Question

If the charge of electron $e$ mass of electron $m$, speed of light in vacuum $c$ and Planck's constant $h$ are taken as fundamental quantities, then the permeability of vacuum $\mu_0$ can be expressed as

• A. $\frac{h}{mc^2}$
• B. $\frac{hc}{me^2}$
• C. $\frac{h}{ce^2}$
• D. $\frac{mc}{he^2}$

Answer 1

Answer: (C)

We can expressed the permeability of vacuum,
$\mu_{0} \propto e^{a} \,m^{b} \,c^{c} h^{d}$
or $\mu_{0}=k e^{a} \,m^{b} \,c^{c} \,h^{d}\,\,\,\,\,\,\,\dots(i)$
Where, $k$ is a dimensional constant.
As we know that,
Dimension of $\mu_{0} =[M \text { L } T^{-2} A^{-2}],$
$ e =[A T]$,
$m =[M]$,
$c =\left[L T^{-1}\right]$,
and $h=[M L^{2} T^{-1}] $
Putting the dimension of various physical quantities in Eq. (i), we get
$\left.\left[M L T^{-2} A^{-2}\right]=k[A T]^{a}[M]^{b}\left[L T^{-1}\right]^{c}\right]\left[M L^{2} T^{-1}\right]^{d}$
$\left[ M L T ^{-2} A ^{-2}\right]=k[ M ]^{b+d}[ L ]^{c+2 d}[ T ]^{a-c-d}[ A ]^{a}$
Compairing the powers of $M, L, T$ and $A$ on the both sides, we get
$b+d=1 \,\,\,\,\,\,\,\,\dots(i)$
$c+2 d =1 \,\,\,\,\,\,\,\,\,\dots(iii)$
$a-c-d =-2 \,\,\,\,\,\,\,\,\dots(iv)$
$a=-2 \,\,\,\,\,\,\,\,\dots (v)$
After solving the Eqs. (ii), (iii), (iv) and (v), we get
$a=-2, $
$b=0 $,
$c=-1$
and $d=1$,
$\therefore \mu_{0}=\frac{h}{c e^{2}}$
So, the permeability of vacuum $\mu_{0}$ can be expressed as,
$\frac{h}{c e^{2}}$