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Question #43853

Difficulty: 8/10 | 🟢 Free

The Question

A rigid superconducting square loop of mass $m$ , side length $L$ , and self-inductance $L_0$ is hinged freely at its top horizontal edge. It hangs as a physical pendulum in a uniform downward vertical magnetic field $\vec{B} = B_0 \hat{j}$ . The acceleration due to gravity is $g$ (downward). At equilibrium ( $\theta = 0$ ), the loop lies in the vertical plane and the current in the loop is zero. If the loop is given a small angular displacement $\theta$ and released, what is the angular frequency $\omega$ of the resulting simple harmonic motion? (Assume the wires of the loop are thin and uniform).

Options

A.
$\sqrt{ \frac{6 g}{5 L} + \frac{12 B_0^2 L^2}{5 m L_0} }$
B.
$\sqrt{ \frac{6 g}{5 L} + \frac{24 B_0^2 L^2}{5 m L_0} }$
C.
$\sqrt{ \frac{3 g}{2 L} + \frac{3 B_0^2 L^2}{m L_0} }$
D.
$\sqrt{ \frac{6 g}{5 L} - \frac{12 B_0^2 L^2}{5 m L_0} }$

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

Physical Pendulum Inertia|Force on Current-Carrying Loop|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the coupling of rigid body dynamics with electrodynamics, specifically targeting the thermodynamic potential of a superconducting loop.

A common lethal mistake is calculating the magnetic potential energy using $U = -\vec{\mu} \cdot \vec{B}$ , which applies only to a loop with a constant current source.

For a superconducting loop where flux is conserved, the work done by the induced emf must be accounted for, making the correct effective potential energy $U = \frac{1}{2}L_0 I^2$ .

Using the wrong potential yields a magnetic torque exactly twice the correct value (Option B).

Additionally, students often misjudge the direction of the torque, thinking it is anti-restoring because the induced dipole opposes the field (Option D), or incorrectly assume the moment of inertia of a square loop is $mL^2/3$ like a solid plate (Option C).

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Oscillations - Class 11 Part 2
Concept	Physical Pendulum Inertia
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free