Question

The relation between electric field E and magnetic field H in electromagnetic wave is:
A. $$E=H$$
B. $$E={\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}H$$
C. $$E=\sqrt{{\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}}H$$
D. $$E=\sqrt{{\displaystyle \frac{{\epsilon }_o}{{\mu }_o}}}H$$

Answer 1

Hint: The ratio of the magnitudes of electric and magnetic fields equals the speed of light in free space.

Formula used:
In free space, where there is no charge or current, the four Maxwell’s equations are of the following form:
$$\begin{gathered} \overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{E}=0 \\ \overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{H}=0 \\ \overrightarrow{\mathrm{\nabla }}\times \overrightarrow{E}=-{\mu }_o{\displaystyle \frac{\partial \overrightarrow{H}}{\partial t}} \\ \overrightarrow{\mathrm{\nabla }}\times \overrightarrow{H}={\epsilon }_o{\displaystyle \frac{\partial \overrightarrow{E}}{\partial t}} \end{gathered}$$

Complete step by step solution:
Consider the plane wave equations of electric wave and magnetic wave:
$$\begin{gathered} \overrightarrow{E}(\overrightarrow{r},t)=\overrightarrow{E_o}{e}^{j(\overrightarrow{k}\cdot \overrightarrow{r}-\omega t)} \\ \overrightarrow{H}(\overrightarrow{r},t)=\overrightarrow{H_o}{e}^{j(\overrightarrow{k}\cdot \overrightarrow{r}-\omega t)} \end{gathered}$$
Where $$\overrightarrow{E_o}$$ and $$\overrightarrow{H_o}$$ are complex amplitudes, which are constants in space and time, $$\overrightarrow{k}$$ is the wave vector determining the direction of propagation of the wave. $$\overrightarrow{k}$$ is defined as
$$\overrightarrow{k}={\displaystyle \frac{2\pi }{\lambda }}\stackrel{\wedge }{n}={\displaystyle \frac{\omega }{c}}\stackrel{\wedge }{n}$$
Where $$\stackrel{\wedge }{n}$$ is the unit vector along the direction of propagation.
Substituting the plane wave solutions in equations $$\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{E}=0$$ and $$\overrightarrow{\mathrm{\nabla }}\cdot \overrightarrow{H}=0$$ respectively:
  $$\overrightarrow{k}\cdot \overrightarrow{E}=0$$ and $$\overrightarrow{k}\cdot \overrightarrow{H}=0$$
Thus, $$\overrightarrow{E}$$ and $$\overrightarrow{H}$$ are both perpendicular to the direction of propagation vector $$\overrightarrow{k}$$.
This implies that electromagnetic waves are transverse in nature.
Substituting the plane wave solutions in equations $$\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{E}=-{\mu }_o{\displaystyle \frac{\partial \overrightarrow{H}}{\partial t}}$$ and $$\overrightarrow{\mathrm{\nabla }}\times \overrightarrow{H}={\epsilon }_o{\displaystyle \frac{\partial \overrightarrow{E}}{\partial t}}$$ respectively:
$$\begin{gathered} \overrightarrow{k}\times \overrightarrow{E}={\mu }_o\omega \overrightarrow{H} \\ \overrightarrow{k}\times \overrightarrow{H}=-{\epsilon }_o\omega \overrightarrow{E} \end{gathered}$$
Since $$\overrightarrow{E}$$ is normal to $$\overrightarrow{k}$$, in terms of magnitude,
$$\begin{gathered} kE={\mu }_o\omega H \\ \sqrt{{\epsilon }_o}E=\sqrt{{\mu }_o}H[k^2={\epsilon }_o{\mu }_o{\omega }^2] \\ E=\sqrt{{\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}}H \end{gathered}$$

Therefore, option C is the correct relation between E and H.

Additional information:
$$\overrightarrow{E}$$ and $$\overrightarrow{H}$$ are both perpendicular to the direction of propagation vector $$\overrightarrow{k}$$.
This implies that electromagnetic waves are transverse in nature.
$$\overrightarrow{k}\times \overrightarrow{E}={\mu }_o\omega \overrightarrow{H}$$ implies that $$\overrightarrow{H}$$ is perpendicular to both $$\overrightarrow{k}$$ and $$\overrightarrow{E}$$.
$$\overrightarrow{k}\times \overrightarrow{H}=-{\epsilon }_o\omega \overrightarrow{E}$$ implies that $$\overrightarrow{E}$$ is perpendicular to both $$\overrightarrow{k}$$ and $$\overrightarrow{H}$$.
Thus, the field $$\overrightarrow{E}$$ and $$\overrightarrow{H}$$ are mutually perpendicular and also they are perpendicular to the direction of propagation vector $$\overrightarrow{k}$$.

The velocity of propagation of electromagnetic waves is equal to the speed of light in free space. This indicates that the light is an electromagnetic wave.

Note: The relation obtained between $$\overrightarrow{E}$$ and $$\overrightarrow{H}$$ is only true for plane electromagnetic waves in free space.