Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 15D (Electromagnetism)

Since the magnetic field is $B=\nabla \times A$, Stokes theorem gives, for any oriented surface $S$ bounded by $C$,Under a gauge transformation $A\mapsto A+\nabla \chi$, the integral changes bybecause $C$ is closed. The flux expression is therefore gauge independent.

The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$. Substituting $B=\nabla \times A$ and using the Coulomb gauge $\nabla \cdot A=0$ givesThe free-space Green function of the Laplacian therefore givesFor the thin wire current stated in the question this becomes

Substituting the wire potential from part (b) into the flux formula from part (a) gives Neumann's mutual-inductance formulaInterchanging the two curves proves $L_{12}=L_{21}$.

Parametrize the coaxial circles byThen, with $\theta =\varphi -\psi$ and $R=\sqrt{a^2+b^2+c^2}$,where $q=2ab/R^2$. One angular integration contributes $2\pi$, soSince $ab=q{R}^2/2$, this is