Question

For an electromagnetic wave traveling in free space, the relation between average energy densities due to electric $\left(U_e\right)$and magnetic $\left(U_m\right)$ fields is :

• A. $U_e\ne U_m$
• B. $U_e=U_m$
• C. $U_e>U_m$
• D. $U_e<U_m$

Answer 1

The correct option is B

$U_e=U_m$

Step 1. Given data,

The average energy density of the electric field $=U_e$

The average energy density of the magnetic field $=U_m$

Step 2. Finding the average energy of the electric field, $U_e$

We know that the average energy density of the electric field is given by the expression,

$U_e=\frac{1}{2}{\epsilon }_0{E}^2...\left(1\right)$ (where ${\epsilon }_0$ is the permittivity of free space and $E$ is electric field).

Since the ratio of electric field and magnetic field in an electromagnetic wave in free space is always equal to the speed of light.

Therefore, $\frac{E}{B}=c$ (where $c$ is speed of light).

$\Rightarrow$$E=cB$

And as we know that $\left(c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\right)$, (where ${\epsilon }_0$ is the permittivity of free space and ${\mu }_0$ is the permeability of free space).

$\Rightarrow$$E=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}B$

From the equation $\left(1\right)$ we can write,

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

$\Rightarrow$ $U_e=\frac{1}{2}{\epsilon }_0\times \frac{1}{{\mu }_0{\epsilon }_0}{B}^2$

We know that the average energy density of the magnetic field is given by the expression,

$U_m=\frac{1}{2}\frac{B^2}{{\mu }_0}...\left(2\right)$ (where ${\mu }_0$ is the permeability of free space and $B$ is magnetic field).

$\Rightarrow$ $U_e=\frac{B^2}{2{\mu }_0}$

Step 3. Equating both the equations

Then from equation, $\left(2\right)$ we can write $U_e=U_m$,

Hence, in electromagnetic waves, the average energy density due to electric and magnetic fields are equal.

$\Rightarrow$ $U_e=U_m$

Hence, option B is the correct answer.