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JEE Advanced 2022 · Paper 1, Physics question 7

Consider an LC circuit, with inductance L=0.1L = 0.1 H and capacitance C=10−3C = 10^{-3} F, kept on a plane. The area of the circuit is 1 m21\ \mathrm m^2. It is placed in a constant magnetic field of strength B0B_0 which is perpendicular to the plane of the circuit. At time t=0t = 0, the magnetic field strength starts increasing linearly as B=B0+βtB = B_0 + \beta t with β=0.04 T s−1\beta = 0.04\ \mathrm{T\,s^{-1}}. The maximum magnitude of the current in the circuit is _____ mA.

Answer and worked solution

Official answer: 3.98

Induced emf. The flux grows at a constant rate:
ε=AdBdt=1×0.04=0.04 V.\varepsilon=A\frac{dB}{dt}=1\times0.04=0.04\ \text{V}.
This constant emf drives the LC circuit.

Charge. The equation Lq¨+qC=εL\ddot q+\frac qC=\varepsilon, with q(0)=0q(0)=0 and i(0)=0i(0)=0, gives
q=Cε(1−cos⁡ωt),ω=1LC=100 s−1.q=C\varepsilon(1-\cos\omega t),\qquad\omega=\frac1{\sqrt{LC}}=100\ \text{s}^{-1}.

Maximum current.
imax⁡=ωCε=100×10−3×0.04=4×10−3 A=4 mA.i_{\max}=\omega C\varepsilon=100\times10^{-3}\times0.04=4\times10^{-3}\ \text{A}=4\ \text{mA}.

Topics:LC oscillationsFaraday's and Lenz's laws
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