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(B) $1$. Dimensional formula of permittivity of free space $(\varepsilon_0)$: From Coulomb's Law,$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$

Vedclass · Accepted Answer

$[\varepsilon _0] = M^{-1}L^{-3}T^4I^2$

Vedclass · Answer

(B) $1$. Dimensional formula of permittivity of free space $(\varepsilon_0)$: From Coulomb's Law,$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$. Rearranging,$\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. Dimensions: $[q] = [IT]$,$[F] = [MLT^{-2}]$,$[r] = [L]$. Substituting: $[\varepsilon_0] = \frac{[IT][IT]}{[MLT^{-2}][L^2]} = \frac{I^2 T^2}{ML^3 T^{-2}} = M^{-1} L^{-3} T^4 I^2$. $2$. Dimensional formula of permeability of free space $(\mu_0)$: From the force between two parallel wires,$F = \frac{\mu_0 I_1 I_2 l}{2\pi r}$. Rearranging,$\mu_0 = \frac{F \cdot 2\pi r}{I_1 I_2 l}$. Dimensions: $[F] = [MLT^{-2}]$,$[r] = [L]$,$[I] = [I]$,$[l] = [L]$. Substituting: $[\mu_0] = \frac{[MLT^{-2}][L]}{[I][I][L]} = MLT^{-2}I^{-2}$. Comparing with the options,option $B$ is correct.

Let $[{\varepsilon _0}]$ denote the dimensional formula of the permittivity of free space and $[{\mu _0}]$ denote the dimensional formula of the permeability of free space. If $M = \text{mass}$,$L = \text{length}$,$T = \text{time}$,and $I = \text{electric current}$,then:

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