Question

The electric field part of an electromagnetic wave in medium is represented by
E_x = 0;
$E_y=2.5\frac{N}{C}\cos \left[\left(2\mathrm{\pi }\times {10}^6\frac{\mathrm{rad}}{\mathrm{s}}\right)t-\left(\mathrm{\pi }\times {10}^{-2}\frac{1}{\mathrm{m}}\right)x\right];$
E_z = 0.

The wave is :
(1) moving in x direction or y direction?
(2) Calculate the frequency and wavelength.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Given}\mathrm{that}, \\ {\mathrm{E}}_{\mathrm{y}}=2.5\frac{\mathrm{N}}{\mathrm{C}}\cos \left[\left(2\mathrm{\pi }\times {10}^6\frac{\mathrm{rad}}{\mathrm{s}}\right)\mathrm{t}-\left(\mathrm{\pi }\times {10}^{-2}\frac{1}{\mathrm{m}}\right)\mathrm{x}\right] \\ \mathrm{The}\mathrm{wave}\mathrm{equation}\mathrm{is}{\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_0\cos \left(\mathrm{\omega t}-\mathrm{kx}\right) \\ \mathrm{\omega }=2\mathrm{\pi }\times {10}^6\mathrm{rad}/\mathrm{s}=2\mathrm{\pi \nu } \\ \mathrm{\nu }=\frac{2\mathrm{\pi }\times {10}^6}{2\mathrm{\pi }}={10}^6\mathrm{Hz} \\ \mathrm{k}=\mathrm{\pi }\times {10}^{-2}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \\ \mathrm{\lambda }=\frac{2\mathrm{\pi }}{\mathrm{\pi }\times {10}^{-2}}=200\mathrm{m} \\ \mathrm{Hence},\mathrm{the}\mathrm{wave}\mathrm{is}\mathrm{moving}\mathrm{along}\mathrm{x}-\mathrm{direction}\mathrm{with}\mathrm{frequency}{10}^6\mathrm{Hz} \\ \mathrm{and}\mathrm{wavelength}200\mathrm{m}. \end{gathered}$$