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Question #43869

Difficulty: 8/10 | 🟢 Free

The Question

A uniform thin paramagnetic needle of length $L$ , density $ho$ , and Curie constant $C$ is pivoted freely at its upper end (the origin) and hangs downwards in the $-y$ direction in a uniform gravitational field $\mathbf{g} = -g \hat{j}$ . The needle is in thermal equilibrium at an absolute temperature $T$ . A non-uniform magnetic field $\mathbf{B} = \beta x \hat{i} + (B_0 - \beta y) \hat{j}$ is applied in the region, where $B_0$ and $\beta$ are positive constants. Assuming the magnetic susceptibility $\chi$ is small ( $\chi \ll 1$ ) and isotropic, what is the period $P$ of small oscillations of the needle in the $xy$ -plane? (Here $\mu_0$ is the vacuum permeability).

Options

A.
$2\pi \sqrt{\frac{2 L}{3 \left( g + \frac{C B_0 \beta}{\mu_0 \rho T} \right)}}$
B.
$2\pi \sqrt{\frac{2 L}{3 \left( g + \frac{2 C B_0 \beta}{\mu_0 \rho T} \right)}}$
C.
$2\pi \sqrt{\frac{2 L}{g + \frac{C B_0 \beta}{\mu_0 \rho T}}}$
D.
$2\pi \sqrt{\frac{2 L}{3 g}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Curie’s Law|Dipole Potential Angular Dependence|Physical Pendulum Inertia

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of thermodynamics (Curie's Law), rigid body dynamics, and electromagnetism.

The primary trap lies in the potential energy of an induced dipole.

Students frequently misapply the permanent dipole energy formula $U = -\mathbf{m} \cdot \mathbf{B}$ , forgetting that for a linear paramagnetic material, the magnetization is induced, requiring the inclusion of a $1/2$ factor ( $U = -\frac{1}{2} \mathbf{M} \cdot \mathbf{B}$ ) to account for the work done during magnetization.

Missing this factor leads to an incorrect magnetic restoring torque that is twice as large (Option B).

Additionally, students must correctly identify the physical pendulum's moment of inertia ( $ML^2/3$ ) rather than treating it as a simple pendulum (Option C), and recognize that the non-uniform field gradient creates a genuine restoring torque even though $\mathbf{m} \parallel \mathbf{B}$ locally (Option D).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Magnetism and Matter - Class 12
Concept	Curie’s Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free