Question

A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$. At the position $M N Q$, the speed of the ring is $v$ and the potential difference developed across the ring is

• A. zero
• B. $B v \pi R^{2} / 2$ and $M$ is at higher potential
• C. $\pi R B v$ and $Q$ is at higher potential
• D. $2 R B v$ and $Q$ is at higher potential

Answer 1

Answer: (D)

Rate of decrease of area of the semicircular ring
$-\frac{d A}{d t}=(2 R) v$
According to Faraday's law of induction induced emf
$e=-\frac{d \varphi}{d t}=-B \frac{d A}{d t}=-B(2 R v)$

The induced current in the ring must generate magnetic field in the upward direction. Thus $Q$ is at higher potential