Question 162: EM waves

Question

The equation of the electric field of an electromagnetic wave propagating through free space is given by : $E=\sqrt{377} \sin \left(6.27 	imes 10^3 t-2.09 	imes 10^{-5} xight) \mathrm{N} / \mathrm{C}$
The average power of the electromagnetic wave is $\left(\frac{1}{\alpha}ight) \mathrm{W} / \mathrm{m}^2$. The value of $\alpha$ is
$$ \left(	ext { Take } \sqrt{\frac{\mu_0}{\varepsilon_o}}=377 	ext { in SI units }ight) $$

Accepted Answer

The standard wave equation for the electric field component of an electromagnetic wave is given as $\mathrm{E}= \mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$, where $\mathrm{E}_0$ is the amplitude of electric field, $\omega$ is the angular frequency and k is the wave number.
The give equation for the electric field of electromagnetic wave is $=\sqrt{377} \sin \left(6.27 	imes 10^3 t-2.09 	imesight. \left.10^{-5} \mathrm{x}ight) \mathrm{N} / \mathrm{C}$
So, electric field amplitude is $\mathrm{E}_0=\sqrt{377} \mathrm{~N} / \mathrm{C}$.
The total energy density (energy per unit volume) of an electromagnetic wave is the sum of the electric field energy density ( $\mathrm{u}_{\mathrm{E}}$ ) and magnetic field energy density ( $\mathrm{u}_{\mathrm{B}}$ ).
$$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \mu_0 \mathrm{~B}^2 $$
In an electromagnetic wave traveling in a vacuum, the instantaneous electric and magnetic fields are related by the speed of light (c):
$$ \mathrm{E}=\mathrm{cB} \Rightarrow \mathrm{~B}=\frac{\mathrm{E}}{\mathrm{c}} $$
So, the energy density for magnetic field is,
$$ u_B=\frac{1}{2} \mu_0\left(\frac{E}{c}ight)^2=\frac{E^2}{2 \mu_0 c^2} $$
Since $c=\frac{1}{\sqrt{\mu_0} \epsilon_0} \Rightarrow c^2=\frac{1}{\mu_0 \epsilon_0} \Rightarrow \mu_0 c^2=\frac{1}{\epsilon_0}$.
$$ \mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2 $$
Thus, the energy is equally shared: $\mathrm{u}_{\mathrm{E}}=\mathrm{u}_{\mathrm{B}}$.
The total instantaneous energy density is:
$$ \mathrm{u}=\mathrm{u}_{\mathrm{E}}+\mathrm{u}_{\mathrm{B}}=\frac{1}{2} \epsilon_0 \mathrm{E}^2+\frac{1}{2} \epsilon_0 \mathrm{E}^2=\epsilon_0 \mathrm{E}^2 $$
Since the electric field oscillates sinusoidally $\left(E=E_0 \sin \omega tight)$, the average value of $E^2$ over one cycle is $\frac{E_0^2}{2}$.
$$ \overline{\mathrm{u}}=\epsilon_0\left(\frac{\mathrm{E}_0^2}{2}ight)=\frac{1}{2} \epsilon_0 \mathrm{E}_0^2 $$
Intensity is defined as the energy crossing a unit area per unit time. It is related to the average energy density by the speed of the wave $c$ :
$$ \mathrm{I}=\overline{\mathrm{u}} 	imes \mathrm{c} $$
$$\Rightarrow $$ $$ I=\left(\frac{1}{2} \epsilon_0 E_0^2ight) c $$
The characteristic impedance of free space, $\mathrm{Z}_0=\sqrt{\frac{\mu_0}{\epsilon_0}}$ (approx $377 \Omega$ ).
$I=\frac{1}{2} \epsilon_0 E_0^2\left(\frac{1}{\sqrt{\mu_0 \epsilon_0}}ight)$
$$\Rightarrow $$ $I=\frac{E_0^2}{2}\left(\frac{\epsilon_0}{\sqrt{\mu_0 \epsilon_0}}ight)=\frac{E_0^2}{2} \sqrt{\frac{\epsilon_0}{\mu_0}}=\frac{(\sqrt{377})^2}{2} 	imes \frac{1}{377}=\frac{1}{2}$
$$\Rightarrow $$ $$ \frac{1}{2}=\frac{1}{\alpha} \Rightarrow \alpha=2 $$
Therefore, the value of $\alpha$ is 2.

EM waves

Choose the correct answer: