A non–conducting ring of radius R having uniformly distributed...

$dq = \dfrac{q}{2\piR} . Rd\theta = \dfrac{q}{2\pi} . d\theta$
$di = \dfrac{dq}{T} = \dfrac{qd\theta\omega}{2\pi2\pi}$
$di = \dfrac{q\omega}{4{\pi}^{2}} . d\theta$

The magnetic field at O due to elementary ring is

$dB = \dfrac{{\mu}_{0}di ( Rsin \theta{)}^{2}}{2{R}^{3}}$
$∫ dB = {∫}_{0}^{\pi}\dfrac{{\mu}_{0}\sin^{2} \theta}{2R}\left(\dfrac{q\omega}{4{\pi}^{2}}\right)d\theta$
$B = \dfrac{{\mu}_{0}q\omega}{16\piR}$

The flux of the magnetic field through the small ring

$ϕ = B\pi{a}^{2}$
$ϕ = \pi{a}^{2} . \dfrac{{\mu}_{0}q\omega}{16\piR}$
$ϕ = \dfrac{{\mu}_{0}q\omega{a}^{2}}{16R}$
$| ε | = \left|\dfrac{dϕ}{dt}\right|$
$| ε | = \dfrac{{\mu}_{0}q{a}^{2}}{16R}\alpha$ .
$= 8volt$
$i = \dfrac{8}{1} = 8A$.