JEE Advanced · Physics · 2. Units & Dimensions

Paragraph : In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $[E] a n d [B]$ stand for dimensions of electric and magnetic fields respectively, while $[ε_{0}]$ and $[μ_{0}]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L] a n d [T]$ are dimensions of length and time respectively. All the quantities are given in $S I$ units

Question : The relation between $[ϵ_{0}] a n d [μ_{0}]$ is

A $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^2[T]^{-2}$
B $\left[\mu_0\right]=\left[\varepsilon_0\right][ L ]^{-2}[T]^2$
C $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[L]^2[T]^{-2}$
D $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[L]^{-2}[T]^2$

Verified Solution

Answer & Solution

Correct Answer

(D) $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[L]^{-2}[T]^2$

Step-by-step Solution

Detailed explanation

We have,

C = \frac{1}{\sqrt{μ_{0} ε_{0}}}

∴ [C^{2}] = [\frac{1}{μ_{0} ε_{0}}]

\Rightarrow L^{2} T^{- 2} = \frac{1}{[μ_{0}] [ε_{0}]}

\Rightarrow [μ_{0}] = {[ε_{0}]}^{- 1} {[L]}^{- 2} {[T]}^{2}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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