A copper rod is bent inot 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 `O O'` is perpendicular to the field direction. At initial moment of time `(t = 0)`, plane of semi-circle was normal to the field direction and the semi-circle 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

Options

A. `(1)/(4)pi omega^(2)a^(2)CB cos omega t`
B. `(1)/(2)pi omega^(2)a^(2)CB cos omega t`
C. `(1)/(4)pi omega^(2)a^(2)CB sin omega t`
D. `(1)/(2)pi omega^(2)a^(2)CB sin omega t`

Answer

B

Explanation

When the copper rod is rotated, flux linked with the circuit varies with time
Thereofore, anemf 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 with semicircular part is
`phi_(1) = (1)/(2) pi a^(2)B cos omega t`
Let area of rectangular part of the circuit be `A`.
`:.` Flux linked withthis part is
`phi_(2) = BA`
`:.` Total flux linked with the circuit is
`phi = (1)/(2) pi a^(2) B cos (omega t) + BA`
`:.` Induced emf in the circuit,
`e = -(d phi)/(dt) = (1)/(2) pi omega a^(2) B sin (omega t)`
Since resistance of the circuit is negligible, therefore potential difference across the capactior is equal to induced emf in the circuit.
`:.` Charge on the capacitor at time `t` is `q = Ce`
`= (1)/(2) pi omega a^(2)CB sin (omega t)`
But current `I = (dq)/(dt) = (1)/(2) pi omega^(2) a^(2) CB cos (omega t)`