Question 65: EM waves

Question

The electric field in an electromagnetic wave is given as
$$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}ight) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$$
where $$\omega$$ and $$c$$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $$5 	imes 10^{-4} \mathrm{~m}^{3}$$ will be
(Given $$\varepsilon_{0}=8.85 	imes 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$$ )

Accepted Answer

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:
$$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}ight) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$$
The amplitude $$E$$ is $$20 \mathrm{~N/C}$$.
Now, we can find the energy density:
$$u = \frac{1}{2} \cdot 8.85 	imes 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2$$
$$u = \frac{1}{2} \cdot 8.85 	imes 10^{-12} \cdot 400$$
$$u = 1.77 	imes 10^{-9} \mathrm{J/m^3}$$
The energy contained in a volume of $$5 	imes 10^{-4} \mathrm{m^3}$$ will be:
$$E_	ext{total} = u \cdot V$$
$$E_	ext{total} = 1.77 	imes 10^{-9} \mathrm{J/m^3} \cdot 5 	imes 10^{-4} \mathrm{m^3}$$
$$E_	ext{total} = 8.85 	imes 10^{-13} \mathrm{J}$$

Answer

$$17 \cdot 7 	imes 10^{-13} \mathrm{~J}$$

Answer

$$28 \cdot 5 	imes 10^{-13} \mathrm{~J}$$

Answer

$$8 \cdot 85 	imes 10^{-13} \mathrm{~J}$$

Answer

$$88 \cdot 5 	imes 10^{-13} \mathrm{~J}$$

EM waves

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