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Question #43409

Difficulty: 8/10 | 🟢 Free

The Question

Consider a hypothetical hydrogen-like ion where the binding energy of the electron in its ground state (n=1n=1) is exceptionally large, given by Eb=0.25mec2E_b = 0.25 m_e c^2, where mem_e is the rest mass of a free electron and cc is the speed of light. The electron (charge −e-e) revolves in a circular Bohr orbit. According to relativistic principles, the binding energy manifests as a mass defect, reducing the effective inertial mass of the bound electron. The ion is placed in a uniform external magnetic field of magnitude BB. If the magnetic field vector makes an angle of 30∘30^{\circ} with the orbital angular momentum vector of the electron, what is the magnitude of the magnetic torque experienced by the electron's orbit? (Assume the Bohr quantization condition L=nℏL = n\hbar holds, and the gyromagnetic ratio is determined by the electron's effective mass).

Options

  • A.

    eℏB3me\frac{e \hbar B}{3 m_e}

  • B.

    eℏB4me\frac{e \hbar B}{4 m_e}

  • C.

    eℏB5me\frac{e \hbar B}{5 m_e}

  • D.

    2eℏB3me\frac{2 e \hbar B}{3 m_e}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Torque on Current Loop|Mass Defect vs. Binding Energy|Quantization of Angular Momentum

Why This Question Is Tricky

Common Mistakes

Students often memorize the classical magnetic moment of a Bohr electron (the Bohr magneton, μ=eℏ/2me\mu = e\hbar/2m_e) and apply it blindly.

This question tests the fundamental origin of the gyromagnetic ratio and its dependence on the actual inertial mass of the charge carrier.

By introducing a relativistic binding energy, the electron's effective mass decreases due to the mass defect (meff=me−Eb/c2m_{eff} = m_e - E_b/c^2).

Failing to account for this mass shift leads to the standard classical torque (Option B).

Furthermore, students might incorrectly add the binding energy to the mass (Option C) or forget the cross-product sine term entirely (Option D).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptTorque on Current Loop
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree