JEE Advance - Physics (2022 - Paper 1 Online - No. 7)
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 _________ $m A$.
Answer
4
Explanation
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
Using Kirchhoff's law we can write
$$ \begin{aligned} & E=L \frac{d i}{d t}+\frac{q}{C} \\\\ & L \frac{d i}{d t}=E-\frac{q}{C} \\\\ & \text { Or } \frac{d^2 q}{d t^2}=-\frac{1}{L C}(q-C E) \end{aligned} $$
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 $t=0, q=0 \& i=0$
So $A=C E \& \phi=-\frac{\pi}{2}$
$$ q=C E-C E \cos \omega t $$
so $i=\frac{d q}{d t}=C E \omega \sin \omega t$
So,
$$ i_{\max } =\frac{10^{-3} \times 0.04}{\sqrt{0.1 \times 10^{-3}}} $$
$$ =4 \mathrm{~mA} $$
$$ |E|=\left|\frac{d \phi}{d t}\right| =\frac{d}{d t}\left(\left(B_0+\beta t\right) A\right) $$
$$ =\beta \times A $$
$$ =0.04 \mathrm{~V} $$
So the circuit can be rearranged as
Using Kirchhoff's law we can write
$$ \begin{aligned} & E=L \frac{d i}{d t}+\frac{q}{C} \\\\ & L \frac{d i}{d t}=E-\frac{q}{C} \\\\ & \text { Or } \frac{d^2 q}{d t^2}=-\frac{1}{L C}(q-C E) \end{aligned} $$
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 $t=0, q=0 \& i=0$
So $A=C E \& \phi=-\frac{\pi}{2}$
$$ q=C E-C E \cos \omega t $$
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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