A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction $\stackrel{\rightharpoonup }{B}$ . At the position MNQ the speed of the ring is V, and the potential difference developed across the ring is 
 

Given data:

Radius of the ring R

Horizontal magnetic field B

Speed of the ring is V

Concept used: Induced EMF

Detailed solution:

We know that induced EMF

$\mathrm{E}=\frac{\mathrm{d}\varphi }{\mathrm{dt}}=\mathrm{B}\frac{\mathrm{dA}}{\mathrm{dt}}$

When it is just about to move out:

$\mathrm{dA}=2\mathrm{R}\left(\mathrm{vdt}\right)$

$\frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{Rv}$

Hence, $\mathrm{E}=2\mathrm{BRV}$

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