Electro Magnetic Induction And Alternating Currents Question 404
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
Options:
A) Zero
B) $ Bv\pi R^{2}/2 $ and M is at lower potential
C) $ \pi RBv $ and Q is at higher potential.
D) $ 2RBv $ and Q is at higher potential.
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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, the 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
BETA
