Question

If ${\mu }_0$ is the permeability of free space and ${\epsilon }_0$ is the permittivity of free space, the speed of light in a vacuum is given by

• A. $\sqrt{{\mu }_0{\epsilon }_0}$
• B. $\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$
• C. $\sqrt{\frac{{\epsilon }_0}{{\mu }_0}}$
• D. $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Answer 1

The correct option is D

$\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Step 1: Given data

${\mu }_0$ is the permeability of free space

${\epsilon }_0$ is the permittivity of free space

Step 2: Solution

The relationship between permittivity, permeability and speed of light is,

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Step 3: Derivation

From Maxwell's equations, with no source in free space,

$$\begin{gathered} \nabla ·\overrightarrow{E}=0...\left(1\right) \\ \nabla ·\overrightarrow{B}=0...\left(2\right) \end{gathered}$$

According to Faraday's law,
$\nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}...\left(3\right)$

According to Ampere circuital law,

$\nabla \times \overrightarrow{B}={\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t}...\left(4\right)$

$\overrightarrow{E}$ is the electric field and $\overrightarrow{B}$ is the magnetic field.

Taking the curl of the third and fourth equations, we get

$$\begin{gathered} \nabla \times \left(\nabla \times \overrightarrow{E}\right)=-\frac{\partial }{\partial t}\left(\nabla \times \overrightarrow{B}\right) \\ \nabla \times \left(\nabla \times \overrightarrow{B}\right)={\mu }_0{\epsilon }_0\frac{\partial }{\partial t}\left(\nabla \times \overrightarrow{E}\right) \end{gathered}$$

Substituting equation $\left(3\right)$ and $\left(4\right)$ in the above equations where appropriate,

$$\begin{gathered} \nabla \times \left(\nabla \times \overrightarrow{E}\right)=-{\mu }_0{\epsilon }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}...\left(5\right) \\ \nabla \times \left(\nabla \times \overrightarrow{B}\right)=-{\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{B}}{\partial t}...\left(6\right) \end{gathered}$$

The curl of the curl of a vector can be expressed as,

$\nabla \times \left(\nabla \times \overrightarrow{V}\right)=\nabla \left(\nabla ·\overrightarrow{V}\right)-{\nabla }^2\overrightarrow{V}$

So,
$$\begin{gathered} \begin{array}{ccc}& -{\mu }_0{\epsilon }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}=\nabla \left(\nabla ·\overrightarrow{E}\right)-{\nabla }^2\overrightarrow{E}& \\ \therefore & {\nabla }^2\overrightarrow{E}={\mu }_0{\epsilon }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}& \left[\because \nabla ·\overrightarrow{E}=0\right]\end{array} \\ \begin{array}{ccc}& -{\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{B}}{\partial t}=\nabla \left(\nabla ·\overrightarrow{B}\right)-{\nabla }^2\overrightarrow{B}& \\ \therefore & {\nabla }^2\overrightarrow{B}={\mu }_0{\epsilon }_0\frac{{\partial }^2\overrightarrow{B}}{\partial t^2}& \left[\because \nabla ·\overrightarrow{B}=0\right]\end{array} \end{gathered}$$

The equation of wave motion is given as,

${\nabla }^2\overrightarrow{V}=\frac{1}{v^2}\frac{{\mathrm{\partial }}^2\overrightarrow{V}}{\mathrm{\partial }{t}^2}$

where $v$ is the velocity of the wave.

We know that velocity of electromagnetic waves is $c$. So,

$\begin{array}{cc}& \frac{1}{c^2}={\mu }_0{\epsilon }_0\\ \therefore & c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\end{array}$

Therefore, Option D is correct