A rectangular loop has a sliding connector PQ | vedclass

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

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 internal resistance of the rod is $R$. The circuit consists of the $EMF$ source $\varepsilon$ in series with its internal resistance $R$, which is then connected to two parallel branches, each having a resistance $R$. The equivalent resistance of the two parallel branches is $R_{eq}' = \frac{R 	imes R}{R + R} = \frac{R}{2}$. The total resistance of the circuit is $R_{total} = R + R_{eq}' = R + \frac{R}{2} = \frac{3R}{2}$. The total current $I$ flowing through the rod $PQ$ is $I = \frac{\varepsilon}{R_{total}} = \frac{Blv}{3R/2} = \frac{2Blv}{3R}$. Since the two parallel branches have equal resistance $R$, the current $I$ splits equally between them: $I_1 = I_2 = \frac{I}{2} = \frac{1}{2} 	imes \frac{2Blv}{3R} = \frac{Blv}{3R}$. Thus, $I_1 = I_2 = \frac{Blv}{3R}$ and $I = \frac{2Blv}{3R}$.

$A$ rectangular loop has a sliding connector $PQ$ of length $l$ and resistance $R \, \Omega$. It is moving with a speed $v$ as shown. The setup is placed in a uniform magnetic field directed into the plane of the paper. The three currents $I_1$, $I_2$, and $I$ are:

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