Start Learning Free

Before You Begin

Question #43019

Difficulty: 8/10 | 🟢 Free

The Question

A fixed large circular coil of radius RR carries a steady current I1I_1. A small rigid circular wire loop of mass mm and radius rr (r≪Rr \ll R) carries a steady current I2I_2. The small loop is suspended from a frictionless horizontal axis passing through its highest point and lying in its own plane. The center of the small loop is located on the central axis of the large coil at a distance x=3Rx = \sqrt{3}R from the center of the large coil. The central axis of the large coil is horizontal. In the equilibrium position, the small loop hangs vertically such that the magnetic field of the large coil is perpendicular to the plane of the small loop. The currents are directed such that the magnetic torque opposes any angular deflection. Find the time period TT of small oscillations of the small loop.

Options

  • A.

    2π3mr2mg+μ0πI1I2r8R2 \pi \sqrt{\frac{3 m r}{2 m g + \frac{\mu_0 \pi I_1 I_2 r}{8 R}}}

  • B.

    2π2mr2mg+μ0πI1I2r8R2 \pi \sqrt{\frac{2 m r}{2 m g + \frac{\mu_0 \pi I_1 I_2 r}{8 R}}}

  • C.

    2π4mr2mg+μ0πI1I2r8R2 \pi \sqrt{\frac{4 m r}{2 m g + \frac{\mu_0 \pi I_1 I_2 r}{8 R}}}

  • D.

    2π3mr2mg+μ0πI1I2rR2 \pi \sqrt{\frac{3 m r}{2 m g + \frac{\mu_0 \pi I_1 I_2 r}{R}}}

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 Axis vs. Center|Physical Pendulum Inertia

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of electromagnetism and rigid body dynamics.

The primary trap lies in the calculation of the moment of inertia for the physical pendulum.

Students frequently default to I=mr2I = mr^2 (treating the loop as a point mass) or I=2mr2I = 2mr^2 (incorrectly applying the perpendicular axis theorem).

Because the loop swings out of its plane, the relevant centroidal moment of inertia is about an axis in the plane of the loop (Icm=12mr2I_{cm} = \frac{1}{2}mr^2), making the correct pivot inertia 32mr2\frac{3}{2}mr^2 via the parallel axis theorem.

Additionally, it requires correctly evaluating the Biot-Savart field on the axis of a coil rather than at its center, penalizing students who blindly use the standard B=μ0I/2RB = \mu_0 I / 2R formula.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptMagnetic Field on Axis vs. Center
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree