Orbiting Electron – Magnetic Dipole Moment Formula If an electron of charge magnitude q moves in a circular orbit of radius R with angular velocity ω, what is the magnitude of its orbital magnetic dipole moment?

Introduction / Context:
An orbiting charge constitutes a current loop and therefore has a magnetic dipole moment. This classical result underpins the connection between orbital motion and magnetism (e.g., the Bohr magneton in quantum mechanics). The formula can be derived from basic definitions of current and loop area.

Given Data / Assumptions:

• Charge magnitude q moving uniformly in a circle of radius R.
• Angular velocity ω (radians per second).
• Non-relativistic, classical picture; magnitude only (sign ignored).

Concept / Approach:

The current associated with one charge completing a revolution each period T = 2π/ω is I = q/T = q ω / (2π). The dipole moment of a planar loop is μ = I * A, where A is the loop area. For a circle, A = π R^2. Substituting gives μ = (q ω / (2π)) * (π R^2) = (q ω R^2)/2.

Step-by-Step Solution:

Verification / Alternative check:

Unit check: I in amperes and A in m^2 gives μ in A·m^2, the correct SI unit for magnetic dipole moment. Quantum analog relates μ to angular momentum L with μ = (q/2m) L for a circular orbit, consistent with μ ∝ ω R^2 since L = m ω R^2.

Why Other Options Are Wrong:

• 0.5 q ω R: missing one factor of R; dimensionally inconsistent.
• 0.5 q^2 ω^2 R^2: introduces extraneous powers; incorrect units.
• 0.5 q ω^2 R: wrong power of ω and R.

Common Pitfalls:

Forgetting that magnetic moment depends on area, not just radius; mixing up angular frequency with linear frequency in current calculation.

Final Answer:

0.5 q ω R^2