Q: A plane EM wave travelling along z– direction is described by $\displaystyle E = E_{0} sin(kz – ωt)\hat{j} $ Show that

(i) The average energy density of the wave is given by $u_{av} = \frac{1}{4} \epsilon_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 \epsilon_0 E_0^2 $

Solution : We know that energy density associated with E.F.

$\displaystyle u_E = \frac{1}{2} \epsilon_0 E^2 $

We know that energy density associated with M.F

$\displaystyle u_B = \frac{B^2}{2 \mu_0} $

Thus total energy of EM wave is given by

$\displaystyle u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 $

$\displaystyle u = \frac{1}{2} \epsilon_0 E^2 + \frac{B^2}{2 \mu_0} $

putting the values of E.F. and M.F.

$\displaystyle u = \frac{1}{2} \epsilon_0 E_0^2 sin^2 (kz -\omega t) + \frac{B_0^2}{2 \mu_0} sin^2 (kz -\omega t) $

If we take the average value over a long period of time $\displaystyle sin^2 (kz -\omega t) = \frac{1}{2}$

The above equation can also be written as

$\displaystyle u_{avg} = \frac{1}{4} \epsilon_0 E_0^2 + \frac{B_0^2}{4 \mu_0} $

But u_{E} = u_{B}

$\displaystyle u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 $

or , $\displaystyle u_{avg} = \frac{B_0^2}{2 \mu_0} $

### Intensity :

Energy crossing per unit area per unit time perpendicular to the direction of propogation is called intensity of wave.

**Energy contained in the volume :**

$\displaystyle U = u_{avg} \times Volume $

$\displaystyle U = \frac{1}{2} \epsilon_0 E_0^2 \times (A \times c \Delta t) $

$\displaystyle Intensity = \frac{U}{A \Delta t} = \frac{1}{2} \epsilon_0 E_0^2 c $

$\displaystyle I = \frac{1}{2} \epsilon_0 c E_0^2 $