Question

In terms of basic units mass(M), length(L), time(T) and charge (Q), the dimension of the magnetic permeability of vacuum (${{\mu }_{\text{o}}}$) would be?
$\begin{aligned} & a)\left[ ML{{Q}^{-2}} \right] \\\ & b)\left[ L{{T}^{-1}}{{Q}^{-1}} \right] \\\ & c)\left[ M{{L}^{2}}{{T}^{-1}}{{Q}^{-2}} \right] \\\ & d)\left[ LT{{Q}^{-1}} \right] \\\ \end{aligned}$

Answer 1

In the above question we are asked to express the permeability of free space in terms of basic units. There are many expressions that we can use in order to express the basics of the particular quantity of interests. But we will first express the permeability such that we can easily express it in basic fundamental units. Further, we will express those quantities in terms of basic dimensions and accordingly obtain the required answer.

Formula used:
$v=\dfrac{1}{\sqrt{{{\in }_{\text{o}}}{{\mu }_{\text{o}}}}}$
$i=\dfrac{q}{t}$

Complete step by step answer:
Let us first express the permeability of free space in terms of permittivity of free space and speed of light in that particular medium. Let us say a particular medium has the permittivity of free space as ${{\in }_{\circ }}$ and speed of light as ‘v’. Than the permeability of free space (${{\mu }_{\circ }}$) for vacuum is given by,
$v=\dfrac{1}{\sqrt{{{\in }_{\text{o}}}{{\mu }_{\text{o}}}}}...(1)$
The physical quantity speed has dimension of $\left[ {{M}^{0}}L{{T}^{-1}} \right]$ and permittivity of free space has dimensions of $\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{2}}{{Q}^{2}} \right]$. Now let us express equation 1 in terms of permeability of free space.
$\begin{aligned} & v=\dfrac{1}{\sqrt{{{\in }_{\text{o}}}{{\mu }_{\text{o}}}}} \\\ & {{\in }_{\text{o}}}{{\mu }_{\text{o}}}=\dfrac{1}{{{v}^{2}}} \\\ & \Rightarrow {{\mu }_{\text{o}}}=\dfrac{1}{{{\in }_{\text{o}}}{{v}^{2}}} \\\ \end{aligned}$
Substituting the dimensions of free permittivity and speed in the above equation we get,
$\begin{aligned} & {{\mu }_{\text{o}}}=\dfrac{1}{{{\in }_{\text{o}}}{{v}^{2}}} \\\ & {{\mu }_{\text{o}}}=\dfrac{1}{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{2}}{{Q}^{2}} \right]{{\left[ {{M}^{0}}L{{T}^{-1}} \right]}^{2}}} \\\ & \Rightarrow {{\mu }_{\text{o}}}={{\left[ {{M}^{-1}}{{L}^{-3}}{{T}^{2}}{{Q}^{2}} \right]}^{-1}}{{\left[ {{M}^{0}}L{{T}^{-1}} \right]}^{-2}} \\\ & \Rightarrow {{\mu }_{\text{o}}}=\left[ {{M}^{1}}{{L}^{3}}{{T}^{-2}}{{Q}^{-2}} \right]\left[ {{M}^{0}}{{L}^{-2}}{{T}^{2}} \right] \\\ & \Rightarrow {{\mu }_{\text{o}}}=\left[ {{M}^{1}}{{L}^{1}}{{T}^{0}}{{Q}^{-2}} \right] \\\ & \therefore {{\mu }_{\text{o}}}=\left[ ML{{Q}^{-2}} \right] \\\ \end{aligned}$

So, the correct answer is “Option A”.

Note:
It is to be noted that one can use any expression which contains permeability of free space in order to determine the dimension of it. One of the common expressions that one can use is Biot-Savart’s law in order to determine the dimensions of permeability of free space. One can derive the units of permittivity of free space by using the expression for force given by Coulomb’s law using the similar procedure as given in the solution.