#### A plane EM wave travelling along z-direction is described by $E=E_{0}\sin (kz-\omega t)\hat{i}\; and\; B=B_{0}\sin (kz-\omega t)\hat{j}$ show that,i) the average energy density of the wave is given by$u_{av}=\frac{(u_{E}+u_{B})}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$ii) the time - averaged intensity of the wave is given by$I_{av}=u_{av}c=\frac{1}{2} (\varepsilon _{0}E^{2})c$

i) The energy density due to electric field E is

$u_{\varepsilon }=\frac{1}{2}(\varepsilon _{0}E^{2})$

The energy density due to magnetic field B is

$u_{B}=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )$

The average energy density of the wave is given by :

$u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$

ii) $u_{av}=\frac{\left ( u_{E}+u_{B} \right )}{2}=\frac{1}{4}.\varepsilon _{0}E^{2}+\frac{1}{4}.\frac{B^{2}}{\mu _{0}}$

Assuming $u_{E}=u_{B},$

$u_{av}=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )=\frac{1}{2}\left ( \frac{B^{2}}{\mu _{0}} \right )$

We know that $c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}=\left ( \frac{E_{0}}{B_{0}} \right )$

The time -averaged intensity of the wave is given as

$I_{av}=u_{av}c=\frac{1}{2}\left ( \varepsilon _{0}E^{2} \right )c$