JEE PYQ: Units & Measurements - Question ID 5e2adbe30f23 (JEE Main 2023)

Core Formula & Concept:

In electromagnetism, two fundamental constants appear in Maxwell’s equations:

• $\mu_{0}$ – the permeability of free space (magnetic constant).
• $\epsilon_{0}$ – the permittivity of free space (electric constant).

From Maxwell’s theory, the speed of light in vacuum $c$ is related to these constants by the identity: $c = \frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}$ Squaring both sides gives: $c^{2} = \frac{1}{\mu_{0} \epsilon_{0}}$ Therefore, the quantity $\displaystyle \frac{1}{\mu_{0} \epsilon_{0}}$ is dimensionally equivalent to the square of a speed, i.e., $\displaystyle \frac{L^{2}}{T^{2}}$.

Step-by-Step Derivation:

Step 1: Write down the dimensions of $\mu_{0}$ and $\epsilon_{0}$.

From the Biot–Savart law and Coulomb’s law, we know: $[\mu_{0}] = \frac{[F]}{[I]^{2}} = \frac{MLT^{-2}}{A^{2}} = MLT^{-2}A^{-2}$ $[\epsilon_{0}] = \frac{[Q]^{2}}{[F][L]^{2}} = \frac{A^{2}T^{2}}{MLT^{-2}L^{2}} = M^{-1}L^{-3}T^{4}A^{2}$

Step 2: Compute the dimension of the product $\mu_{0} \epsilon_{0}$.

Multiply the dimensions: $[\mu_{0} \epsilon_{0}] = [\mu_{0}] \cdot [\epsilon_{0}] = (MLT^{-2}A^{-2}) \cdot (M^{-1}L^{-3}T^{4}A^{2}) = M^{1-1}L^{1-3}T^{-2+4}A^{-2+2} = M^{0}L^{-2}T^{2}A^{0} = L^{-2}T^{2}$

Step 3: Find the dimension of $\displaystyle \frac{1}{\mu_{0} \epsilon_{0}}$.

Taking the reciprocal of the above result: $\left[\frac{1}{\mu_{0} \epsilon_{0}}\right] = \frac{1}{L^{-2}T^{2}} = L^{2}T^{-2}$ This matches the dimension of speed squared, $\displaystyle \frac{L^{2}}{T^{2}}$.

Step 4: Match with the given options.

Option C is $\displaystyle \frac{L^{2}}{T^{2}}$, which is exactly the dimension we derived.

Common Traps & Exam Tip:

Many students confuse the dimensions of $\mu_{0}$ and $\epsilon_{0}$ or forget to take the reciprocal. A quick sanity check is to recall that $\displaystyle \frac{1}{\mu_{0} \epsilon_{0}} = c^{2}$, whose dimension is clearly $\displaystyle \frac{L^{2}}{T^{2}}$. This shortcut can save precious time in the exam.