 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/9a/c/solution

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

Each field has only an azimuthal component and is independent of

r

and

ϕ

(\nabla \times A)_{r} = \frac{1}{r s i n θ} \frac{\partial}{\partial θ} (sin θ A_{ϕ}), (\nabla \times A)_{θ} = - \frac{1}{r} \frac{\partial}{\partial r} (r A_{ϕ}),

and its

ϕ

component vanishes. The half-angle identities

sin θ tan \frac{θ}{2} = 1 - cos θ, sin θ (- cot \frac{θ}{2}) = - (1 + cos θ)

show in both cases that

\nabla \times A_{+} = \nabla \times A_{-} = \frac{e _{r}}{r ^{2}} = \frac{x}{r ^{3}} .

Apply part (a) with

g (r) = r^{- 3}

. Then

\nabla \cdot \frac{x}{r ^{3}} = r (- 3 r^{- 4}) + 3 r^{- 3} = 0

away from the origin, explicitly confirming that each resulting curl has zero divergence.

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