Question

In the given figure, the four rods have $\lambda$ resistance per unit length. The arrangement is kept in a magnetic field of constant magnitude $B$ and directed perpendicular to the plane of the figure. Initially, the sides as shown form a square. Now each wire starts moving with constant velocity $v$ towards the opposite wire.


Calculate the current flowing in the circuit.

• A. Zero
• B. $\frac{Bv}{\lambda }$
• C. $\frac{Bv}{4\lambda }$
• D. $\frac{4Bv}{\lambda }$

Answer 1

The correct option is B $\frac{Bv}{\lambda }$

Since, each rod is moving with velocity $v$, so the emf induced across each rod will be the same in magnitude and is given by,

$e=Bvl.....\left(1\right)$

The equivalent circuit can be drawn as,


Here, $r$ is the resistance of each rod.

$\therefore r=\lambda l.....\left(2\right)$

The current flowing in the circuit is,

$i=\frac{e_{eq}}{r_{eq}}$

$\Rightarrow i=\frac{[e+e+e+e]}{(r+r+r+r)}=\frac{e}{r}[\because \text{It is a series combination}]$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow i=\frac{Bvl}{\lambda l}$

$\therefore i=\frac{Bv}{\lambda }$

Hence, option $\left(B\right)$ is the correct answer.