JEE PYQ: Units & Measurements - Question ID 12d996648d4b (JEE Main 2025)

Core Formula & Concept:

In electromagnetism, the magnetic field $B$ and the permeability of free space $\mu_0$ are related through the fundamental force law on a current-carrying conductor. The key formula is the magnetic force per unit length on a wire carrying current $I$ placed in a magnetic field $B$:

$\frac{F}{l} = I B \sin\theta$

where $F$ is the force, $l$ is the length of the wire, $I$ is the current, and $\theta$ is the angle between the wire and the field. For dimensional analysis, we take $\sin\theta = 1$, so:

$F = I l B$

From this, we can express $B$ in terms of force, current, and length:

$B = \frac{F}{I l}$

The permeability of free space $\mu_0$ appears in the Biot-Savart law and Ampère’s circuital law. Its dimensions are derived from the force between two parallel current-carrying wires:

$F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$

where $d$ is the separation between the wires. Solving for $\mu_0$:

$\mu_0 = \frac{2 \pi d F}{I_1 I_2 l}$

Thus, the dimensions of $\mu_0$ can be expressed in terms of force, current, and length.

Step-by-Step Derivation:

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

Step 1: Determine the dimensions of $B$.

From $F = I l B$, we rearrange to get:

$B = \frac{F}{I l}$

The dimensions of force $F$ are $\mathrm{MLT}^{-2}$, current $I$ is $\mathrm{A}$, and length $l$ is $\mathrm{L}$. Thus:

$[B] = \frac{\mathrm{MLT}^{-2}}{\mathrm{A} \cdot \mathrm{L}} = \mathrm{MT}^{-2}\mathrm{A}^{-1}$ Step 2: Determine the dimensions of $\mu_0$.

From the force between two parallel wires:

$F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$

Rearranging for $\mu_0$:

$\mu_0 = \frac{2 \pi d F}{I_1 I_2 l}$

The dimensions of $d$ and $l$ are both $\mathrm{L}$, and $I_1$ and $I_2$ are both $\mathrm{A}$. Thus:

$[\mu_0] = \frac{\mathrm{L} \cdot \mathrm{MLT}^{-2}}{\mathrm{A} \cdot \mathrm{A} \cdot \mathrm{L}} = \frac{\mathrm{ML}^2 \mathrm{T}^{-2}}{\mathrm{A}^2} = \mathrm{MLT}^{-2}\mathrm{A}^{-2}$ Step 3: Compute the dimensions of $\frac{B}{\mu_0}$.

Now, divide the dimensions of $B$ by the dimensions of $\mu_0$:

$\left[\frac{B}{\mu_0}\right] = \frac{[B]}{[\mu_0]} = \frac{\mathrm{MT}^{-2}\mathrm{A}^{-1}}{\mathrm{MLT}^{-2}\mathrm{A}^{-2}}$

Simplify the expression by canceling common terms:

• $\mathrm{M}$ cancels out.
• $\mathrm{T}^{-2}$ cancels out.
• $\mathrm{A}^{-1}$ in the numerator and $\mathrm{A}^{-2}$ in the denominator becomes $\mathrm{A}^{1}$.
• $\mathrm{L}$ in the denominator remains as $\mathrm{L}^{-1}$.

Thus:

$\left[\frac{B}{\mu_0}\right] = \mathrm{L}^{-1} \mathrm{A}$ Step 4: Match with the given options.

The derived dimensions $\mathrm{L}^{-1} \mathrm{A}$ correspond to option B.

Common Traps & Exam Tip:

Students often confuse the dimensions of $B$ and $\mu_0$, leading to incorrect cancellations. Common mistakes include:

• Forgetting that $\mu_0$ has dimensions involving $\mathrm{A}^{-2}$, not $\mathrm{A}^{-1}$. This leads to incorrect simplification of the ratio $\frac{B}{\mu_0}$.
• Misapplying the formula for magnetic force, such as using $F = qvB$ (which involves charge and velocity) instead of $F = I l B$. While both are correct, the latter is more straightforward for dimensional analysis in this context.
• Overcomplicating the problem by introducing unnecessary constants like $\epsilon_0$ or $c$, which are irrelevant here.

Exam Tip: Always start with the simplest formula that relates the quantities in question. For $B$ and $\mu_0$, the force on a current-carrying wire is the most direct path to their dimensions. Double-check each step of dimensional cancellation to avoid sign errors or missing terms.