Question

The dimensions of $\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$ where the symbols have their usual meaning are :
A. $\left[ LT \right]$
B. $\left[ {{L}^{2}}{{T}^{2}} \right]$
C. $\left[ {{L}^{2}}{{T}^{-2}} \right]$
D. $\left[ L{{T}^{-1}} \right]$`

Answer 1

Using the formula for electrostatic force between two charges and calculate the dimensional formula for the electric permittivity of free space (${{\varepsilon }_{0}}$). To find the dimensional formula of the permeability of free space (${{\mu }_{0}}$), you may use a formula for magnetic field at the centre of a current carrying coil.

Formula used:
${{F}_{e}}=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$
$B=\dfrac{{{\mu }_{0}}i}{2R}$
${{c}^{2}}=\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$

Complete step by step answer:
Let us first understand the given symbols. ${{\mu }_{0}}$ is called magnetic permeability of free space and ${{\varepsilon }_{0}}$ is electric permittivity of free space.The electric permittivity of free space is used in the formula for the electrostatic force between two charges. If there are two charges ${{q}_{1}}$ and ${{q}_{2}}$ are separated by a distance d, then the electrostatic force between the two charges is given as,
${{F}_{e}}=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$ …. (i).
From (i) we get that ${{\varepsilon }_{0}}=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{d}^{2}}{{F}_{e}}}$.
The dimensional formula of charge is $\left[ AT \right]$.
The dimensional formula of length is $\left[ L \right]$.
The dimensional formula of force is $\left[ ML{{T}^{-2}} \right]$.
Therefore, the dimensional formula of ${{\varepsilon }_{0}}$ is $\dfrac{\left[ AT \right]\left[ AT \right]}{\left[ {{L}^{2}} \right]\left[ ML{{T}^{-2}} \right]}=\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]$.

To find the dimensional formula of ${{\mu }_{0}}$, we will use a formula for magnetic field at the centre of a current carrying coil i.e. $B=\dfrac{{{\mu }_{0}}i}{2R}$, where B is the magnitude of the magnetic field, i is current in the coil and R is the radius of the coil.
Therefore,
${{\mu }_{0}}=\dfrac{2BR}{i}$
The dimensional formula of R is $\left[ L \right]$.The dimensional formula of i is $\left[ A \right]$. The dimensional formula of B is $\left[ M{{T}^{-2}}{{A}^{-1}} \right]$.
Therefore, the dimensional formula of ${{\mu }_{0}}$ is $\dfrac{\left[ M{{T}^{-2}}{{A}^{-1}} \right]\left[ L \right]}{\left[ A \right]}=\left[ ML{{T}^{-2}}{{A}^{-2}} \right]$.
This means that the dimensional formula of the term $\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$ is equal to $\dfrac{1}{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}} \right]\left[ ML{{T}^{-2}}{{A}^{-2}} \right]}=\dfrac{1}{\left[ {{M}^{-0}}{{L}^{-2}}{{T}^{2}}{{A}^{0}} \right]}=\left[ {{L}^{2}}{{T}^{-2}} \right]$.

Hence, the correct option is C.

Note: If you know the relation between the permeability and permittivity of free space, then this is a very easy question. The relation between the permeability and permittivity of free space is given as,
${{c}^{2}}=\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$
Here, c is the speed of light in vacuum.
We know that the dimensional formula for speed is $\left[ L{{T}^{-1}} \right]$. Therefore, the dimensional formula for ${{c}^{2}}$ is $\left[ {{L}^{2}}{{T}^{-2}} \right]$.This means that the dimensional formula of $\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$ is $\left[ {{L}^{2}}{{T}^{-2}} \right]$.