Question

The energy of electromagnetic wave in vacuum is given by the relation

• A. $\frac{E^2}{2{\epsilon }_0}+\frac{B^2}{2{\mu }_0}$
• B. $\frac{1}{2}{\epsilon }_0{E}^2+\frac{1}{2}{\mu }_0{B}^2$
• C. $\frac{E^2+B^2}{c}$
• D. $\frac{1}{2}{\epsilon }_0{E}^2+\frac{B^2}{2{\mu }_0}$

Answer 1

The correct option is D $\frac{1}{2}{\epsilon }_0{E}^2+\frac{B^2}{2{\mu }_0}$

$\frac{1}{2}{\epsilon }_0{E}_0^2$ is electric energy density.
$\frac{B^2}{2{\mu }_0}$ is magnetic energy density.
So total energy $=\frac{1}{2}{\epsilon }_0{E}_0^2+\frac{B^2}{2{\mu }_0}$