question_answer

A thin semicircular conducting ring of radius R is fallin 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

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

Correct Answer: D

Solution :

[d] 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