Question

Unit of ${E_0}/{H_0}$ is:
A. $A$
B. $\Omega $
C. $V$
D. $V{m^{ - 2}}$

Answer 1

Hint:In this question, we are asked to find the unit of the ratio of the electric field to the magnetic field of the free space or the vacuum. For this, we will use the relation of magnetic induction with both the electric and magnetic field. After that, we will equate both these relations to find the required ratio. Then we will use the dimensional equations of the quantities obtained in the ratio to determine its unit.

Formulas used:
\[{B_0} = {\mu _0}{H_0}\]
where, \[{B_0}\] is the electromagnetic induction, \[{\mu _0}\] is the vacuum permeability and \[{H_0}\] is the magnetic field.
\[{B_0} = {\mu _0}{\varepsilon _0}c{E_0}\]
where, \[{B_0}\] is the electromagnetic induction, \[{\mu _0}\] is the vacuum permeability, \[{\varepsilon _0}\] is the permittivity of free space or vacuum, \[c\] is the speed of electromagnetic wave and \[{E_0}\] is the electric field.

Complete step by step answer:
We know that the relation between the electromagnetic induction and the magnetic field is given by: \[{B_0} = {\mu _0}{H_0}\]
Also, the relation between the electromagnetic induction and the electric field is given by:
\[{B_0} = {\mu _0}{\varepsilon _0}c{E_0}\]
Now, to find the ratio, we will equate both these equations.
\[{\mu _0}{H_0} = {\mu _0}{\varepsilon _0}c{E_0} \\
\Rightarrow \dfrac{{{E_0}}}{{{H_0}}} = \dfrac{1}{{{\varepsilon _0}c}} \\ \]
The dimensional equation of permittivity of the free space is: \[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}\]
The dimensional equation of speed of electromagnetic wave is: \[{L^1}{T^{ - 1}}\]
Therefore, the dimension of
\[\dfrac{1}{{{\varepsilon _0}c}} = \dfrac{1}{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^1}{T^{ - 1}}]}} \\
\Rightarrow\dfrac{1}{{{\varepsilon _0}c}}= \dfrac{1}{{[{M^{ - 1}}{L^{ - 2}}{T^3}{A^2}]}} \\
\therefore\dfrac{1}{{{\varepsilon _0}c}}= [{M^1}{L^2}{T^{ - 3}}{A^{ - 2}}]\]
This dimensional equation is the same as that of the resistance and the unit of resistance is ohm $\Omega $. Thus, the unit of ${E_0}/{H_0}$ is $\Omega $.

Hence, option B is the right answer.

Note:Here, we have determined the unit of ${E_0}/{H_0}$. We can also find its value by the obtained equation \[\dfrac{{{E_0}}}{{{H_0}}} = \dfrac{1}{{{\varepsilon _0}c}}\]
We know that \[{\varepsilon _0} = 8.85 \times {10^{ - 12}}\dfrac{{{C^2}}}{{N{m^2}}}\] and \[c = 3 \times {10^8}\dfrac{m}{s}\]
\[ \Rightarrow \dfrac{{{E_0}}}{{{H_0}}} = \dfrac{1}{{8.85 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}} \\
\therefore\dfrac{{{E_0}}}{{{H_0}}}= 376.6 \approx 377\Omega \]