As shown in the above figure, a uniform magnetic field is directed out of the plane of a conducting ring of radius r. The ring is in a pure rolling motion on a horizontal surface with the velocity of the centre of the ring being $v.$ Now, find the potential difference between the centre and the point of contact, $V_{\mathrm{O}}-V_{\mathrm{A}}$.

For every pure rolling case, we can always consider it as instantaneous pure rotation about an axis passing through the point of contact, i.e, in this case point A.

Now we can easily apply the concepts of motional EMF in case of a rotating rod. The rotational induced EMF in OA is given by 

From Lenz's law, current will flow in anticlockwise direction in the loop, i.e., from point A to point O. Therefore point A will be at a greater potential compared to point O. 

Hence, ${\mathrm{V}}_{\mathrm{O}}-{\mathrm{V}}_{\mathrm{A}}=-\frac{{\mathrm{B\omega r}}^2}{2}=-\frac{\mathrm{Bvr}}{2}$

$(\because v=r\omega )$