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Question #43258

Difficulty: 8/10 | 🟢 Free

The Question

A solid non-conducting sphere of radius RR has a uniform positive volume charge density ρ\rho. The sphere rotates with a constant angular velocity ωsk^\omega_s \hat{k} about the z-axis. An electron (mass mm, charge −e-e) orbits in a stable circular path of radius rr (r≪Rr \ll R) in the equatorial xyxy-plane, centered at the origin, with an orbital angular velocity ωk^\omega \hat{k} (where ω>0\omega > 0). Assume the magnetic field produced by the rotating sphere is uniform inside and equal to its value at the center. It is given that the parameters satisfy the relations eρ3mϵ0=2Ω2\frac{e \rho}{3 m \epsilon_0} = 2 \Omega^2 and eμ0ρωsR23m=Ω\frac{e \mu_0 \rho \omega_s R^2}{3 m} = \Omega, where Ω\Omega is a positive constant. What is the magnitude of the magnetic dipole moment of the orbiting electron?

Options

  • A.

    eΩr2e \Omega r^2

  • B.

    12eΩr2\frac{1}{2} e \Omega r^2

  • C.

    12eΩr2\frac{1}{\sqrt{2}} e \Omega r^2

  • D.

    2eΩr22 e \Omega r^2

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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

Magnetic Dipole Moment of Revolving Electron|Non-conducting Charged Sphere|Potential of Concentric Shells

Why This Question Is Tricky

Common Mistakes

This question tests the simultaneous application of Gauss's Law, the Biot-Savart Law for rotating charge distributions, and the Lorentz force.

The primary trap lies in correctly identifying the direction of the magnetic force on the negative charge.

Students often forget the negative sign in F⃗B=−e(v⃗×B⃗)\vec{F}_B = -e(\vec{v} \times \vec{B}), leading them to subtract the magnetic force from the electrostatic force instead of adding it.

This sign error results in the incorrect quadratic equation ω2+Ωω−2Ω2=0\omega^2 + \Omega \omega - 2\Omega^2 = 0, yielding ω=Ω\omega = \Omega and leading directly to the distractor option B.

Additionally, it tests whether students remember the factor of 1/21/2 in the magnetic moment formula μ=12evr\mu = \frac{1}{2}evr (distractor D) and whether they completely ignore the internal magnetic field generated by the rotating sphere (distractor C).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptMagnetic Dipole Moment of Revolving Electron
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree