JEE PYQ: Units & Measurements - Question ID 0f37ea8dff55 (JEE Main 2025)

In electromagnetism, the two fundamental constants of free space are:

• Permeability of free space, $\mu_0$, which relates magnetic flux density to magnetic field strength.
• Permittivity of free space, $\epsilon_0$, which relates electric displacement to electric field strength.

A key result from Maxwell’s equations is that electromagnetic waves propagate in free space with speed $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}.$ This speed $c$ has the dimension of length per time, i.e. $[c] = \mathrm{L\,T^{-1}}$.

Our goal is to find the dimension of the product $\mu_0 \epsilon_0$ and then invert it to obtain the dimension of $\frac{1}{\mu_0 \epsilon_0}$.

Step 1 – Write down the dimensions of $\mu_0$ and $\epsilon_0$.

From the definitions:

• $\mu_0$ appears in the force law between two current-carrying wires: $F = \frac{\mu_0 I_1 I_2 \ell}{2\pi r}.$ Force $F$ has dimension $\mathrm{M\,L\,T^{-2}}$, current $I$ has $\mathrm{I}$, length $\ell$ and $r$ have $\mathrm{L}$. Solving for $\mu_0$ gives $[\mu_0] = \frac{\mathrm{M\,L\,T^{-2}}}{\mathrm{I^2}} = \mathrm{M\,L\,T^{-2}\,I^{-2}}.$
• $\epsilon_0$ appears in Coulomb’s law: $F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}.$ Charge $q$ has dimension $\mathrm{I\,T}$, so $[\epsilon_0] = \frac{\mathrm{I^2\,T^2}}{\mathrm{M\,L^3\,T^{-2}}} = \mathrm{M^{-1}\,L^{-3}\,T^4\,I^2}.$

Step 2 – Compute the dimension of the product $\mu_0 \epsilon_0$.

Multiply the two dimensions: $[\mu_0 \epsilon_0] = \bigl(\mathrm{M\,L\,T^{-2}\,I^{-2}}\bigr) \bigl(\mathrm{M^{-1}\,L^{-3}\,T^4\,I^2}\bigr) = \mathrm{M^{1-1}\,L^{1-3}\,T^{-2+4}\,I^{-2+2}} = \mathrm{L^{-2}\,T^2}.$

Step 3 – Invert the product to find $\frac{1}{\mu_0 \epsilon_0}$.

Taking the reciprocal of $\mathrm{L^{-2}\,T^2}$ gives $\left[\frac{1}{\mu_0 \epsilon_0}\right] = \mathrm{L^{2}\,T^{-2}} = \frac{\mathrm{L^2}}{\mathrm{T^2}}.$

Step 4 – Match with the given options.

The dimension $\mathrm{L^2/T^2}$ corresponds exactly to option B.

Students often confuse the dimensions of $\mu_0$ and $\epsilon_0$, especially the signs of the exponents. A frequent mistake is to forget that $\mu_0$ carries $\mathrm{I^{-2}}$ while $\epsilon_0$ carries $\mathrm{I^2}$, so the current dimensions cancel out in the product. Always double-check the exponent arithmetic when multiplying dimensions.

Quick sanity check: Since $c^2 = \frac{1}{\mu_0 \epsilon_0}$, and $[c] = \mathrm{L\,T^{-1}}$, squaring gives $[c^2] = \mathrm{L^2\,T^{-2}}$. This immediately confirms that $\frac{1}{\mu_0 \epsilon_0}$ must have dimension $\mathrm{L^2/T^2}$.