Exams › JEE Advanced › Physics
An LC circuit with inductance L = 0.1 H and capacitance C = 10⁻³ F lies in a plane. The loop area is 1 m² and is placed in a magnetic field perpendicular to its plane. Starting at t = 0, the field increases linearly as B = B0 + (beta)t with beta = 0.04 T/s. Find the maximum magnitude of the current in the circuit (in mA).
- 0.4 mA
- 4 mA
- 40 mA
- 0.04 mA
Correct answer: 4 mA
Solution
A constant EMF e = A*(dB/dt) = A*beta acts on the series LC loop. The charge oscillates about the value Ce. The maximum current in undamped LC oscillation driven by a step EMF is I_max = e*sqrt(C/L).
Related JEE Advanced Physics questions
⚔️ Practice JEE Advanced Physics free + battle 1v1 →