JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Electromagnetic Induction

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 T s^{- 1}

. The maximum magnitude of the current in the circuit is ____

m A

Answer: 4

Solution

The induced emf is:

ε = A \frac{d B}{d t} = 1 \times 0.04 = 0.04

V

This charges the capacitor to

q_{0} = C ε

.

In an LC circuit, maximum current occurs when all energy is magnetic:

\frac{q _{0}^{2}}{2 C} = \frac{1}{2} L I_{0}^{2}

I_{0} = \frac{q _{0}}{L C} = \frac{C ε}{L C} = \frac{C}{L} \times ε

I_{0} = \frac{1 0 ^{- 3}}{0.1} \times 0.04 = 0.01 \times 0.04 = 0.1 \times 0.04 = 0.004 A = 4 m A

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer