Question

Let $\left[{\in }_0\right]$ denote the dimensional formula of the permittivity of the vacuum and $\left[{\mu }_0\right]$ that of the permeability of the vacuum. If $M=$ mass, $L=$ length, $T=$ time and $I=$ electric current, then:
Given $F=\frac{1}{4\pi {\in }_0}.\frac{q_1{q}_2}{r^2}$, where $q_1$ and $q_2$ are charges and r is the distance. Also,
$c=\frac{1}{\sqrt{{\in }_0{\mu }_0}}$, where $c$ is speed.

• A. $\left[{\in }_0\right]=\left[M^{-1}{L}^{-3}{T}^{2}I\right]$
• B. $\left[{\in }_0\right]=\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]$
• C. $\left[{\mu }_0\right]=\left[ML{T}^{-2}{I}^{-2}\right]$
• D. $\left[{\mu }_0\right]=\left[M{L}^2{T}^{-1}I\right]$

Answer 1

Answer: (B), (C)

$\left[{\mu }_0\right]=\left[ML{T}^{-2}{I}^{-2}\right]$

$F=\frac{1}{4\pi {\in }_0}.\frac{q_1{q}_2}{r^2}\Rightarrow \left[{\in }_0\right]=\frac{\left[q_1\right]\left[q_2\right]}{\left[F\right]\left[r^2\right]}=\frac{[IT{]}^2}{\left[ML{T}^{-2}\right]\left[L^2\right]}$
$=\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]$
Speed of light, $c=\frac{1}{\sqrt{{\in }_0{\mu }_0}}$
$\therefore \left[{\mu }_0\right]=\frac{1}{\left[{\in }_0\right][c{]}^2}\Rightarrow \left[{\mu }_0\right]=\frac{1}{\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]{\left[L{T}^{-1}\right]}^2}$