Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

Total energy density of electromagnetic waves in vacuum is given by the relation:

Options:

$\frac{1}{2} . \frac{E^2}{\varepsilon_0}+\frac{B^2}{2 \mu_0}$

$\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \mu_0 B^2$

$\frac{E^2+B^2}{c}$

$\frac{1}{2} \varepsilon_0 E^2+\frac{B^2}{2 \mu_0}$

Correct Answer:

$\frac{1}{2} \varepsilon_0 E^2+\frac{B^2}{2 \mu_0}$

Explanation:

The energy in EM waves is divided equally between the electric and magnetic fields.

The total energy per unit volume is u = ue + um

$=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\mu_0}$