Question

During the propagation of electromagnetic waves in a medium:
A. Electric energy density is equal to the magnetic energy density
B. Both electric and magnetic energy densities are zero
C. Electric energy density is double of the magnetic energy density
D. Electric energy density is half of the magnetic energy density

Answer 1

Hint: The given question is based on the concept of the electromagnetic waves. Using the properties of the components of the electromagnetic waves, that is, the electric field and the magnetic field and the mathematical relation of the same, we will find the answer to this question.

Formulae used:
$$E_0=c{B}_0$$
$$c={\displaystyle \frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}}$$
$${\mu }_E={\displaystyle \frac{1}{2}}{\epsilon }_0{E}_0^2$$
$${\mu }_B={\displaystyle \frac{1}{2}}{{\displaystyle \frac{B_0}{{\mu }_0}}}^2$$

Complete answer:
The electromagnetic waves are the composition of the electric and the magnetic fields. The oscillating electric and magnetic fields produce the electromagnetic waves. The direction of the propagation of these waves is perpendicular to the direction of the electric and magnetic fields.

The relation between the electric field and the magnetic field is given as follows:

$$E_0=c{B}_0$$
Where $$E_0$$ is the electric field, $c$ is the speed of light in vacuum and $$B_0$$ is the magnetic field.

This speed of light is given as follows:

$$c={\displaystyle \frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}}$$
Where $${\mu }_0$$ is the permeability of free space and $${\epsilon }_0$$ is the permittivity of the free space.

The electric energy density is represented by the mathematical expression as given follows:
$${\mu }_E={\displaystyle \frac{1}{2}}{\epsilon }_0{E}_0^2$$
Where $${\epsilon }_0$$ is the permittivity of the free space and $$E_0$$ is the electric field.

The magnetic energy density is represented by the mathematical expression as given follows.
$${\mu }_B={\displaystyle \frac{1}{2}}{{\displaystyle \frac{B_0}{{\mu }_0}}}^2$$
Where $${\mu }_0$$ is the permeability of free space and $$B_0$$ is the magnetic field.

Now, we will derive the relation between the electric and the magnetic energy densities.

$$\begin{array}{rl}& {\mu }_E={\displaystyle \frac{1}{2}}{\epsilon }_0{E}_0^2\\ & \Rightarrow {\mu }_E={\displaystyle \frac{1}{2}}{\epsilon }_0{c}^2{B}_0^2\end{array}$$

Now substitute the expression of the velocity of light.

$$\begin{array}{rl}& {\mu }_E={\displaystyle \frac{1}{2}}{\epsilon }_0\left({\displaystyle \frac{1}{{\mu }_0{\epsilon }_0}}\right)B_0^2\\ & \Rightarrow {\mu }_E={\displaystyle \frac{1}{2}}{\displaystyle \frac{B_0^2}{{\mu }_0}}\end{array}$$

The RHS part of the above expression represents the magnetic energy density.

$${\mu }_E={\mu }_B$$

$$\therefore$$ The electric energy density is equal to the magnetic energy density

As the value of the electric energy density is equal to the magnetic energy density, thus, the option (A) is correct.

Note:
As the electromagnetic waves are the composition of both the fields, that is, the electric field and the magnetic field, the properties of these fields affect the waves. Thus, the mathematical representation that relates the parameters such as, electric field, magnetic field, the electric field density, the magnetic field, the speed of light should be known to solve this type of problems.