Physics - Unit & Dimension Question with Solution | TestHub

We know that, velocity of light $c$ is terms of ${ϵ}_0$ and ${\mu }_0$ is given by

$$\begin{gathered} c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}} \\ \Rightarrow c^2=\frac{1}{{\mu }_0{\epsilon }_0} \end{gathered}$$

$$\begin{gathered} c^2=\frac{1}{{\mu }_0^2}\frac{{\mu }_0}{{\epsilon }_0} \\ \Rightarrow c=\frac{1}{{\mu }_0}\sqrt{\frac{{\mu }_0}{{\epsilon }_0}} \\ \Rightarrow \sqrt{\frac{{\mu }_0}{{\epsilon }_0}}=c{\mu }_0 ...\left(\mathrm{i}\right) \end{gathered}$$

Dimension of the magnetic field, $B=\frac{\tau }{NIA}=\frac{\left[{\mathrm{ML}}^{2 }{\mathrm{T}}^{-2}\right]}{\left[A\right]\left[{\mathrm{L}}^2\right]}$

$=\left[{\mathrm{MA}}^{-1}{\mathrm{T}}^{-2}\right]$

Again, $B={\mu }_{0}nI$

$\therefore {\mu }_0=\frac{B}{nl}=\frac{\left[{\mathrm{MA}}^{-1}{\mathrm{T}}^{-2}\right]}{\left[{\mathrm{L}}^{-1}\right]\left[A\right]}=\left[{\mathrm{MLA}}^{-2}{\mathrm{T}}^{-2}\right]$

${\mu }_0=\left[{\mathrm{MLA}}^{-2}{\mathrm{T}}^{-2}\right] ...\left(\mathrm{ii}\right)$

and $C=\left[{\mathrm{LT}}^{-1}\right] ...\left(\mathrm{iii}\right)$

From Equations. (i), (ii) and (iii), we get

$\sqrt{\frac{{\mu }_0}{{ϵ}_0}}=\left[{\mathrm{LT}}^{-1}\right]\left[{\mathrm{MLA}}^{-2}{\mathrm{T}}^{-2}\right]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-3}{\mathrm{A}}^{-2}\right]$