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.
