Microscopic current loop: magnetic moment and angular velocity Assertion (A): For an electron of charge magnitude e revolving in a circular orbit of radius R with angular velocity ω, the orbital magnetic dipole moment is μ = 0.5 * e * ω * R^2. Reason (R): Orbital angular momentum and magnetic dipole moment are always equal in magnitude.

Difficulty: Easy

Correct Answer: A is true but R is false

Explanation:


Introduction / Context:
A moving charge constitutes a current loop and therefore produces a magnetic dipole moment. The classical electron-in-orbit model, though simplified, builds intuition for relationships among current, area, magnetic moment μ, and orbital angular momentum L.



Given Data / Assumptions:

  • Electron of charge magnitude e circulates in a circle of radius R with angular velocity ω.
  • We consider magnitudes (signs from negative charge can be treated separately).
  • Nonrelativistic classical picture is sufficient for this relation.


Concept / Approach:
The effective current is i = q * f = q * (ω / 2π). The loop area is A = π R^2. Thus, μ = i * A = (q ω / 2π) * π R^2 = (q ω R^2) / 2. For electron magnitude e, μ = 0.5 * e * ω * R^2 (direction opposite to L due to negative charge). Angular momentum is L = m * v * R = m * ω * R^2. μ and L are proportional, not equal: μ = (q / 2m) * L (Bohr magneton concept).



Step-by-Step Solution:
Compute current: i = e * ω / (2π).Compute area: A = π R^2.Magnetic moment: μ = i * A = (e ω R^2) / 2.Compare with L: L = m ω R^2 ⇒ μ = (e / 2m) * L (proportional, not equal).



Verification / Alternative check:
In atomic physics, μ/L ratio equals e/(2m), leading to the Bohr magneton μB = e ħ / (2m) for one quantum of angular momentum ħ. The equality claim is therefore incorrect.



Why Other Options Are Wrong:
A is correct by direct calculation; R is false because it asserts equality instead of proportionality; the other combinations contradict these facts.



Common Pitfalls:
Ignoring the factor 1/2 from current–frequency relation; forgetting the electron’s negative charge reverses μ direction relative to L.



Final Answer:
A is true but R is false

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