- Home
- Standard 12
- Physics
The electric field of a plane electromagnetic wave is given by $\vec E = {E_0}\hat i\,\cos \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)$ The corresponding magnetic field $\vec B$ is then given by
$\vec B = \frac{{{E_0}}}{C}\hat j\,\sin \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)$
$\vec B = \frac{{{E_0}}}{C}\hat k\,\sin \,\left( {kz} \right)\,\cos \,\left( {\omega t} \right)$
$\vec B = \frac{{{E_0}}}{C}\hat j\,\cos \,\left( {kz} \right)\,\sin \,\left( {\omega t} \right)$
$\vec B = \frac{{{E_0}}}{C}\hat j\,\sin \,\left( {kz} \right)\,\sin \,\left( {\omega t} \right)$
Solution
$\because \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}} \| \overrightarrow{\mathrm{v}}$
Given that wave is propagating along positive $z$ -axis and $\overrightarrow{\mathrm{E}}$ along positive $x$ -axis. Hence $\overrightarrow{\mathrm{B}}$ along $\mathrm{y}$ -axis.
From Maxwell equation
$\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{E}}=-\frac{\partial \mathrm{B}}{\partial \mathrm{t}}$
i.e. $\frac{\partial E}{\partial Z}=-\frac{\partial B}{d t}$ and $B_{0}=\frac{E_{0}}{C}$
