First slide
Magnetic flux and induced emf
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

Moderate
Solution

Rate of decrease of area of the semicircular ring  dAdt=(2R)V
According to Faraday's law of induction induced emf e=dt=BdAdt=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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