 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-1/16b/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For the charge-generated electric field, use the electric potential

ϕ (r, t) = \frac{1}{4 π ε _{0}} \int \frac{ρ ( r ^{'} , t )}{∣ r - r ^{'} ∣} d V^{'}, E = - \nabla ϕ .

The vector potential is time independent, so its usual additional contribution

- \partial_{t} A

is zero. The continuity equation now gives

\partial_{t} E = - \frac{1}{4 π ε _{0}} \nabla \int \frac{ρ ˙ ( r ^{'} , t )}{s} d V^{'} = \frac{1}{4 π ε _{0}} \nabla \int \frac{\nabla ^{'} \cdot J ( r ^{'} )}{s} d V^{'} .

Thus the second term in part (iii) is exactly the displacement current contribution

μ_{0} ε_{0} \partial_{t} E

, and

\nabla \times B = μ_{0} J + μ_{0} ε_{0} \partial_{t} E .

This is the Ampère-Maxwell equation. The constructed field also has zero curl, consistent with the time-independent magnetic field and Faraday's law. A static addition to the electric field does not change this conclusion; the integral construction selects the fields sourced by the given densities rather than an independently added radiation field.

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