Question

Let $\left[{\epsilon }_0\right]$ denote the dimensional formula of the permittivity of the vacuum and $\left[{\mu }_0\right]$ denote that of permeability of the vacuum. Find correct option if
$I$ = electric current.

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

Answer 1

The correct option is C $\left[{\mu }_0\right]=\left[ML{T}^{-2}{I}^{-2}\right]$

As we know that from Coulomb’s law
$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

${\epsilon }_0=\frac{\left[q_1\right]\left[q_2\right]}{4\pi \left[F\right]\left[r^2\right]}=\frac{[IT{]}^2}{\left[ML{T}^{-2}\right]\left[L^2\right]}$

And $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_{{}_0}}}$
${\mu }_0=\frac{1}{\left[{\epsilon }_0\right]\left[c^2\right]}$
${\mu }_0=\frac{1}{\left[M^{-1}{L}^{-3}{T}^4{I}^2\right]\left[L{T}^{-1{]}^2}\right]}]$

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

Answer is c