JEE Advanced 2022 — Paper 1 — Question 25

$\phi=\left(\mathrm{B}_{0}+\beta \mathrm{t}\right) \mathrm{A}$

$\left|\varepsilon_{\text {ind }}\right|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\beta \mathrm{A}$

Applying kVL in the equivalent circuit diagram of the loop.

$$\beta A - L\frac{di}{dt} - \frac{q}{C} = 0$$

Also, $i=\frac{d q}{d t}$ From (i) and (ii) $L \frac{d^{2} i}{d t^{2}}=-\frac{i}{c}$

$\mathrm{i}=\mathrm{i}_{\mathrm{m}} \sin \omega \mathrm{t} \omega=\frac{1}{\sqrt{L C}}$

$\int_{0}^{q} d q=i_{m} \int_{0}^{t} \sin (\omega t) d t$

$\mathrm{q}=\frac{\mathrm{i}_{\mathrm{m}}}{\omega}(1-\cos \omega \mathrm{t})$

when $\mathrm{i}=\mathrm{i}_{\mathrm{m}} \Rightarrow \sin \omega \mathrm{t}=1 \Rightarrow \cos \omega \mathrm{t}=0$

$q=\frac{i_{m}}{\omega}=\beta A C$

$\mathrm{I}_{\mathrm{m}}=\omega \beta \mathrm{AC}=\beta \mathrm{A} \sqrt{\frac{C}{L}}=4 \mathrm{~mA}$