Question

Let $\left[{\epsilon }_o\right]$ denote the dimensional formula of the permittivity of the vacuum, and $\left[{\mu }_o\right]$ that of the permeability of the vacuum. If M$=$mass, L$=$length, T$=$time and I$=$ electric current.

• A. $\left[{\mu }_o\right]=M^{-1}{L}^{-3}{T}^{2}I$
• B. $\left[{\epsilon }_o\right]=M^{-1}{L}^{-3}{T}^4{I}^2$
• C. $\left[{\mu }_o\right]=ML{T}^{-2}{I}^{-2}$
• D. $\left[{\mu }_o\right]=M{L}^2{T}^{-1}I$

Answer 1

Answer: (B), (C)

The correct options are
B $\left[{\epsilon }_o\right]=M^{-1}{L}^{-3}{T}^4{I}^2$
C $\left[{\mu }_o\right]=ML{T}^{-2}{I}^{-2}$

Dimension formula of ${\epsilon }_0$

${\epsilon }_0=\frac{1}{4\pi F}\frac{v_1{v}_2}{r^2}(F=ML{T}^{-2}){\epsilon }_0=\frac{1}{ML{T}^{-2}}\frac{LAT\times AT}{l^2}=M^{-1}{L}^{-3}{t}^4{A}^2$

Dimension of ${\mu }_0$

$N/A^2$ or $Wb{A}^{-1}{m}^{-1}=\left[M\right]\left[L\right]\left[T{]}^{-2}\right[A{]}^{-2}$