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(C) The induced electromotive force $(EMF)$ in the moving rod $PQ$ is given by $\varepsilon = Bvl = 3\ T 	imes 5\ m/s 	imes 2\ m = 30\ V$.The rod $P

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$I_1 = I_2 = 1\ A, I = 2\ A$

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(C) The induced electromotive force $(EMF)$ in the moving rod $PQ$ is given by $\varepsilon = Bvl = 3\ T 	imes 5\ m/s 	imes 2\ m = 30\ V$.The rod $PQ$ acts as a battery of $30\ V$ with an internal resistance of $10\ \Omega$. The circuit consists of two parallel branches,each with a resistance of $10\ \Omega$,connected in series with the rod's internal resistance.The equivalent resistance of the two parallel branches is $R_p = \frac{10\ \Omega 	imes 10\ \Omega}{10\ \Omega + 10\ \Omega} = 5\ \Omega$.The total resistance of the circuit is $R_{total} = R_{internal} + R_p = 10\ \Omega + 5\ \Omega = 15\ \Omega$.The total current $I$ flowing through the rod $PQ$ is $I = \frac{\varepsilon}{R_{total}} = \frac{30\ V}{15\ \Omega} = 2\ A$.Since the two parallel branches have equal resistance ($10\ \Omega$ each),the current $I$ splits equally between them: $I_1 = I_2 = \frac{I}{2} = \frac{2\ A}{2} = 1\ A$.

$A$ rectangular loop has a sliding connector $PQ$ of length $2\ m$ and resistance $10\ \Omega$. It is moving with a speed $5\ m/s$ as shown in the figure. The set-up is placed in a uniform magnetic field of $3\ T$ directed into the plane of the paper. Find the three currents $I_1$,$I_2$,and $I$.

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