Q.
A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction $\overrightarrow{B},$ figure. At the position $MNQ$, the speed of the ring is $v.$ The potential difference developed across the ring is
A) Zero
B) $\frac{1}{2}\ B\ v\pi\ R^{2},\ and\ M\ $is at a higher potential
C) $\pi\ R\ B\ v,\ and\ Q$ is at a higher potential
D) $2\ R\ B\ v,\ and\ Q$ is at a higher potential
Explanation
(d) The emf induced in a conductor does not depend o its shape, but only on its end points, $M\ \ and\ Q$ in this case. Thus the conductor is equivalent to an imaginary straight conductor of $l = MQ = 2R$. Therefore, potential difference developed across the ring =$Blv = B(2R)v.$ And the direction of induced current is from $Q\ to\ M.Therefore,\ Q\ $is at higher potential.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.