JEE Main 2024 — Electromagnetic Waves Question with Solution

The average energy density of an electromagnetic wave is given by the formula:

$$U=\frac{1}{2}{\epsilon }_0{E}^2+\frac{1}{2}\frac{B^2}{{\mu }_0}$$

For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:

$$U_E=\frac{1}{2}{\epsilon }_0{E}^2$$

Given that the electric field amplitude ($E$) is $50{\mathrm{Vm}}^{-1}$ and the permittivity of free space (${\epsilon }_0$) is $8.85\times 10^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$, we can calculate the energy density due to the electric field as follows:

$$U_E=\frac{1}{2}\times 8.85\times 10^{-12}\times (50{)}^2=1.10625\times 10^{-8}J/{\mathrm{m}}^3$$

Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.