Question

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

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

Answer 1

Answer: (B), (C)

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

$\begin{array}{c}\text{Unit of}{\epsilon }_0=\frac{C^2}{N-m^2}\\ \left({\epsilon }_0\right]=\frac{I^2{T}^4}{M{L}^3}=\left(M^{-1}{L}^{-3}{T}^4{I}^2\right]\\ \end{array}$
We know that ${\mu }_0{ϵ}_0$= $\frac{1}{c^2}$ (where $c$ is the velocity of light)
Using this we can say $\left[{\mu }_0\right]$= $\left[ML{T}^{-2}{I}^{-2}\right]$