Question

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

• A. Zero
• B. $\frac{B \pi R^{2}}{2}$ and $M$ is at a higher potential
• C. $\pi BRv$ and $Q$ is at a higher potential
• D. 2 $RBv$ and $Q$ is at a higher potential

Answer 1

Answer: (D)

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,
$e_{M N Q}=e_{M Q}=Bvl=Bv\left(\right.2R\left.\right)$ $\left[\right.l=MQ=2R\left]\right.$

$V_{Q}-V_{M}=v\left(B\right)\left(2 R\right)$
Therefore, the potential difference developed across the ring is $2RBv$ with Q at a higher potential.