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Question #77498

Difficulty: 5/10 | 🟢 Free

The Question

A small bar magnet of mass mm and magnetic dipole moment MM falls under gravity along the axis of a fixed horizontal circular conducting ring of radius aa and electrical resistance RR. The magnetic dipole moment remains aligned with the vertical axis (zz-axis) throughout the motion. The self-inductance of the ring is negligible. At a specific height z=a/2z = a/2 above the center of the ring, the magnet is observed to fall with zero acceleration. What is the velocity vv of the magnet at this instant? (Assume the magnetic field of the magnet is that of an ideal point dipole, and the acceleration due to gravity is gg.)

Options

  • A.

    3125mgRa4576μ02M2\frac{3125 m g R a^4}{576 \mu_0^2 M^2}

  • B.

    128mgRa49μ02M2\frac{128 m g R a^4}{9 \mu_0^2 M^2}

  • C.

    3125mga4576μ02M2R\frac{3125 m g a^4}{576 \mu_0^2 M^2 R}

  • D.

    255mgRa224μ0M\frac{25\sqrt{5} m g R a^2}{24 \mu_0 M}

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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

Relative Velocity in 1D|Lenz’s Law Energy Conservation|Induced Electric Fields

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This problem tests the student's ability to model velocity-dependent inductive drag by coupling Faraday's Law with Newtonian dynamics.

A common trap is failing to square the flux gradient when equating mechanical power to electrical power (Fmv=I2RF_m v = I^2 R), which leads to dimensionally incorrect linear dependencies (Option D).

Another trap is misapplying Ohm's law by placing resistance in the denominator of the velocity expression (Option C), or evaluating the flux gradient at an incorrect position like z=az=a (Option B).

The core physical principle requires recognizing that the back-EMF acts as a non-conservative damping force Fm∝vF_m \propto v, necessitating a strict energy-conservation approach to find the terminal state.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMotion in a Straight Line - Class 11 Part 1
ConceptRelative Velocity in 1D
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree