JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

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

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Solution

EMF driving the loop: $ε = \frac{d Φ}{d t} = β A = 0.04 \times 1 = 0.04$ V (constant).

Kirchhoff's voltage law: $ε = L \frac{d i}{d t} + \frac{q}{C}$ .

Differentiating once and using $i = d q / d t$ gives $\frac{d ^{2} q}{d t ^{2}} = - \frac{1}{L C} (q - C ε)$ , which is SHM about the equilibrium charge $q_{e q} = C ε$ with angular frequency $ω = \frac{1}{L C}$ .

With initial conditions $q (0) = 0$ and $i (0) = 0$ : $q (t) = C ε (1 - cos ω t)$ , so $i (t) = C ε ω sin ω t$ .

Maximum current: $i_{m a x} = C ε ω = \frac{C ε}{L C} = ε \frac{C}{L} = 0.04 \frac{1 0 ^{- 3}}{0.1} = 0.04 \times 0.1 = 4 \times 1 0^{- 3}$ A.

$i_{m a x} = 4$ mA.

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