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

$\varphi =\pi a^{2}B$

$e=\pi a^2.\frac{dB}{dt}=\pi a^2\alpha$

Let R be the resistance of the ring. Then current in the ring is $i=\frac{e}{R}$.

Consider a small element $dI$ on the ring

Emf induced in the element  $de=\left(\frac{e}{2\pi a}\right).dI$

Resistance of the element $dR=\left(\frac{R}{2\pi a}\right).dI$

$\therefore$ Potential difference across the element

$=de-idR$

$=\left(\frac{e}{2\pi a}\right)dI-\left(\frac{e}{R}\right)\left(\frac{R}{2\pi a}\right)dI=0$

$\therefore$ All points on the ring are at same potential.

Further $\oint \overrightarrow{E}.\overrightarrow{dl}=\frac{d\varphi }{dt}$ Or $EI=\pi a^2.\frac{dB}{dt}$

Or $E\left(2\pi a\right)=\pi a^2\alpha$

$\therefore$$E=\frac{a\alpha }{2}$