$(a)\;\text{All points on ring will not be at same potential} \\(b)\;\text{emf induced the ring is $\large\frac{\pi}{2}$$ R^2 \beta$ }\\ (c)\;\text {electric intensity at any point is zero} \\ (d)\;E= \large\frac{R\beta}{2}$

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$\phi= \pi R^2B$

$e= \pi R^2 \large\frac{dB}{dt}$

$\quad =\pi R^2 \beta$

current $=\large\frac{e}{r}$

Considering $dl$ a small element of ring.

$de=\bigg(\large\frac{e}{2 \pi R}\bigg)$$dl$

$de=\bigg(\large\frac{r}{2 \pi R}\bigg)$$dl$

potential difference $= de-idr$

$\qquad= \large\frac{e}{2 \pi R}$$ dl - \frac\large{e}{r} \frac{r}{2 \pi R}$$dR=0$

$\therefore$ all points in the ring are same potential

$\oint E.\bar{dl} =\large\frac{d \phi}{dt} $

$El= \pi R^2 \large\frac{dB}{dt}$

$E 2\pi R= \pi R^2 \beta$

$E=\large\frac{\beta R}{2}$

Hence d is the correct answer.

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