A copper rod is bent into a semi-circle of radius a and at ends straight parts are bent along diameter of the semi-circle and are passed through fixed, smooth, and conducting ring O and O' as shown in figure. A capacitor having capacitance C is connected to the rings. The system is located in a uniform magnetic field of induction B such that axis of rotation OO' is perpendicular to the field direction. At initial moment of time (t = 0), plane of semi-circle was no(nal to the field direction and the semicircle is set in rotation with constant angular velocity $\omega$. Neglect the resistance and inductance of the circuit. The current flowing through the circuit as function of time is

When the copper rod is rotated, flux linked with the circuit varies with time. 

Therefore, an emf is induced in the circuit- 

At time t, plane of semi-circle makes angle $\omega t$ with the plane of rectangular part of the circuit. Hence, component of the magnetic induction normal to plane of semi-circle is equal to $B \cos \omega t$

Flux linked with semicircular part is

${\varphi }_1=\frac{1}{2}\pi a^{2}B \cos \omega t$

Let area of rectangular part of the circuit be A.
$\therefore$ Flux linked with this part is

${\varphi }_2=BA$

$\therefore$ Total flux linked with the circuit is

$\varphi =\frac{1}{2}\pi a^{2}B \cos \omega t+BA$

$\therefore$Induced emf in the circuit,

$e=-\frac{d\varphi }{dt}=\frac{1}{2}\pi \omega a^{2}B \sin \left(\omega t\right)$

Since resistance of the circuit is negligible, therefore, potential difference across the capacitor is equal to induced emf in the
circuit.
Charge on the capacitor at time t is q = Ce

$=\frac{1}{2}\pi \omega a^{2}CB \sin \left(\omega t\right)$

But current $I=\frac{dq}{dt}=\frac{1}{2}\pi {\omega }^2{a}^{2}CB \cos \left(\omega t\right)$