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

• 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

Answer 1

The correct option is D 2RBV and Q is at higher potential

Rate of decrease of area of the semicircular ring $-\frac{dA}{dt}=\left(2R\right)V$
According to Faraday's law of induction induced emf
$e=-\frac{d\varphi }{dt}=-B\frac{dA}{dt}=-B\left(2RV\right)$


The induced current in the ring must generate magnetic field in the upward direction. Thus Q is at higher potential.