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. (a) `(1)/(4)piomega^(2)a^(2)Cbcos omegat`
B. ( b) `(1)/(2)piomega^(2)a^(2)CBcos omegat`
C. ( c) `(1)/(4)piomega^(2)a^(2)CBsin omegat`
D. (d) `(1)/(2)piomega^(2)a^(2)CBsin omegat`

Answer

B

Explanation

(b) 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 `omegat` with the plane of rectangular part of the circuit. Hence, comonent of the magnetic inducetion normal to plane of semi-circle is equal to `B cos omegat`.
`phi_(1) = (1)/(2) pia^(2)Bcos omegat`
Let area of rectangular part of the circuit be `A`.
`:.` Flux linked with this part is
`phi_(2) = BA`
`:.` Total flux linked with the circuit is
`phi = (1)/(2) pia^(2)Bcos (omegat) + BA`
`:.` Induced emf in the circuit,
`e = -(dphi)/(dt) = (1)/(2) piomega a^(2)Bsin(omegat)`
Since resistance of the circuit is negligible, therefore, potential difference acroos the capacitor is equal to induced emf in the circuit.
`:.` Change on the capacitor at time `t` is `q = Ce`
`= (1)/(2) pi omega a^(2)CBsin(omegat)`
But current `I = (dq)/(dt) = (1)/(2)piomega^(2) a^(@)CBcos(omegat)`