Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 15D (Electromagnetism)

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Take the curl of the integral from part (a). Since $J(r^{\prime })$ is independent of $r$,This is the Biot-Savart law.

Parametrize the loop by $r^{\prime }=R{\hat{e}}_r(\varphi )$ and $d{\ell }^{\prime }=R{\hat{e}}_{\varphi }(\varphi )d\varphi$. At the observation point $r=z{\hat{e}}_z$, the Biot-Savart law for a line current givesThe components perpendicular to the axis cancel by rotational symmetry, while the axial component of the cross product is $R^{2}d\varphi$. ThereforeThe direction is fixed by the right-hand rule and reverses when the electric current reverses. At the centre,This is the on-axis magnetic field of a circular current loop.

Measure $z$ from the centre of the loop of radius $R$, toward the second loop. Between the loops the two axial fields have opposite directions. Their magnitudes areEquating them and cancelling common factors givesTaking the power $2/3$ yields $(D-z{)}^2=4z^2$. Since $0<z<D$, we have $D-z=2z$, and thereforeThus the magnetic-field cancellation between oppositely driven coaxial loops occurs one third of the separation from the smaller loop.