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$(C)$ The moving rod $PQ$ acts as a motional $EMF$ source with $\varepsilon = Blv$. The resistance of the rod is $R$. The circuit consists of the rod

Vedclass · Accepted Answer

$I_1 = I_2 = \frac{Blv}{3R}, I = \frac{2Blv}{3R}$

Vedclass · Answer

$(C)$ The moving rod $PQ$ acts as a motional $EMF$ source with $\varepsilon = Blv$. The resistance of the rod is $R$. The circuit consists of the rod $PQ$ in series with two parallel branches, each of resistance $R$. The equivalent resistance of the two parallel branches is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$. The total resistance of the circuit is $R_{eq} = R + R_p = R + \frac{R}{2} = \frac{3R}{2}$. The total current $I$ flowing through the rod $PQ$ is $I = \frac{\varepsilon}{R_{eq}} = \frac{Blv}{3R/2} = \frac{2Blv}{3R}$. Since the two parallel branches have equal resistance $R$, the current $I$ divides equally between them. Therefore, $I_1 = I_2 = \frac{I}{2} = \frac{1}{2} 	imes \frac{2Blv}{3R} = \frac{Blv}{3R}$.

$A$ 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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