\( - \frac{{dA}}{{dt}} = (2R)\;V\)
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.