A rectangular loop has a sliding connector PQ | vedclass

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(C) The motional electromotive force $(EMF)$ induced in the sliding rod $PQ$ is given by $\varepsilon = Bv\ell$.Substituting the given values: $\varep

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

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(C) The motional electromotive force $(EMF)$ induced in the sliding rod $PQ$ is given by $\varepsilon = Bv\ell$.Substituting the given values: $\varepsilon = 3\ T 	imes 5\ m/s 	imes 2\ m = 30\ V$.The rod $PQ$ acts as a battery of $30\ V$ with internal resistance $10\Omega$. This battery is connected to two parallel branches,each containing a $10\Omega$ resistor.The total resistance of the circuit is $R_{eq} = 10\Omega + (10\Omega \parallel 10\Omega) = 10\Omega + 5\Omega = 15\Omega$.The total current $I$ flowing through the rod $PQ$ is $I = \frac{\varepsilon}{R_{eq}} = \frac{30\ V}{15\Omega} = 2\ A$.Since the two parallel branches have equal resistance ($10\Omega$ each),the current $I$ splits equally into $I_1$ and $I_2$.Therefore,$I_1 = I_2 = \frac{I}{2} = \frac{2\ A}{2} = 1\ A$.Thus,$I_1 = 1\ A, I_2 = 1\ A, I = 2\ A$.

$A$ 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. The set-up is placed in a uniform magnetic field $3\ T$ directed into the plane of the paper. Find the three currents $I_1$,$I_2$,and $I$.

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