Question

In an electromagnetic wave, the amplitudes of magnetic field ${H_0}$ and ${E_0}$ in free space are related as:
A) ${H_0} = {E_0}$
B) ${H_0} = \dfrac{{{E_0}}}{c}$
C) ${H_0} = {E_0}\sqrt {{\mu _0}{\varepsilon _0}} $
D) ${H_0} = {E_0}\sqrt {\dfrac{{{\varepsilon _0}}}{{{\mu _0}}}} $

Answer 1

Hint: Before we proceed into the problem, it is important to know the definitions of permittivity and permeability.
Permittivity is the measure of the electric polarizability of a dielectric and Permeability is the measure of resistance of the material against the formation of magnetic field lines through the materials.

Complete step by step answer:
The dielectric is defined as the electric insulator that can be polarised by applying an electric field. That means that whenever a dielectric medium is placed in an external electric field, the electric charges do not flow through the material like it happens in a normal electric conductor, but the charge carries a slight shift from their equilibrium positions. This results in dielectric polarization. In other words, the positive and negative charges re-align themselves along the direction of the electric field.
The permittivity is a measure of this particular property.
This means that the materials which have a higher permittivity polarizes to a greater extent in the presence of an external electric field compared to a material with lower permittivity.
The field created due to the polarisation –
$D = \varepsilon E$
The permittivity of free space, also called vacuum permittivity, is the ratio of $\dfrac{D}{E}$ in space.
The value is –
${\varepsilon _0} = 8.854 \times {10^{ - 12}}F{m^{ - 1}}$
The permeability is the measure of resistance of the material against the formation of magnetic field lines through the materials. The absolute value of the permeability is –
$\Rightarrow {\mu _0} = 4\pi \times {10^{ - 7}}H{m^{ - 1}}$
If we multiply these two quantities, we get –
$\Rightarrow {\varepsilon _0}{\mu _0} = 8.854 \times {10^{ - 12}} \times 4\pi \times {10^{ - 7}} = 111.26 \times {10^{ - 19}}$
Inverting,
$\Rightarrow \dfrac{1}{{{\varepsilon _0}{\mu _0}}} = \dfrac{1}{{111.26 \times {{10}^{ - 19}}}} = 8.98 \times {10^{16}}$
This value is almost equal to the square of the speed of light.
Hence, we have –
$\Rightarrow \dfrac{1}{{{\varepsilon _0}{\mu _0}}} = {c^2}$
$\Rightarrow c = \sqrt {\dfrac{1}{{{\varepsilon _0}{\mu _0}}}} $
The electromagnetic wave consists of transverse electric field waves and transverse magnetic field waves travelling at right angles to each other, such that the ratio of the amplitudes of magnetic field and electric field is equal to the velocity of the wave, which is the standard value of velocity of light in vacuum.
If ${H_0}$ and ${E_0}$ are the amplitudes of the magnetic field and electric field of an electromagnetic wave, we have –
$\Rightarrow \dfrac{{{H_0}}}{{{E_0}}} = \dfrac{1}{c}$
Substituting the value of c from the above equation –
$\Rightarrow \dfrac{{{H_0}}}{{{E_0}}} = \dfrac{1}{{\sqrt {\dfrac{1}{{{\varepsilon _0}{\mu _0}}}} }}$
$ \Rightarrow \dfrac{{{H_0}}}{{{E_0}}} = \sqrt {{\varepsilon _0}{\mu _0}} $
$\therefore {H_0} = {E_0}\sqrt {{\mu _0}{\varepsilon _0}} $

Hence, the correct option is Option C.

Note: Here, the magnetic field is denoted by H. There is another magnetic field denoted by B. Both B and H are used to denote magnetic fields with the same units, but B represents the magnetic field density related to magnetic flux, while H represents the magnetic field intensity, which talks mainly about the strength of the poles.