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

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

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.