QJEEAdv22Ph254d: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph254d Solution: Consider an LC circuit, with inductance and capacitance , kept…

Question

Verified Solution • 2025–26 Session moderate

Consider an LC circuit, with inductance

L=0.1 \,\mathrm{H}

and capacitance

C=10^{-3} \mathrm{~F}

, kept on a plane. The area of the circuit is

1 \mathrm{~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}+\beta t

with

\beta=0.04 \,\mathrm{T\,s}^{-1}

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

m A

.

Accepted Answer

Emf induced in the circuit is

|E|=\left|\frac{d \phi}{d t}\right| =\frac{d}{d t}\left(\left(B_0+\beta t\right) A\right)

=\beta \times A

=0.04 \mathrm{~V}

So the circuit can be rearranged as

JEE Advanced 2022 Paper 1 Online Physics - Electromagnetic Induction Question 5 English Explanation

Using Kirchhoff's law we can write

\begin{aligned}&amp; E=L \frac{d i}{d t}+\frac{q}{C} \\\\&amp; L \frac{d i}{d t}=E-\frac{q}{C} \\\\&amp; \text { Or } \frac{d^2 q}{d t^2}=-\frac{1}{L C}(q-C E)\end{aligned}

Using SHM concept we can write

q=C E+A \sin (\omega t+\phi)\left(\text { where } \omega=\frac{1}{\sqrt{L C}}\right)

at

t=0, q=0 \&amp; i=0

So

A=C E \&amp; \phi=-\frac{\pi}{2}

q=C E-C E \cos \omega t

so

i=\frac{d q}{d t}=C E \omega \sin \omega t

So,

i_{\max } =\frac{10^{-3} \times 0.04}{\sqrt{0.1 \times 10^{-3}}}

=4 \mathrm{~mA}

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