Question

Let $[{\epsilon }_0]$ denote the dimensional formula of the permittivity of the vacuum and $[{\mu }_0]$ that of the permeability of the vacuum. If $M$ = mass, $L$ = length, $T$ = time and $I$ = electric current.

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

Answer 1

Answer: (B), (C)

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