QJEEM23Phb55c: JEE Main 2023 Physics - Class 12 JEEM23Phb55c Solution: The electric field in an electromagnetic wave is given as where …

To find the energy contained in a volume of an electromagnetic wave, we need to calculate the energy density and then multiply it by the volume.

For an electromagnetic wave, the energy density $u$ is given by:

$u = \frac{1}{2} \varepsilon_0 E^2$

where $\varepsilon_0$ is the vacuum permittivity and $E$ is the electric field amplitude.

First, let's find the amplitude of the electric field. In the given equation:

$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$

The amplitude $E$ is $20 \mathrm{~N/C}$ .

Now, we can find the energy density:

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2$

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 400$

$u = 1.77 \times 10^{-9} \mathrm{J/m^3}$

The energy contained in a volume of $5 \times 10^{-4} \mathrm{m^3}$ will be:

$E_\text{total} = u \cdot V$

$E_\text{total} = 1.77 \times 10^{-9} \mathrm{J/m^3} \cdot 5 \times 10^{-4} \mathrm{m^3}$

$E_\text{total} = 8.85 \times 10^{-13} \mathrm{J}$