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

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 $e=-\frac{d\phi }{dt}=-B\frac{dA}{dt}=-B\left(2Rv\right)$

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