Question

An electromagnetic wave in vacuum has the electric and magnetic field $\vec{E}$ and $\vec{B}$, which are always perpendicular to each other. If the direction of polarization is given by $\vec{X}$ and that of wave propagation by $\vec{k}$,then:
A. $\vec{X} \| {\vec{B}}$ and $\vec{k}\parallel \vec{B}\times \vec{E} $
B. $\vec{X}\| {\vec{E}}$ and $\vec{k}\parallel \vec{E}\times \vec{B}$
C. $\vec{X} \| {\vec{B}}$ and $\vec{k} \parallel \vec{E}\times \vec{B}$
D. $\vec{X} \| {\vec{E}}$ and $\vec{k} \parallel \vec{B}\times \vec{E}$

Answer 1

Hint: We have an electromagnetic wave in vacuum. It is said that the electric field and the magnetic field are perpendicular to each other. We know that the electromagnetic wave is a transverse wave. Therefore the properties of transverse waves are applicable to the electromagnetic wave.

Complete answer:
In the question we are given an electromagnetic wave in vacuum.
It is said that the magnetic field and electric fields are always perpendicular to each other and it is represented as $\vec{B}$ and $\vec{E}$ respectively.
We know that electromagnetic waves are transverse in nature.
We know that the direction of polarization in a transverse wave is parallel to the electric field.
Therefore here we can say that,
$\Rightarrow \vec{X}\left\| {\vec{E}} \right.$, were ‘$\vec{X}$’ represents the direction of polarization.
We know that the equation for magnetic field intensity is given as,
$\vec{H}=\dfrac{\vec{k}\times \vec{E}}{\mu \omega }$, were ‘$\vec{k}$’ is the direction of wave propagation, ‘$\mu $’ is the magnetic permeability.
We have the equation for magnetic field as,
$\vec{B}=\mu \vec{H}$
Substituting the known value in the above equation, we get
$\Rightarrow \vec{B}=\mu \dfrac{\vec{k}\times \vec{E}}{\mu \omega }$
$\Rightarrow \vec{B}=\dfrac{\vec{k}\times \vec{E}}{\omega }$
Similarly the equation for electric field is given as,
$\vec{E}=-\left( \dfrac{\vec{k}\times \vec{H}}{\varepsilon \omega } \right)$, here ‘$\varepsilon $’ is the permittivity of free space.
From the equation for the magnetic field, we can see that $\vec{B}\bot \vec{k}$ and from the equation for the electric field we can see that $-\vec{E}\bot \vec{k}$.
From this we can conclude that ‘$\vec{k}$’ is perpendicular to both the electric and magnetic field.
So if we take the cross product of electric and magnetic field, i.e. $\vec{E}\times \vec{B}$, the direction of propagation of wave will be parallel to this, i.e.
$\Rightarrow \vec{k}\parallel \vec{E}\times \vec{B}$
Therefore we have,
$\vec{X}\left\| {\vec{E}} \right.\text{and }\vec{k}\parallel \vec{E}\times \vec{B}$

So, the correct answer is “Option B”.

Note:
When the particles of a wave are displaced perpendicular to the direction of propagation of that wave, then such a wave is called a transverse wave.
Electromagnetic waves are transverse waves. This wave does not require a medium to travel, i.e. they can even travel through vacuum. Maxwell showed that the speed of an electromagnetic wave in vacuum is $3\times {{10}^{8}}m/s$.