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Question #83227

Difficulty: 8/10 | 🟢 Free

The Question

A uniform non-magnetic metal rod of length $L$ , cross-sectional area $A$ , Young's modulus $Y$ , and coefficient of linear expansion $\alpha$ is placed along the x-axis. Its left end is rigidly fixed at $x=0$ . A small magnetic dipole of magnetic moment $\vec{\mu} = -\mu \hat{i}$ (where $\mu > 0$ ) is rigidly attached to the free right end of the rod at $x=L$ . The system is placed in a non-uniform magnetic field given by $\vec{B} = B_0 \left[ 1 + \frac{\lambda}{2} (x-L)^2 \right] \hat{i}$ for $x \ge L$ , where $B_0$ and $\lambda$ are positive constants. The rod is heated uniformly such that its temperature increases by $\Delta T$ . Assuming the magnetic dipole remains aligned with the x-axis and the rod does not buckle, what is the total potential energy stored in the system (the sum of the elastic strain energy in the rod and the work done against the magnetic force)?

Options

A.
$\frac{\mu B_0 \lambda Y A L^2 \alpha^2 \Delta T^2}{2(\mu B_0 \lambda L + Y A)}$
B.
$\frac{\mu B_0 \lambda Y^2 A^2 L^2 \alpha^2 \Delta T^2}{2(\mu B_0 \lambda L + Y A)^2}$
C.
$\frac{1}{2} \mu B_0 \lambda L^2 \alpha^2 \Delta T^2$
D.
$\frac{\mu^2 B_0^2 \lambda^2 Y A L^3 \alpha^2 \Delta T^2}{2(\mu B_0 \lambda L + Y A)^2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Thermal Stress Calculation|First Law Application|Energy Stored in Inductor (Analogy)

Why This Question Is Tricky

Examiner Levers

Visualization Stress

Requires mental rotation or 3D mapping (Optics/Geometry/Vectors).

Difficulty Boost : +4

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the 'Work-Energy Equivalence' trap in the thermodynamics of solids.

Students frequently misapply the First Law of Thermodynamics by assuming that the work done by the expanding rod against the external magnetic field (Option B) is the only mechanical energy stored in the system.

They correctly calculate the actual displacement $\delta$ but neglect the elastic strain energy stored inside the rod's lattice due to it being prevented from reaching its full thermal length $L(1+\alpha \Delta T)$ .

By mapping the magnetic field gradient to an equivalent Hookean spring, the problem requires students to recognize that the total mechanical energy is analogous to two springs in series, demanding the inclusion of both the external work and the internal elastic potential energy.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Thermal Stress Calculation
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free