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

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TestHub · Accepted Answer

Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i. e.,

$e_{M N Q} = e_{M Q} = B v l = B v (2 R)$ $[l = MQ = 2 R]$

$V_{Q} - V_{M} = v (B) (2 R)$

Therefore, the potential difference developed across the ring is $2 R B v$ with Q at a higher potential.

Physics - EMI/AC Question with Solution | TestHub

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