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. $\frac{h}{m{c}^2}$
• B. $\frac{hc}{m{e}^2}$
• C. $\frac{h}{c{e}^2}$
• D. $\frac{mc}{h{e}^2}$

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 ${\mu }_0=[M\text{ L }{T}^{-2}{A}^{-2}],$
$e=[AT]$,
$m=[M]$,
$c=\left[L{T}^{-1}\right]$,
and $h=[M{L}^2{T}^{-1}]$
Putting the dimension of various physical quantities in Eq. (i), we get
$\left[ML{T}^{-2}{A}^{-2}\right]=k[AT{]}^a[M{]}^b{\left[L{T}^{-1}\right]}^c]{\left[M{L}^2{T}^{-1}\right]}^d$
$\left[ML{T}^{-2}{A}^{-2}\right]=k[M{]}^{b+d}[L{]}^{c+2d}[T{]}^{a-c-d}[A{]}^a$
Compairing the powers of $M,L,T$ and $A$ on the both sides, we get
$b+d=1\dots (i)$
$c+2d=1\dots (iii)$
$a-c-d=-2\dots (iv)$
$a=-2\dots (v)$
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,
$\frac{h}{c{e}^2}$