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Question #43427

Difficulty: 8/10 | 🟢 Free

The Question

An electron of mass mm (in kg\text{kg}) and charge −e-e (in C\text{C}) is constrained to move along a microscopic circular metallic ring of radius RR (in m\text{m}). The ring is initially at temperature T0T_0 (in K\text{K}) and has a coefficient of linear thermal expansion α\alpha (in K−1\text{K}^{-1}). The electron is in the nn-th stationary state, satisfying Bohr's angular momentum quantization condition. If the temperature of the ring is increased by a small amount ΔT\Delta T (in K\text{K}, such that αΔT≪1\alpha \Delta T \ll 1), and the electron remains in the same nn-th quantum state, which of the following correctly gives the fractional change in its orbital magnetic moment (Δμμ)\left(\frac{\Delta \mu}{\mu}\right) and the fractional change in its equivalent orbital current (ΔII)\left(\frac{\Delta I}{I}\right)?

Options

  • A.

    Δμμ=0\frac{\Delta \mu}{\mu} = 0, ΔII=−2αΔT\frac{\Delta I}{I} = -2\alpha \Delta T

  • B.

    Δμμ=2αΔT\frac{\Delta \mu}{\mu} = 2\alpha \Delta T, ΔII=0\frac{\Delta I}{I} = 0

  • C.

    Δμμ=αΔT\frac{\Delta \mu}{\mu} = \alpha \Delta T, ΔII=−αΔT\frac{\Delta I}{I} = -\alpha \Delta T

  • D.

    Δμμ=2αΔT\frac{\Delta \mu}{\mu} = 2\alpha \Delta T, ΔII=−2αΔT\frac{\Delta I}{I} = -2\alpha \Delta T

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Magnetic Moment of Orbiting Electron|Variation of Density with Temperature|Expansion of Cavities in Solids

Why This Question Is Tricky

Common Mistakes

Students often misapply classical macroscopic concepts to quantum states.

A common trap is assuming the electron's speed vv or angular velocity ω\omega remains constant as the cavity expands, leading to an incorrect variation in angular momentum LL.

By failing to apply Bohr's quantization condition (L=nℏL = n\hbar), which dictates that LL and thus the magnetic moment μ=e2mL\mu = \frac{e}{2m}L are strictly conserved for a given state nn, they incorrectly calculate a non-zero fractional change for μ\mu.

Furthermore, they fail to realize that to conserve LL, the velocity must decrease, causing the orbital current II to scale as R−2R^{-2} rather than R−1R^{-1} or remaining constant.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMoving Charges and Magnetism - Class 12
ConceptMagnetic Moment of Orbiting Electron
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree