Question
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
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 :
$$\left( {\frac{h}{{c{e^2}}}} \right)$$
Solution :
Let $${\mu _0}$$ related with $$e,m,c$$ and $$h$$ as follows.
$$\eqalign{
& {\mu _0} = k{e^a}{m^b}{c^c}{h^d} \cr
& \left[ {ML{T^{ - 2}}{A^{ - 2}}} \right] = {\left[ {AT} \right]^a}{\left[ M \right]^b}{\left[ {L{T^{ - 1}}} \right]^c}{\left[ {M{L^2}{T^{ - 1}}} \right]^d} \cr
& = \left[ {{M^{b + d}}{L^{c + 2d}}{T^{a - c - d}}{A^a}} \right] \cr} $$
On comparing both sides we get
$$\eqalign{
& a = - 2,b = 0,c = - 1,d = 1 \cr
& \therefore \left[ {{\mu _0}} \right] = \left[ {\frac{h}{{c{e^2}}}} \right] \cr} $$