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Question #43272

Difficulty: 8/10 | 🟢 Free

The Question

A thin non-conducting rod of length $L$ is placed along the positive $x$ -axis with its ends at $x = 0$ and $x = L$ . The rod possesses a non-uniform magnetic moment density (magnetic moment per unit length) given by $\vec{\lambda}(x) = \alpha x \hat{i}$ , where $\alpha$ is a positive constant with units of Ampere. Find the magnetic field $\vec{B}$ at a point $P$ located on the $x$ -axis at $x = d$ (where $d > L$ ). Assume the surrounding medium is vacuum with permeability $\mu_0$ .

Options

A.
$\frac{\mu_0 \alpha L^2}{4\pi d(d-L)^2} \hat{i}$
B.
$\frac{\mu_0 \alpha L^2}{4\pi d^3} \hat{i}$
C.
$\frac{\mu_0 \alpha L^2}{4\pi (d - L/2)^3} \hat{i}$
D.
$\frac{\mu_0 \alpha L^2}{4\pi (d - 2L/3)^3} \hat{i}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Magnetic Field on Axial vs Equatorial Line|Continuous Charge Distribution

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply the superposition principle to a continuous, non-uniform distribution of magnetic dipoles.

A common trap for JEE Advanced students is to calculate the total magnetic moment of the rod ( $M_{total} = \int \alpha x dx = \alpha L^2 / 2$ ) and then incorrectly apply the standard point-dipole formula $B = (\mu_0 / 4\pi)(2M_{total} / r^3)$ .

Distractors are generated by placing this lumped total magnetic moment at the origin, the geometric center ( $L/2$ ), or the 'center of magnetic moment' ( $2L/3$ ).

The standard $1/r^3$ approximation strictly fails for extended sources at distances comparable to their dimensions, forcing rigorous integration of infinitesimal dipole contributions.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Magnetism and Matter - Class 12
Concept	Magnetic Field on Axial vs Equatorial Line
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free