Question

If electronic charge $e$, electron mass $m$, speed of light in vaccum $c$ and Plank's constant $h$ are taken as fundamental quantities, the permeability of vaccum ${\mu }_0$ can be expressed in units of
Hint: $\left[h\right]=\left[M{L}^2{T}^{-1}\right],\left[{\mu }_0\right]=\left[ML{T}^{-2}{A}^{-2}\right]$

• A. $\left(\frac{h}{m{e}^2}\right)$
• B. $\left(\frac{hc}{m{e}^2}\right)$
• C. $\left(\frac{h}{c{e}^2}\right)$
• D. $\left(\frac{m{c}^2}{h{e}^2}\right)$

Answer 1

The correct option is C $\left(\frac{h}{c{e}^2}\right)$

Let ${\mu }_0$ related with $e,m,c$ and $h$ as follows
${\mu }_0=k{e}^a{m}^b{c}^c{h}^d$, where $k$ is any constant.
$\Rightarrow \left[ML{T}^{-2}{A}^{-2}\right]=\left[AT{]}^a\right[M{]}^b\left[L{T}^{-1}{]}^c\right[M{L}^2{T}^{-1}{]}^d$
$\Rightarrow \left[ML{T}^{-2}{A}^{-2}\right]=\left[M^{(b+d)}{L}^{(c+2d)}{T}^{(a-c-d)}{A}^{\left(a\right)}\right]$
Now, comparing both sides and solving the equations obtained we get
$a=-2,b=0,c=-1,d=1$
Thus, the relation will be $\left[{\mu }_0\right]=\left[\frac{h}{c{e}^2}\right]$