JEE Advanced 2022 Paper 1 · Q07 · LC Oscillations & EMI

Question

Consider an LC circuit, with inductance L = 0.1 H and capacitance C = 10⁻³ F, kept on a plane. The area of the circuit is 1 m². It is placed in a constant magnetic field of strength B₀ which is perpendicular to the plane of the circuit. At time t = 0, the magnetic field strength starts increasing linearly as B = B₀ + βt with β = 0.04 T s⁻¹. The maximum magnitude of the current in the circuit is ___ mA.

SolveKar AI · Accepted Answer

Induced EMF from changing flux: ε = −A·dB/dt = −1·β = −0.04 V (constant). For an LC circuit driven by a constant EMF starting at t = 0 with q(0) = 0, i(0) = 0, the equation is L(di/dt) + q/C = ε. Solution oscillates: q(t) = Cε(1 − cos ωt) with ω = 1/√(LC). Current i(t) = dq/dt = Cε·ω·sin ωt. Maximum current i_max = Cεω = Cε/√(LC) = ε√(C/L) = 0.04·√(10⁻³/0.1) = 0.04·√(0.01) = 0.04·0.1 = 0.004 A = 4 mA. Units check: [V]·√(F/H) = [V]·√(s²/Ω²·... ) = [V]/[Ω] = [A] ✓.