Question

In an electromagnetic wave, the average energy density associated with the magnetic field is?

• A. $\frac{{{\mathrm{Li}}_0}^2}{2}$
• B. $\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}$
• C. $\frac{{\mathrm{\mu }}_0{\mathrm{B}}^2}{2}$
• D. $\frac{{\mathrm{\mu }}_0}{2{\mathrm{B}}^2}$

Answer 1

The correct option is B

$\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}$

Energy density :

In an electromagnetic wave, the amplitude is known to be the maximum field strength of the electric field and magnetic field, and the energy of a wave is determined by the amplitude of the wave.

The energy per unit volume is denoted as energy density (u) which is the sum of energy density from the electric field and magnetic field.
The expression for energy density can be shown as:
$u={\displaystyle \frac{1}{2}}{\epsilon }_0{E}^2+{\displaystyle \frac{1}{2{\mu }_0}}{B}^2$ (1)

Where,$u$ = energy density, ${\epsilon }_0$=permittivity of free space, ${\mu }_0$=vacuum permeability,$B$ = magnetic field and $E$ = electric field.

Expression for energy density of electric field:
The average energy density due to the electric field can be given as:
$u=\frac{1}{2}{\epsilon }_0{E}^2$ (2)

Relation between electric and magnetic field:

For electromagnetic the relation between electric and magnetic field can be given as:

$$\begin{gathered} \mathrm{E}=\mathrm{cB} \\ \Rightarrow \mathrm{B}=\frac{\mathrm{E}}{\mathrm{c}} \end{gathered}$$

Where, $B$ = magnetic field and $E$ = electric field

$\mathrm{c}=\mathrm{speed}\mathrm{of}\mathrm{light}=3\times {10}^8$

and ${\mathrm{c}}^2=\frac{1}{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}$

Average energy density of magnetic field:

Putting the given values in equation(1), we can calculate average energy density of magnetic field as:

$$\begin{gathered} \mathrm{u}\left(\mathrm{B}\right)=\left(\frac{1}{2}\right){\mathrm{\epsilon }}_0{\left(\mathrm{cB}\right)}^2 \\ =\left(\frac{1}{2}\right){\mathrm{\epsilon }}_0\left(\frac{1}{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}\right){\mathrm{B}}^2 \\ =\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0} \end{gathered}$$
Hence, the average energy density associated with the magnetic field is $\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0}$. Hence, option B is correct.