Question

A non-conducting ring of radius $R$ having uniformly distributed charge $Q$ starts rotating about $x-x^{\prime }$ axis passing through diameter with an angular acceleration $\alpha$ , as shown in the figure. Another small conducting ring having radius $a$ $\left(a\ll R\right)$ is kept fixed at the centre of the bigger ring is such a way that axis $x{x}^{\prime }$ is passing through its centre and perpendicular to its plane. If the resistance of the small ring is $r=1\Omega ,$ find the induced current in it in $ampere$ .

(Given $q=\frac{16\times 10^2}{{\mu }_0}C,$ $R=1m,a=0.1m$ , $\alpha =8rad{s}^{-2}$ )

Answer 1

Answer: 8

$dq=\frac{q}{2\pi R}\cdot Rd\theta =\frac{q}{2\pi }\cdot d\theta$
$di=\frac{dq}{T}=\frac{qd\theta \omega }{2\pi 2\pi }$
$di=\frac{q\omega }{4{\pi }^2}\cdot d\theta$
$dB=\frac{{\mu }_{0}di(R\sin \theta {)}^2}{2R^3}$
$\int dB={\int }_0^{\pi }\frac{{\mu }_0{\sin }^2\theta }{2R}\left(\frac{q\omega }{4{\pi }^2}\right)d\theta$
$B=\frac{{\mu }_{0}q\omega }{16\pi R}$
$\varphi =B\pi a^2$
$\varphi =\pi a^2\cdot \frac{{\mu }_{0}q\omega }{16\pi R}$
$\varphi =\frac{{\mu }_{0}q\omega a^2}{16R}$
$|\epsilon |={\left|\frac{d\varphi }{dt}\right|}_2$
$\epsilon \mid =\frac{{\mu }_{0}q{a}^2}{16R}\alpha$
$=8$ volt
$i=\frac{8}{1}=8A$