Electro Magnetic Induction And Alternating Currents Question 310

Question: 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]

Options:

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

Show Answer

Answer:

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.



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