Q121 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Physics
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\(\frac{h c}{m e^2}\)
- Correct answerC\(\frac{h}{c e^2}\)
- D\(\frac{m c^2}{h e^2}\)
Solution
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, \(\frac{h}{c e^2}\).
Hence, the correct answer is option: 'C'.