 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/36b/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

R = ∣ x ∣

R = x / R

, and

t_{R} = t - R / c

. To leading nonzero multipole expansion order, the loop is a time-dependent magnetic dipole with

m (t) = π r^{2} I (t) n .

Expanding the retarded electromagnetic potential across the small loop gives

A (x, t) = \frac{μ _{0}}{4 π} [\frac{m ( t _{R} ) \times R}{R ^{2}} + \frac{m ˙ ( t _{R} ) \times R}{c R}] + O (\frac{r ^{3}}{R ^{3}}) .

The

R^{- 2}

term is the near magnetic-dipole field and the

R^{- 1}

term is the magnetic dipole radiation field. Substituting the Fourier series for the current gives

A (x, t) = \frac{μ _{0} r ^{2}}{4} (n \times R) n = 0 \sum \infty I_{n} [\frac{sin ( nω t _{R} )}{R ^{2}} + \frac{nω}{c R} cos (nω t_{R})] .

The

n = 0

summand vanishes for the stated sine series.

 New to topics? Read the docs here!