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Question #43440

Difficulty: 8/10 | 🟢 Free

The Question

An electron of mass mm (in kg\text{kg}) and charge −e-e (in C\text{C}) revolves counterclockwise in the xyxy-plane in a circular orbit of radius RR (in m\text{m}) with a constant angular speed ω\omega (in rad s−1\text{rad s}^{-1}) around the origin. A closed spherical surface SS of radius r0r_0 (in m\text{m}, where r0≫Rr_0 \gg R) is centered at the origin. Let Φtotal\Phi_{\text{total}} (in Wb\text{Wb}) be the net magnetic flux through the entire surface SS, and ∣Φupper∣|\Phi_{\text{upper}}| (in Wb\text{Wb}) be the magnitude of the magnetic flux through the upper hemisphere (z>0z > 0) of SS. Which of the following correctly represents Φtotal\Phi_{\text{total}} and ∣Φupper∣|\Phi_{\text{upper}}|? (Assume vacuum with permeability μ0\mu_0 in T m A−1\text{T m A}^{-1})

Options

  • A.

    Φtotal=0\Phi_{\text{total}} = 0, ∣Φupper∣=μ0eωR24r0|\Phi_{\text{upper}}| = \frac{\mu_0 e \omega R^2}{4 r_0}

  • B.

    Φtotal=0\Phi_{\text{total}} = 0, ∣Φupper∣=μ0eωR22r0|\Phi_{\text{upper}}| = \frac{\mu_0 e \omega R^2}{2 r_0}

  • C.

    Φtotal=μ0eωR22r0\Phi_{\text{total}} = \frac{\mu_0 e \omega R^2}{2 r_0}, ∣Φupper∣=μ0eωR24r0|\Phi_{\text{upper}}| = \frac{\mu_0 e \omega R^2}{4 r_0}

  • D.

    Φtotal=μ0eωR24r0\Phi_{\text{total}} = \frac{\mu_0 e \omega R^2}{4 r_0}, ∣Φupper∣=0|\Phi_{\text{upper}}| = 0

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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

Gauss’s Law for Magnetism|Magnetic Dipole Moment of Revolving Electron|Variable Force Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of Gauss's Law for Magnetism and the integration of a non-uniform magnetic dipole field.

A common trap is assuming that a localized current loop behaves analogously to a net electric charge, leading students to incorrectly guess a non-zero total flux.

Furthermore, calculating the hemispherical flux forces students to correctly apply the dipole field formula in spherical coordinates and perform a surface integral.

Alternatively, astute students can use the divergence theorem to equate the hemispherical flux to the flux through the equatorial plane outside the loop, testing their transition from calculus-heavy brute force to conceptual physical intuition.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMagnetism and Matter - Class 12
ConceptGauss’s Law for Magnetism
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree