Question

Multiple Correct Answers Type
Let [${ϵ}_0$] denote the dimensional formula of the permittivity of vacuum and $\left[{\mu }_0\right]$ that of the permeability of vacuum. If M = mass, L = length, T = time, and I = electric current, then

• A. $\left[{ϵ}_0\right]=M^{-1}{L}^{-3}{T}^2{I}^2$
• B. $\left[{ϵ}_0\right]=M^{-1}{L}^{-3}{T}^4{I}^2$
• C. $\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}$
• D. $\left[{\mu }_0\right]=M{L}^2{T}^{-1}I$

Answer 1

Answer: (B), (C)

$\left[{ϵ}_0\right]=M^{-1}{L}^{-3}{T}^4{I}^2$
$\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}$

$F=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

${ϵ}_0=M^{-1}{L}^{-3}{T}^4{I}^2$

${\mu }_0=\frac{F}{qvnI}$

${\mu }_0=M^1{L}^1{T}^{-2}{I}^{-2}$