JEE Challenger
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PhyEasyAdv 2018 • Paper 1

Dimensional Relation Between Free Space Permittivity and Permeability

Comprehension Passage

In electromagnetic theory, electric and magnetic phenomena are interconnected. Consequently, the dimensions of electric and magnetic quantities are also related to one another. In the following questions, [E][E] and [B][B] represent the dimensions of electric and magnetic fields, respectively, whereas [ϵ0][\epsilon_0] and [μ0][\mu_0] designate the dimensions of the permittivity and permeability of free space, respectively. Furthermore, [L][L] and [T][T] denote the dimensions of length and time, respectively. All quantities are expressed in SI units.

Determine the correct dimensional relationship between [ϵ0][\epsilon_0] and [μ0][\mu_0]:

Options

A

[μ0]=[ϵ0][L]2[T]−2[\mu_0] = [\epsilon_0] [L]^2 [T]^{-2}

B

[μ0]=[ϵ0][L]−2[T]2[\mu_0] = [\epsilon_0] [L]^{-2} [T]^2

C

[μ0]=[ϵ0]−1[L]2[T]−2[\mu_0] = [\epsilon_0]^{-1} [L]^2 [T]^{-2}

D

[μ0]=[ϵ0]−1[L]−2[T]2[\mu_0] = [\epsilon_0]^{-1} [L]^{-2} [T]^2

Correct

Topics & Concepts

Units and MeasurementsElectromagnetic Waves

Step-by-Step Solution

To determine the correct dimensional relationship between the permittivity of free space [ϵ0][\epsilon_0] and permeability of free space [μ0][\mu_0], we use the relation for the speed of light in vacuum cc:

c=1μ0ϵ0c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}

Squaring both sides and solving for μ0\mu_0: c2=1μ0ϵ0  ⟹  μ0=ϵ0−1c−2c^2 = \frac{1}{\mu_0 \epsilon_0} \implies \mu_0 = \epsilon_0^{-1} c^{-2}

Taking the dimensions on both sides: [μ0]=[ϵ0]−1[c]−2[\mu_0] = [\epsilon_0]^{-1} [c]^{-2}

Since cc represents the speed of light, its dimensions are: [c]=[L][T]−1[c] = [L][T]^{-1}

Therefore, [c]−2=([L][T]−1)−2=[L]−2[T]2[c]^{-2} = \left([L][T]^{-1}\right)^{-2} = [L]^{-2}[T]^2

Substituting this into the dimensional formula gives: [μ0]=[ϵ0]−1[L]−2[T]2[\mu_0] = [\epsilon_0]^{-1} [L]^{-2} [T]^2

Thus, the correct option is (D).

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