Question 369: Electromagnetic Induction

Question

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
$$
|E|=\left|\frac{d \phi}{d t}ight|  =\frac{d}{d t}\left(\left(B_0+\beta tight) Aight) $$
$$
 =\beta 	imes A $$
$$
 =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} \\
& 	ext { 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(	ext { where } \omega=\frac{1}{\sqrt{L C}}ight)
$$
at $t=0, q=0 \& i=0$
So $A=C E \& \phi=-\frac{\pi}{2}$
$$
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} 	imes 0.04}{\sqrt{0.1 	imes 10^{-3}}} $$
$$ =4 \mathrm{~mA}

$$

Electromagnetic Induction

Integer / Numeric Answer Type