The uniform magnetic field perpendicular to the plane of a conducting ring of radius a changes at the rate of $\mathrm{\alpha }$, then

$\begin{array}{l}\mathrm{\varphi }={\mathrm{\pi a}}^2\mathrm{B}\\ \mathrm{e}={\mathrm{\pi a}}^2\frac{\mathrm{dB}}{\mathrm{dt}}={\mathrm{\pi a}}^2\mathrm{\alpha }\\ \mathrm{E}2\mathrm{\pi a}=\mathrm{e}\Rightarrow \mathrm{E}=\frac{{\mathrm{\pi a}}^2\mathrm{\alpha }}{2\mathrm{\pi a}}=\frac{\mathrm{\alpha a}}{2}\end{array}$

Let R be the resistance of the ring. Then current in the ring is i = e/R
Consider a small element $d\ell$. on the ring,

emf induced in the element, $\mathrm{de}=\left(\frac{\mathrm{e}}{2\mathrm{\pi a}}\right)\mathrm{dl}$

resistance of the element, $\mathrm{dR}=\left(\frac{\mathrm{R}}{2\mathrm{\pi a}}\right)\mathrm{dl}$

$\therefore$Potential difference across the element 

             $\begin{array}{l}=\mathrm{de}-\mathrm{idR}\\ =\left(\frac{\mathrm{e}}{2\mathrm{\pi a}}\right)\mathrm{dl}-\left(\frac{\mathrm{e}}{\mathrm{R}}\right)\left(\frac{\mathrm{R}}{2\mathrm{\pi a}}\right)\mathrm{dl}=0\end{array}$