Q154 / 160 · AP EAMCET 2024 (20 May Shift 1) · Physics
The average value of electric energy density in an electromagnetic wave is.
[where $\mathrm{E}_0$ is peak value]
- A$\frac{\epsilon_0 \mathrm{E}_{\mathrm{rms}}^2}{4}$
- B$\frac{1}{2} \epsilon_0 \mathrm{E}_0^2$
- C$\frac{1}{2} \epsilon_0 \mathrm{E}_0$
- Correct answerD$\frac{1}{4} \epsilon_0 \mathrm{E}_0^2$
Solution
The energy density in EM wave is
$\mathrm{u}=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^2=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}_0^2 \sin ^2(\omega \mathrm{t}-\mathrm{kx})$
$\therefore \quad$ Average energy density,
$ \lt \mathrm{u}\gt=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \lt \sin ^2(\omega \mathrm{t}-\mathrm{kx})\gt=\left(\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2\right) \times \frac{1}{2}$
$=\frac{1}{4} \epsilon_0 \mathrm{E}_0^2$
Hence, the correct answer is option: 'D'.