Electromagnetic Induction and Alternating Current 2 Question 4
5. A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $\mathbf{B}$. At the position $M N Q$ the speed of the ring is $v$ and the potential difference developed across the ring is
$(1996,2 M)$
(a) zero
(b) $B v \pi R^{2} / 2$ and $M$ is at higher potential
(c) $\pi B R v$ and $Q$ is at higher potential
(d) $2 R B v$ and $Q$ is at higher potential
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Answer:
Correct Answer: 5. (d)
Solution:
- Induced motional emf in $M N Q$ is equivalent to the motional emf in an imaginary wire $M Q$ i.e.
$$ e _{M N Q}=e _{M Q}=B v l=B v(2 R) \quad[l=M Q=2 R] $$
Therefore, the potential difference developed across the ring is $2 R B v$ with $Q$ at higher potential.