The energy density of an electromagnetic wave in a vacuum is given by the relation:

  • A
    $\frac{E^2}{2\varepsilon_0} + \frac{B^2}{2\mu_0}$
  • B
    $\frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2}\mu_0 B^2$
  • C
    $\frac{E^2 + B^2}{c}$
  • D
    $\frac{1}{2}\varepsilon_0 E^2 + \frac{B^2}{2\mu_0}$

Explore More

Similar Questions

An electromagnetic wave is travelling in $x$-direction with electric field vector given by $\vec{E}_{y} = E_{0} \sin(kx - \omega t) \hat{j}$. The correct expression for the magnetic field vector is:

The electric field in a plane electromagnetic wave is given by $E = 50 \sin(500x - 10 \times 10^{10}t) \, V/m$. The velocity of the electromagnetic wave in this medium is: (Given $C = \text{speed of light in vacuum}$)

The figure shows the electric field $\overrightarrow{E}$ of an $EM$ wave at a certain point and a certain instant. The wave is transporting energy in the negative $z$ direction ($-z$ direction). What is the direction of the magnetic field $\overrightarrow{B}$ of the wave at that point and instant?

Write the equation for the energy density of electromagnetic waves.

Electromagnetic waves are transverse in nature is evident by

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo