Question

A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in horizontal magnetic induction $\overrightarrow{B}$. At the position $MNQ$, the speed of the ring is $V$, and the potential difference developed across the ring is

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

Answer 1

The correct option is D $2RBV$ and $Q$ is at higher potential

$E=\frac{d\varphi }{dt}=B\frac{dA}{dt}$

When it is just about to move out:

$dA=2R\left(vdt\right)$

$\frac{dA}{dt}=2Rv$

Hence, $E=2BRV$