question_answer

A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ, the speed of the ring is V and the potential difference developed across the ring is [IIT JEE 1996]

A)            Zero

B)            \[B\nu \pi {{R}^{2}}/2\] and M is at higher potential

C)            \[\pi RBV\]and Q is at higher potential

D)            2RBV and Q is at higher potential

Correct Answer: D

Solution :

                   Rate of decrease of area of the semicircular ring \[-\frac{dA}{dt}=(2R)\ V\] According to Faraday's law of induction induced emf \[e=-\frac{d\varphi }{dt}=-\ B\frac{dA}{dt}=-\ B\ (2RV)\] The induced current in the ring must generate magnetic field in the upward direction. Thus Q is at higher potential.