Question

A rectangular loop has a sliding connector $PQ$ of length $\ell$ and resistance $R\Omega$ and it is moving with speed $v$ as shown. The setup is placed in a uniform magnetic field $(B)$ going into the plane of paper. The current in connector $PQ$ is $\frac{B\ell v}{nR}$. Find $n$.

Answer 1

2

Answer 2

The motional EMF induced in the sliding connector PQ is given by $\mathcal{E} = B\ell v$. The two vertical resistors of resistance $2R$ each are connected in parallel, so their equivalent resistance is $R_{parallel} = \frac{2R \times 2R}{2R + 2R} = R$. The total resistance of the circuit is $R_{total} = R_{PQ} + R_{parallel} = R + R = 2R$. According to Ohm's law, the current in PQ is $I = \frac{\mathcal{E}}{R_{total}} = \frac{B\ell v}{2R}$. Given current is $\frac{B\ell v}{nR}$. Equating the two expressions for current: $\frac{B\ell v}{2R} = \frac{B\ell v}{nR}$, which implies $2R = nR$, so $n=2$.