Question

In SI units, the dimensions of $\sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}$ is
$\begin{aligned} & A.{{A}^{-1}}TM{{L}^{3}} \\\ & B.{{A}^{2}}{{T}^{3}}{{M}^{-1}}{{L}^{-2}} \\\ & C.A{{T}^{2}}{{M}^{-1}}{{L}^{-1}} \\\ & D.A{{T}^{-3}}M{{L}^{\dfrac{3}{2}}} \\\ \end{aligned}$

Answer 1

First of all the terms given for the dimensional analysis should be known well. Here ${{\varepsilon }_{0}}$ is the electrical permittivity of the material and ${{\mu }_{0}}$ is the magnetic permeability of the material. The dimensions of these terms should be found first in order to calculate the dimension of the mentioned term in the question.

Complete step by step answer:
Electrical permittivity is given by the equation,
${{\varepsilon }_{0}}=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi F{{r}^{2}}}$
From this we can derive the dimensional formula of the permittivity,
${{\varepsilon }_{0}}=\dfrac{\left[ AT \right]\left[ AT \right]}{\left[ ML{{T}^{-2}} \right]\left[ {{L}^{2}} \right]}$
Simplifying,
$\left[ {{\varepsilon }_{0}} \right]=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]$
Now let us look at magnetic permeability of a material,
The magnetic permeability is given by the equation,
Magnetic Permeability = Magnetic flux density / [Magnetic field strength]
That is,
${{\mu }_{0}}=\left[ M{{L}^{0}}{{T}^{-2}}{{A}^{-1}} \right]\left[ {{M}^{0}}{{L}^{-1}}{{T}^{0}}{{A}^{1}} \right]$
Simplifying this will give the dimensional formula for permeability,
That is,
${{\mu }_{0}}=\left[ M{{L}^{1}}{{T}^{-2}}{{A}^{-2}} \right]$
Now as per the question, we have to find the dimensional formula of $\sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}$
That is,
By substituting the values in it will results in,
$\sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}={{\left( \dfrac{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]}{\left[ M{{L}^{1}}{{T}^{-2}}{{A}^{-2}} \right]} \right)}^{\dfrac{1}{2}}}$
Simplifying this will give us the answer for the question,

& \sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}={{\left[ {{M}^{-2}}{{L}^{-4}}{{T}^{6}}{{A}^{4}} \right]}^{\dfrac{1}{2}}} \\\ & \sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}=\left[ {{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{A}^{2}} \right] \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** The ability of a substance in order to store electrical potential energy under the influence of an electric field calculated by the ratio of the capacitance of a capacitor with the substance as dielectric to its capacitance with vacuum as dielectric. It is also known as dielectric constant. Magnetic permeability is described as the ratio of the magnetic induction to the magnetic intensity. It is found to be a scalar quantity and is mentioned using the symbol. Magnetic permeability is used to calculate a material's resistance to the magnetic field or to measure the degree to which magnetic field can penetrate through a substance.