Question

The relation between electric field E and magnetic field H in electromagnetic wave is:
A. $E=H$
B. $E=\dfrac{{{\mu }_{o}}}{{{\varepsilon }_{o}}}H$
C. $E=\sqrt{\dfrac{{{\mu }_{o}}}{{{\varepsilon }_{o}}}}H$
D. $E=\sqrt{\dfrac{{{\varepsilon }_{o}}}{{{\mu }_{o}}}}H$

Answer 1

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:

$$\overset{\to }{\mathop{\nabla }}\,\cdot \overset{\to }{\mathop{E}}\,=0 \\\ \overset{\to }{\mathop{\nabla }}\,\cdot \overset{\to }{\mathop{H}}\,=0 \\\ \overset{\to }{\mathop{\nabla }}\,\times \overset{\to }{\mathop{E}}\,=-{{\mu }_{o}}\dfrac{\partial \overset{\to }{\mathop{H}}\,}{\partial t} \\\ \overset{\to }{\mathop{\nabla }}\,\times \overset{\to }{\mathop{H}}\,={{\varepsilon }_{o}}\dfrac{\partial \overset{\to }{\mathop{E}}\,}{\partial t} \\\$$

Complete step by step solution:
Consider the plane wave equations of electric wave and magnetic wave:

$$\overset{\to }{\mathop{E}}\,\left( \overset{\to }{\mathop{r}}\,,t \right)=\overset{\to }{\mathop{{{E}_{o}}}}\,{{e}^{j(\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{r}}\,-\omega t)}} \\\ \overset{\to }{\mathop{H}}\,\left( \overset{\to }{\mathop{r}}\,,t \right)=\overset{\to }{\mathop{{{H}_{o}}}}\,{{e}^{j(\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{r}}\,-\omega t)}} \\\$$

Where $\overset{\to }{\mathop{{{E}_{o}}}}\,$ and $\overset{\to }{\mathop{{{H}_{o}}}}\,$ are complex amplitudes, which are constants in space and time, $\overset{\to }{\mathop{k}}\,$ is the wave vector determining the direction of propagation of the wave. $\overset{\to }{\mathop{k}}\,$ is defined as
$\overset{\to }{\mathop{k}}\,=\dfrac{2\pi }{\lambda }\overset{\wedge }{\mathop{n}}\,=\dfrac{\omega }{c}\overset{\wedge }{\mathop{n}}\,$
Where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector along the direction of propagation.
Substituting the plane wave solutions in equations $\overset{\to }{\mathop{\nabla }}\,\cdot \overset{\to }{\mathop{E}}\,=0$ and $\overset{\to }{\mathop{\nabla }}\,\cdot \overset{\to }{\mathop{H}}\,=0$ respectively:
$\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{E}}\,=0$ and $\overset{\to }{\mathop{k}}\,\cdot \overset{\to }{\mathop{H}}\,=0$
Thus, $\overset{\to }{\mathop{E}}\,$ and $\overset{\to }{\mathop{H}}\,$ are both perpendicular to the direction of propagation vector $\overset{\to }{\mathop{k}}\,$.
This implies that electromagnetic waves are transverse in nature.
Substituting the plane wave solutions in equations $\overset{\to }{\mathop{\nabla }}\,\times \overset{\to }{\mathop{E}}\,=-{{\mu }_{o}}\dfrac{\partial \overset{\to }{\mathop{H}}\,}{\partial t}$ and $\overset{\to }{\mathop{\nabla }}\,\times \overset{\to }{\mathop{H}}\,={{\varepsilon }_{o}}\dfrac{\partial \overset{\to }{\mathop{E}}\,}{\partial t}$ respectively:

$$\overset{\to }{\mathop{k}}\,\times \overset{\to }{\mathop{E}}\,={{\mu }_{o}}\omega \overset{\to }{\mathop{H}}\, \\\ \overset{\to }{\mathop{k}}\,\times \overset{\to }{\mathop{H}}\,=-{{\varepsilon }_{o}}\omega \overset{\to }{\mathop{E}}\, \\\$$

Since $\overset{\to }{\mathop{E}}\,$ is normal to $\overset{\to }{\mathop{k}}\,$, in terms of magnitude,

$$\text{ }kE={{\mu }_{o}}\omega H \\\ \sqrt{{{\varepsilon }_{o}}}E=\sqrt{{{\mu }_{o}}}H\text{ }\\!\\![\\!\\!\text{ }{{k}^{2}}={{\varepsilon }_{o}}{{\mu }_{o}}{{\omega }^{2}}] \\\ \text{ }E=\sqrt{\dfrac{{{\mu }_{o}}}{{{\varepsilon }_{o}}}}H \\\$$

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

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

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 $\overset{\to }{\mathop{E}}\,$ and $\overset{\to }{\mathop{H}}\,$ is only true for plane electromagnetic waves in free space.