Questions

# 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

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a
14πω2a2CB cos ωt
b
12πω2a2CB cos ωt
c
14πω2a2CB sin ωt
d
12πω2a2CB sin ωt

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detailed solution

Correct option is 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 ωt 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 cosωtFlux linked with semicircular part isϕ1=12πa2B cos ωtLet area of rectangular part of the circuit be A.∴ Flux linked with this part isϕ2=BA∴ Total flux linked with the circuit isϕ=12πa2B cosωt+BA∴Induced emf in the circuit,e=-dϕdt=12πωa2B sinωtSince resistance of the circuit is negligible, therefore, potential difference across the capacitor is equal to induced emf in thecircuit.Charge on the capacitor at time t is q = Ce=12πωa2CB sinωtBut current I=dqdt=12πω2a2CB cosωt