Question

The average energy-density of electromagnetic wave given by E=(50N//C) sin (omegat-kx) will be nearly:

Answer 1

Given:

Equation of electromagnetic wave :

$E = 50 \sin( \omega t - kx)$

To find:

Average energy density of the wave.

Calculation:

Energy density associated with electric field in free space is equal to :

$\therefore \: u_{E} = \dfrac{1}{2} \epsilon_{0} {E}^{2}$

Energy density associated with magnetic field in free space is equal to :

$\therefore \: u_{B} = \dfrac{1}{2} \dfrac{1}{ \mu_{0}} {B}^{2}$

Now , we know that the magnetic field energy density and the electric field energy density are equal in an Electromagnetic wave.

So, total energy density :

$\therefore \: u = u_{E} + u_{B}$

$= > \: u = u_{E} + u_{E} = u_{B} + u_{B}$

$= > \: u =2 u_{E} = 2u_{B}$

$\boxed{ = > \: u = \epsilon_{0} {E}^{2} = \dfrac{ {B}^{2} }{\mu_{0}} }$

HOPE IT HELPS.