ$. This combination is connected in parallel with the two external resistors,each of resistance $R$. First,calculate the equivalent resistance of the two external resistors connected in parallel: $R_{p} = \frac{R 	imes R}{R + R} = \frac{R}{2}$. Now,the total resistance of the circuit is the sum of the resistance of the wire $PQ$ and the equivalent parallel resistance: $R_{eq} = R + R_{p} = R + \frac{R}{2} = \frac{3 R}{2}$. Finally,the current $I$ flowing through the wire $PQ$ is given by Ohm's law: $I = \frac{\varepsilon}{R_{eq}} = \frac{B l v}{3 R / 2} = \frac{2 B l v}{3 R}$.

$A$ rectangular loop circuit has a sliding wire $PQ$ as shown in the figure. The loop is placed in a magnetic field $B$,perpendicular to its plane. The resistance of the wire $PQ$ is $R$. If the wire moves with constant velocity $v$,then find the current flowing in the wire $PQ$?

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