Question

A plane EM wave travelling alone z-direction is described by $\overrightarrow{E}=E_0\sin (kz-\omega t)\hat{i}$ and $\overrightarrow{B}=B_0\sin (kz-\omega t)\hat{j}$.

• A. The average energy density of the wave is given by $u_{av}=\frac{1}{4}{\epsilon }_0{E}_0^2+\frac{1}{4}\frac{B_0^2}{{\mu }_0}.$
• B. The time averaged intensity of the wave is given by $I_{av}=\frac{1}{2}{\in }_{0}c{E}_0^2$
• C. Both (a) and (b)
• D. None of these

Answer 1

Answer: (C)

Both (a) and (b)

(i) The e.m. wave carries energy which is due to electric field vector and magnetic field vector. In e.m. wave $E$ and $B$ vary from point to point and from moment to moment. Let $E$ and $B$ be their time average. The energy density due to the electric field $E$ is

$u_E=\frac{1}{2}{\in }_0{E}^2$

The energy density due to magnetic field $B$ is $uB=\frac{1}{2}\frac{B^2}{{\mu }_0}$

Total average energy density of em wave

$u_{av}=uE+uB=\frac{1}{2}{\in }_0{E}^2+\frac{1}{2}\frac{B^2}{{\mu }_0}$ ...(i)

Let the em wave be propagation along z-direction. The electric field vector and magnetic field vectors be represented by

$E=E_0\sin (kz-wt)$

$B=B_0\sin (kz-wt)$

The time average value of $E^2$ over complete cycle $=E_0^2/2$

and time average value of $B^2$ over complete cycle $=B_0^2/2$

$\therefore u_{av}=\frac{1}{2}{\in }_0\frac{E_0^2}{2}+\frac{1}{2}{\mu }_0\left(\frac{B_0^2}{2}\right)=\frac{1}{4}{\in }_0{E}_0^2+\frac{B_0^2}{4{\mu }_0}$

(ii) we know that $E_0=c{B}_0$ and $c=\frac{1}{\sqrt{{\mu }_0{\in }_0}}$.

$\therefore \frac{1}{4}\frac{B_0^2}{{\mu }_0}=\frac{1}{4}\frac{E_0^2/c^2}{{\mu }_0}=\frac{E_0^2}{4{\mu }_0}\times {\mu }_0{\in }_0=\frac{1}{4}{\in }_0{E}_0^2$

$\therefore uB=uE$

Hence, $u_{av}=\frac{1}{4}{\in }_0{E}_0^2+\frac{1}{4}\frac{B_0^2}{{\mu }_0}=\frac{1}{4}{\in }_0{E}_0^2+\frac{1}{4}{\in }_0{E}_0^2=\frac{1}{4}{\in }_0{E}_0^2=\frac{1}{2}\frac{B_0^2}{{\mu }_0}$

Time average intensity of the wave

$I_{av}=u_{av}c=\frac{1}{2}\in E_0^{2}c=\frac{1}{2}{\in }_{0}c{E}_0^2$
Hence, the correct option is $\left(C\right)$