We can expressed the permeability of vacuum,
\(\mu_0 \propto e^a m^b c^c h^d\)
or \(\mu_0=k e^a m^b c^c h^d\) ...(i)
Where, \(k\) is a dimensional constant.
As we know that,
Dimension of
\(\begin{aligned}
\mu_0 & =\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right], \\
e & =[\mathrm{AT}], \\
m & =[\mathrm{M}], \\
c & =\left[\mathrm{L} \mathrm{T}^{-1}\right], \\
h & =\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right],
\end{aligned}\)
and
Putting the dimension of various physical quantities in Eq. (i), we get
\(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right]=k[\mathrm{~A} \mathrm{~T}]^a\left[\mathrm{M}^b\left[\mathrm{~L} \mathrm{~T}^{-1}\right]^c\right]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]^d\)
\(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right]=k[\mathrm{M}]^{b+d}[\mathrm{~L}]^{c+2 d}[\mathrm{~T}]^{a-c-d}[\mathrm{~A}]^a\)
Comparing the powers of \(\mathrm{M}, \mathrm{L}, \mathrm{T}\) and \(\mathrm{A}\) on the both sides, we get
\(\begin{aligned}
b+d & =1 \quad \ldots (ii) \\
c+2 d & =1 \quad \ldots (iii) \\
a-c-d & =-2 \quad \ldots (iv) \\
a & =-2 \quad \ldots (v)
\end{aligned}\)
After solving the Eqs. (ii), (iii), (iv) and (v), we get
\(\begin{aligned}
a & =-2, \\
b & =0, \\
c & =-1, \\
\text{and } d & =1, \\
\therefore \mu_0 & =\frac{h}{c e^2}
\end{aligned}\)
So, the permeability of vacuum \(\mu_0\) can be expressed as, \(\frac{h}{c e^2}\).