Question

The figure shows a smooth pair of thick metallic rails connected across a battery of emf $\epsilon$ having a negligible internal resistance. A wire $ab$ of length $l$ and resistance $r$ can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field $B$. At any instant, the wire is given a small velocity $v$ towards the right. Find the current in the arrangement at this instant.

• A. $\frac{Bvl}{r}$
• B. $\frac{Bvl+\epsilon }{r}$
• C. $\frac{\epsilon }{r}$
• D. $\frac{\epsilon -Bvl}{r}$

Answer 1

The correct option is D $\frac{\epsilon -Bvl}{r}$


Wire $ab$ is moving in a magnetic field. Therefore, an emf will be developed across the wire.

The induced emf, $E=Bvl...\left(1\right)$

From the right-hand rule, we can say that, $V_b>V_a$.

Therefore, the equivalent circuit is -


The equivalent emf of the circuit is,

$E_{eq}=\epsilon -E$

$\therefore$ The current flowing through the wire $ab$ is,

$I=\frac{\epsilon -E}{r}$

$\Rightarrow I=\frac{\epsilon -Bvl}{r}\left[\text{From 1}\right]$