Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

The rms value of the electric field of the light coming from the sun is 720 NC-1. The average total energy density of the electromagnetic wave is:

Options:

$3.3 \times 10^{-3} \mathrm{Jm}^{-3}$

$4.58 \times 10^{-6} \mathrm{Jm}^{-3}$

$6.37 \times 10^{-9} \mathrm{Jm}^{-3}$

$81.35 \times 10^{-12} \mathrm{Jm}^{-3}$

Correct Answer:

$4.58 \times 10^{-6} \mathrm{Jm}^{-3}$

Explanation:

Total average energy density of electromagnetic wave is

u = $\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0} \mathrm{~B}_{\mathrm{rms}}^2$

$=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0}\left(\frac{\mathrm{E}_{\mathrm{rms}}^2}{\mathrm{c}^2}\right) \quad\left(∵ \mathrm{B}_{\mathrm{rms}}=\frac{\mathrm{E}_{\mathrm{rms}}}{\mathrm{c}}\right)$

$=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2 \mu_0} \mathrm{E}_{\mathrm{rms}}^2 \varepsilon_0 \mu_0$

$=\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2+\frac{1}{2} \varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2=\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2$

$=\left(8.85 \times 10^{-12}\right) \times(720)^2=4.58 \times 10^{-6} \mathrm{Jm}^{-3}$