JEE PYQ: Units & Measurements - Question ID 39d4452db19d (JEE Main 2022)

Core Formula & Concept:

In electromagnetism, the magnetic field $B$ and the permeability of free space $\mu_0$ are related through fundamental physical laws. The key concept here is dimensional analysis, where we express physical quantities in terms of their base dimensions: mass ($M$), length ($L$), time ($T$), and electric current ($A$).

The magnetic field $B$ is defined via the Lorentz force law: $\vec{F} = q (\vec{v} \times \vec{B})$ From this, the dimensions of $B$ can be derived as: $[B] = \frac{[F]}{[q][v]} = \frac{M L T^{-2}}{(A T)(L T^{-1})} = M T^{-2} A^{-1}$

The permeability of free space $\mu_0$ appears in Ampère’s law and the Biot-Savart law. Its dimensions are derived from the force between two current-carrying wires: $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$ Rearranging for $\mu_0$: $[\mu_0] = \frac{[F] [r]}{[I]^2 [L]} = \frac{M L T^{-2} \cdot L}{A^2 \cdot L} = M L T^{-2} A^{-2}$

Step-by-Step Derivation:

We need to find the dimensions of $\frac{B^2}{\mu_0}$. Let’s proceed step-by-step:

1. Write the dimensions of $B$ and $\mu_0$: $[B] = M T^{-2} A^{-1}, \quad [\mu_0] = M L T^{-2} A^{-2}$
2. Compute $[B^2]$: $[B^2] = [B]^2 = (M T^{-2} A^{-1})^2 = M^2 T^{-4} A^{-2}$
3. Divide $[B^2]$ by $[\mu_0]$: $\left[\frac{B^2}{\mu_0}\right] = \frac{[B^2]}{[\mu_0]} = \frac{M^2 T^{-4} A^{-2}}{M L T^{-2} A^{-2}}$
4. Simplify the expression by canceling common terms: $= \frac{M^2}{M} \cdot \frac{T^{-4}}{T^{-2}} \cdot \frac{A^{-2}}{A^{-2}} \cdot \frac{1}{L} = M^{2-1} T^{-4+2} A^{0} L^{-1} = M L^{-1} T^{-2}$

Thus, the dimensions of $\frac{B^2}{\mu_0}$ are $[M L^{-1} T^{-2}]$, which matches option C.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Incorrect dimensions for $B$ or $\mu_0$: Some confuse the dimensions of $B$ with those of magnetic flux ($\phi = B \cdot A$), leading to errors. Always derive $[B]$ from the Lorentz force law.
• Forgetting to square $B$: The question involves $B^2$, not $B$. Squaring the dimensions is crucial.
• Miscounting exponents during division: When dividing $[B^2]$ by $[\mu_0]$, students may incorrectly subtract exponents (e.g., $T^{-4} / T^{-2} = T^{-2}$ instead of $T^{-2}$). Double-check each step.
• Ignoring current ($A$) dimensions: Since $B$ and $\mu_0$ both involve current, their $A$ terms cancel out. Missing this leads to incorrect options like D.

Exam Tip: Always verify dimensions by cross-checking with known formulas. For example, $\frac{B^2}{\mu_0}$ has the same dimensions as energy density (energy per unit volume), which is $M L^{-1} T^{-2}$. This consistency check can help confirm the answer.