Solution:

Ans. At an instant $t$, charge $q$ on the capacitor and the current $i$ are given by :

$$ \begin{aligned} & q(t)=q_{0} \cos \omega t \ & i(t)=-q_{0} \omega \sin \omega t \end{aligned} $$

Energy stored in the capacitor at time $t$ is

$$ U_{K}=\frac{1}{2} C V^{2}=\frac{1}{2} \frac{q^{2}}{C}=\frac{q_{0}^{2}}{2 C} \cos ^{2}(\omega t) \quad 1 $$

Energy stored in the inductor at time $t$ is

$$ \begin{aligned} U_{M} & =\frac{1}{2} L i^{2} \ & =\frac{1}{2} L q_{0}^{2} \omega^{2} \sin ^{2}(\omega t) \ & =\frac{q_{0}^{2}}{2 C} \sin ^{2}(\omega t) \quad(\because \omega=1 / \sqrt{L C}) \mathbf{1} \end{aligned} $$

Sum of energies

$$ U_{E}+U_{M}=\frac{q_{0}^{2}}{2 C}\left(\cos ^{2} \omega t+\sin ^{2} \omega t\right) $$

$$ =\frac{q_{0}^{2}}{2 C} $$

This sum is constant in time as $q_{0}$ and $C$, both are time-independent.

[CBSE Marking Scheme 2017]

[AT Q. 2. Obtain the expression for the energy density of magnetic field $B$ produced in the inductor.

U] [Delhi Comptt. 2016]

Ans. Instantaneous Induced $e m f$ in an inductor when current changes through it

$$ e=-L \frac{d I}{d t} $$

Hence, instantaneous applied voltage

$$ e=V=L \frac{d I}{d t} $$

Work done, $d W=V . d q=$ VIdt

$$ \therefore \quad d W=\text { LIdI } $$

Long Answer Type Questions

$$ \begin{aligned} \Rightarrow \quad \int d W & =\int_{0}^{I} L I d I \ W & =\frac{1}{2} L I^{2} \end{aligned} $$

Energy density, $\quad U=\frac{\text { total energy stored }}{\text { volume }} \quad 1 / 2$

$$ U=\frac{\left(\frac{1}{2}\right) L I^{2}}{A l}=\frac{\frac{1}{2}(L I) I}{A l} $$

$$ \text { Flux }=N B A=L I $$

and

$$ B=\frac{\mu_{0} N I}{l} \Rightarrow I=\frac{B l}{\mu_{0} N} \quad 1 / 2 $$

$\therefore \quad \quad \quad \quad \quad=\frac{\frac{1}{2}(N B A) \cdot \frac{B l}{\mu_{0} N}}{A l}=\frac{B^{2}}{2 \mu_{0}} 1 / 2$

[CBSE Marking Scheme 2016]

(5 marks each)