Question

If electron charge e, electron mass m, speed of light in vacuum c and Planck's constant h are taken as fundamental constant h are taken as fundamental quantities, the permeability of vacuum ${\mu }_0$ can be expressed in units of
A. $({\displaystyle \frac{m{c}^2}{h{e}^2}})$
B. $({\displaystyle \frac{h}{m{e}^2}})$
C. $({\displaystyle \frac{hc}{m{e}^2}})$
D. $({\displaystyle \frac{h}{c{e}^2}})$

Answer 1

Hint: To solve this problem, first write the dimensions of permeability of vacuum in terms of other fundamental quantities i.e. e, c, h and m. Substitute the dimensions of all the above mentioned fundamental quantities. Consider exponents of these fundamental quantities to be a, b, c and d. Now, compare the powers on both the sides and find the value of a, b, c and d. Substitute these values back in the first equation. Then, write the dimensions in equation form. This will give the permeability of vacuum in terms of mentioned fundamental quantities.

Complete step by step answer:
The dimensions of permeability of vacuum in terms of other fundamental quantities can be written as,
$[{\mu }_0]=[e^a{c}^b{h}^c{m}^d]$ ...(1)
Dimensions of ${\mu }_0$ are $[ML{T}^{-2}{A}^{-2}]$.
Dimensions of e are $[AT]$.
Dimensions of c are $[L{T}^{-1}]$.
Dimensions of h are $[M{L}^2{T}^{-1}]$.
Dimensions of m are $[M]$.
Substituting all the values in the equation. (1) we get,
$[ML{T}^{-2}{A}^{-2}]={[AT]}^a{[L{T}^{-1}]}^b{[M{L}^2{T}^{-1}]}^c{[M]}^d$
$[ML{T}^{-2}{A}^{-2}]=[A^a{L}^{b+2c}{T}^{a-b-c}{M}^{c+d}]$
Equating the powers on both the sides we get,
$c+d=1$ ...(2)
$b+2c=1$ ...(3)
$a-b-c=-2$ ...(4)
$a=-2$ ...(5)
Substituting value of a in equation. (4) we get,
$b+c=0$ ...(6)
Subtracting equation. (6) and (3) we get,
$c=1$
Substituting value of a and c in equation. (4) we get,
$b=-1$
Now, substituting the value of c in the equation. (2) we get,
$d=0$
Substituting these values in the equation. (1) we get,
$[{\mu }_0]=[e^{-2}{c}^{-1}{h}^1{m}^0]$
$\Rightarrow [{\mu }_0]=[e^{-2}{c}^{-1}{h}^1]$
$\Rightarrow {\mu }_0={\displaystyle \frac{h}{c{e}^2}}$
Hence, the permeability of vacuum ${\mu }_0$ can be expressed in units of $({\displaystyle \frac{h}{c{e}^2}})$.

So, the correct answer is “Option D”.

Note: To solve these types of questions, students must know the dimensions of questions or the units of each quantity. Dimensions of a quantity are basically based on the unit of the quantity. So, if students know the unit of a quantity they can derive the dimension of that respective physical or fundamental quantity.