Electro Magnetic Induction And Alternating Currents Question 404

Question: 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

Options:

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

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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 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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