Question

Let $ [ \varepsilon_0 ] $ denote the dimensional formula of the permittivity of the vacuum and $ [ \mu_0 ] $ that of the permeability of the vacuum. If $M$ = mass, $L$ = length, $T$ = time and $I$ = electric current.

• A. $ [ \varepsilon_0 ] = [ M^{ - 1} L^{ - 3} T^2 I ] $
• B. $ [ \varepsilon_0 ] = [ M^{ - 1} \, L^{ - 3} \, T^4 \, I^2 ] $
• C. $ [ \mu_0 ] = [ ML \, T^{ - 2} I^{ - 2} ] $
• D. $ [ \mu_0 ] = [ ML^2 \, T^{ - 1} \, I ] $

Answer 1

Answer: (B), (C)

$F =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}} $
${\left[\varepsilon_{0}\right] } =\frac{\left[q_{1}\right]\left[q_{2}\right]}{[F]\left[r^{2}\right]}=\frac{\left[ IT ^{2}\right.}{\left[ MLT ^{-2}\right]\left[ L ^{2}\right]}$
$=\left[ M ^{-1} L ^{-3} T ^{4} I ^{2}\right]$
Speed of light, $c=\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}$
$\therefore \left[\mu_{0}\right] =\frac{1}{\left[\varepsilon_{0}\right][c]^{2}}=\frac{1}{\left[M^{-1} L^{-3} T ^{4} I ^{2}\right]\left[ LT ^{-1}\right]^{2}}$
$=\left[ MLT ^{-2} I ^{-2}\right]$