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If electronic charge $e$, electron mass $m$, speed of light in vacuum $c$ and Planck 's constant $h$ are taken as fundamental quantities, the permeability of vacuum $\mu _0$ can be expressed in units of
$\left( {\frac{h}{{m{e^2}}}} \right)$
$\left( {\frac{{hc}}{{m{e^2}}}} \right)$
$\left( {\frac{h}{{c{e^2}}}} \right)$
$\left( {\frac{{m{c^2}}}{{h{e^2}}}} \right)$
Solution
Let $\mu_{0}$ related with $e, m, c$ and $h$ as follows.
$\mu_{0}=k e^{a} m^{b} c^{c} h^{d}$
${\left[M L T^{-2} A^{-2}\right]=[A T]^{a}[M]^{b}\left[L T^{-1}\right]^{c}\left[M L^{2} T^{-1}\right]^{d}}$
$=\left[M^{b+d} L^{c+2 d} T^{a-c-d} A^{a}\right]$
On comparingboth sides we get
$a=-2 \quad \ldots(i)$
$b+d=1 \ldots(i i)$
$c+2 d=1 \ldots(i i i)$
$a-c-d=-2 \ldots(i v)$
By equations $(i),(i i),(i i i)$ and $(i v)$ we get,
$a=-2, b=0, c=-1, d=1$
$\left[\mu_{0}\right]=\left[\frac{h}{c e^{2}}\right]$
