The equation of the electric field of an electromagnetic wave propagating through free space is give

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 $\_\_\_\_$
$\left(ight.$ Take $\sqrt{\frac{\mu_0}{\varepsilon_o}}=377$ in SI units $)$

Professor PC Thomas Classes · Answer

The average power per unit area (intensity) of an electromagnetic wave is given by $I = \frac{E_0^2}{2\mu_0 c}$. We know that the speed of light is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$, thus $\mu_0 c = \mu_0 \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \sqrt{\frac{\mu_0}{\epsilon_0}}$. We are given $E_0 = \sqrt{377} 	ext{ N/C}$ and $\sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \Omega$. Substituting these yields $I = \frac{(\sqrt{377})^2}{2 	imes 377} = \frac{377}{2 	imes 377} = \frac{1}{2} 	ext{ W/m}^2$. Comparing this to $\frac{1}{\alpha} 	ext{ W/m}^2$, we get $\alpha = 2$.