Question

A conducting ring of radius $2R$ rolls on a sooth horizontal conducting surface as shown in figure. A uniform horizontal magnetic field $B$ is perpendicular to the plane of the ring. The potential of $A$ with respect to $O$ is

• A. $2BvR$
• B. $\frac{1}{2}BvR$
• C. $8BvR$
• D. $4BvR$

Answer 1

The correct option is D $4BvR$

The rolling motion can be simplified to rotational motion of the ring about the instantaneous axis of rotation about point O. The angular speed of the rotation is $\omega =\frac{v}{r}=\frac{v}{2R}$

Hence the potential difference across points A and O is $\frac{B\omega l^2}{2}=\frac{B\left(\frac{v}{2R}\right)(4R{)}^2}{2}=4BvR$