Question

A conducting straight wire $PQ$ of length $l$ is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a velocity $v$. There exists a uniform horizontal magnetic field $B$ in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire $PQ$ at the position shown in the figure will be

• A. $Bvl$
• B. $2Bvl$
• C. $3Bvl/2$
• D. $\text{Zero}$

Answer 1

The correct option is A $Bvl$


We know, in pure rolling bottom-most point is at rest.
$\therefore \omega =\frac{v}{R}=\frac{v}{l/2}=\frac{2v}{l}$

$\epsilon =\frac{B\omega l^2}{2}=\frac{B\left(\frac{2v}{l}\right)l^2}{2}=Bvl$