Question

A fully charged capacitor C with charge ${{\text{q}}_{0}}$ is connected to a coin of self-inductance L at t = 0. The time at which, the energy is stored equally between the electric and the magnetic fields is
(A) $\sqrt{\text{LC}}$
(B) $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{ }\sqrt{\text{LC}}$
(C) $\text{ }\!\!\pi\!\!\text{ }\sqrt{\text{LC}}$
(D) $\text{2 }\!\!\pi\!\!\text{ }\sqrt{\text{LC}}$

Answer 1

Hint: A inductor is usually a coil of wire that sits up an alternating magnetic field around it when an alternating current flows through it. Inductance is the property of an inductor that opposes the charge in current maximum energy in inductance is
$\text{L}=\dfrac{1}{2}\text{ L}{{\text{I}}^{2}}$
Capacitors store energy on their conductive plates in the form of an electrical charge. Maximum energy in capacitor is
$=\dfrac{{{\text{q}}^{2}}_{0}}{2\text{C}}$ .

Complete step by step solution
Consider a capacitor C having charge${{\text{q}}_{0}}$ and maximum energy in C is
$\dfrac{{{{{q}}^{2}}}}{2\text{C}}$
Where
 ${{\text{q}}_{\max }}=\text{max}\times \text{charge}$
C = charge on capacitor
In a induction ‘L’, the maximum energy is given by $\dfrac{1}{2}\text{ L}{{\text{I}}^{2}}_{\text{max}}$
L = self-inductance
I = maximum current
In LC oscillation, the energy is transferred to L to O or C to L energy will be equal when,
Energy in capacitor = Energy in inductor
$\dfrac{{{\text{q}}^{2}}}{2\text{C}}=\dfrac{1}{2}\text{L}{{\text{I}}^{2}}$ …… (1)
As initially charge is maximum
$\text{q}={{\text{q}}_{0}}$ coswt
$\begin{align}
  & \text{I}=\dfrac{\text{dq}}{\text{dt}} \\
 & \Rightarrow \dfrac{\text{d}}{\text{dt}}\left( {{\text{q}}_{0}}\text{ cos wt} \right)=-\text{w}{{\text{q}}_{0}}\text{ sin cot} \\
\end{align}$
Putting value in equation (1)
\[\begin{align}
  & \dfrac{{{\left( {{\text{q}}_{0}}\text{cos wt} \right)}^{2}}}{2\text{C}}=\dfrac{1}{2}\text{L}{{\left[ -\text{w}{{\text{q}}_{0}}\text{ sinwt} \right]}^{2}} \\
 & \dfrac{{{\left( {{\text{q}}_{0}}\text{cos wt} \right)}^{2}}}{{{\left( \text{w}{{\text{q}}_{0}}\sin \text{wt} \right)}^{2}}}=\text{LC} \\
\end{align}\]
Squaring both sides
$\begin{align}
  & \dfrac{{{\text{q}}_{0}}\text{cos wt}}{\text{w}{{\text{q}}_{0}}\text{sin wt}}=\sqrt{\text{LC}} \\
 & \dfrac{\text{tan wt}}{\text{w}}=\sqrt{\text{LC}} \\
 & \dfrac{\text{w}}{\text{tan wt}}=\dfrac{1}{\sqrt{\text{LC}}} \\
\end{align}$
In LC oscillation,$\text{w}=\dfrac{1}{\sqrt{\text{LC}}}$
Therefore
$\begin{align}
  & \text{tan wt}=1 \\
 & \Rightarrow \text{wt}=\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4} \\
 & \text{ t}=\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4\text{w}} \\
\end{align}$
Put the value of w
$\text{t}=\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4}\sqrt{\text{LC}}$
$\therefore $ This is the time at which the energy is stored equally between the electric and the magnetic field.
$\therefore $ Option (B) is correct.

Note
Keep in mind that oscillation of an LC circuit is a cyclic interchange between electric energy stored in the capacitor and magnetic energy stored in the inductor. Also remember that frequency of oscillation is expressed by the expression$\text{w}=\dfrac{1}{2\text{ }\!\!\pi\!\!\text{ }}\sqrt{\text{LC}}$. Natural frequency of an LC-circuit is 1, 25,000 cycles per second.