A plane $EM$ wave travelling along $z-$ direction is described$\vec E = {E_0}\,\sin \,(kz - \omega t)\hat i$ and $\vec B = {B_0}\,\sin \,(kz - \omega t)\hat j$. Show that

$(i)$ The average energy density of the wave is given by $U_{av} = \frac{1}{4}{ \in _0}E_0^2 + \frac{1}{4}.\frac{{B_0^2}}{{{\mu _0}}}$

$(ii)$ The time averaged intensity of the wave is given by  $ I_{av}= \frac{1}{2}c{ \in _0}E_0^2$ વડે આપવામાં આવે છે.

$(i)$ In electromagnetic waves due to electric field vector and magnetic field vector waves carry energy $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ varies with time and position.

- Let $\mathrm{E}$ and $\mathrm{B}$ is average value at given time,

$U_{E}=\frac{1}{2} \epsilon_{0} E^{2} \text { and }$

energy density due to magnetic field $B$,

$\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu_{0}}$

Total average energy density of $EM$ waves,

$\mathrm{U}_{\text {average }}=\mathrm{U}_{\mathrm{E}}+\mathrm{U}_{\mathrm{B}}=\frac{1}{2} \mathrm{E}_{0} \mathrm{E}^{2}+\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu_{0}}$

Consider$ EM$ wave propagating in $z$-direction hence electric field and magnetic field will be represented as

$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \sin (k z-\omega t) \text { and }$

$\overrightarrow{\mathrm{B}}=\mathrm{B}_{0} \sin (k z-\omega t)$

Average value of $\mathrm{E}^{2}$ over one cycle (periodic time)

$\left\langle\mathrm{E}^{2}\right\rangle=\frac{\mathrm{E}_{0}^{2}}{2}$

Hence $U_{\text {average }}=\frac{1}{2} E_{0} \frac{E_{0}^{2}}{2}+\frac{1}{2} \mu_{0}\left(\frac{B_{0}^{2}}{2}\right)$

So, $U_{\text {avearge }}=\frac{1}{4}\left[\epsilon_{0} E_{0}^{2}+\frac{B_{0}^{2}}{4 \mu_{0}}\right]$

`$(ii)$ Now $\mathrm{E}_{0}=c \mathrm{~B}_{0}$ and $c=\frac{1}{\sqrt{\mu_{0} \in_{0}}}$ In equation (1) if $U_{\text {average }}=0$, then

$\frac{1}{4} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}=-\frac{1}{4} \epsilon_{0} \mathrm{E}_{0}^{2}$

$=-\frac{1}{4} \cdot \frac{\mathrm{E}_{0}^{2}}{\mu_{0} c^{2}}\left[\because c=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}\right]$

Neglecting negative sign,

$\frac{1}{4} \frac{\mathrm{B}_{0}^{2}}{\mu_{0}}=\frac{1}{4} \frac{\mathrm{E}_{0}^{2} / c^{2}}{\mu_{0}}=\frac{\mathrm{E}_{0}^{2}}{4 \mu_{0}} \cdot \mu_{0} \in_{0} \quad\left[c^{2}=\frac{1}{\mu_{0} \mathrm{E}_{0}}\right]$