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Question

(C) The moving rod $PQ$ acts as a source of induced $emf$ $\varepsilon = Blv$ with internal resistance $R$. The circuit consists of this $emf$ source

Vedclass · Accepted Answer

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

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(C) The moving rod $PQ$ acts as a source of induced $emf$ $\varepsilon = Blv$ with internal resistance $R$. The circuit consists of this $emf$ source in series with a resistor $R$,which is then connected in parallel to two other branches,each containing a resistor $R$. Let the potential at $Q$ be $V_Q$ and at $P$ be $V_P$. The total resistance of the circuit is the internal resistance $R$ plus the equivalent resistance of the two parallel branches,which is $R/2$. Total resistance $R_{eq} = R + \frac{R 	imes R}{R + R} = R + \frac{R}{2} = \frac{3R}{2}$. The total current $I$ flowing through the rod is $I = \frac{\varepsilon}{R_{eq}} = \frac{Blv}{3R/2} = \frac{2Blv}{3R}$. Since the two parallel branches have equal resistance $R$,the current $I$ 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 $R \ \Omega$ 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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