Orbital magnetic moment of an electron in circular motion An electron (charge −e in sign, magnitude e) revolves in a circular orbit of radius R around a proton with constant angular velocity ω. What is the magnitude of its orbital magnetic dipole moment?

Introduction:
Charged particles in circular motion form a current loop and therefore possess a magnetic dipole moment. This concept underlies atomic magnetism, NMR-like phenomena, and the classical picture of orbital contributions to magnetic behavior.

Given Data / Assumptions:

• Electron follows a circular path of radius R with angular speed ω.
• Classical current-loop model used to estimate magnetic moment.
• Magnitude requested; direction (sign) is not required.

Concept / Approach:

The magnetic dipole moment of a planar loop is μ = I * area. The effective current from a revolving charge is I = q * f, where f = ω / (2π). The area of the circular orbit is A = π R^2. Using q = e (magnitude), the moment magnitude is μ = (e * ω / (2π)) * π R^2 = (e ω R^2) / 2.

Step-by-Step Solution:

Verification / Alternative check:

Dimension check: e (C) * ω (s^-1) * R^2 (m^2) gives C·m^2/s = A·m^2, the correct unit for magnetic moment.

Why Other Options Are Wrong:

Options with R or R^3 have wrong dimensions; factors of 1 or 2 mis-handle the frequency-to-ω conversion; direction/sign does not affect the magnitude requested.

Common Pitfalls:

Using I = e * ω without dividing by 2π; forgetting to multiply by loop area π R^2.

Final Answer:

0.5 e ω R^2