Q 12-08-151JEE MainJEE Main 2021 (31 Aug, Shift 1)Medium
The electric field in an electromagnetic wave is given by $E = (50\ \text{N C}^{-1})\sin\omega\left(t - \dfrac xc\right)$. The energy contained in a cylinder of volume $V$ is $5.5\times10^{-12}$ J. The value of $V$ is ______ $\text{cm}^3$. (given $\epsilon_0 = 8.8\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$)
Numerical value type. Enter your answer.
Answer: 500
Average energy density (electric + magnetic): $u = \dfrac12\epsilon_0E_0^2 = \dfrac12\times8.8\times10^{-12}\times2500 = 1.1\times10^{-8}\ \text{J m}^{-3}$.
$$V = \frac{5.5\times10^{-12}}{1.1\times10^{-8}} = 5\times10^{-4}\ \text{m}^3 = 500\ \text{cm}^3$$
Solution by Sreeraj P, M.Sc Physics