Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 4

A force-free magnetic field obeysequivalently $\mathbf{j}\times \mathbf{B}=0$: its current is parallel to its magnetic field. In a very low-density exterior, material pressure and inertia are too small to balance a finite Lorentz force. Quasistatic force balance therefore drives the magnetic field toward a force-free configuration.

The magnetic energy isUsing the ideal-MHD induction equation and integrating the identity $\mathbf{B}\cdot \nabla \times (\mathbf{u}\times \mathbf{B})$ by parts givesThe volume term vanishes for a force-free magnetic field, leaving

Axisymmetry makes the azimuthal direction geometrically distinct, so the field splits into a meridional poloidal part and an azimuthal toroidal part:The divergence-free condition on the poloidal field permits the poloidal magnetic flux functionor in cylindrical coordinatesThe magnetic flux through a circle of radius $r$ at fixed $z$ isThus, after choosing $\psi =0$ on the axis, $2\pi \psi$ is the enclosed poloidal flux and surfaces of constant $\psi$ are magnetic surfaces.

The toroidal component of $(\nabla \times \mathbf{B})\times \mathbf{B}=0$ givesBoth $\psi$ and $r{B}_{\varphi }$ are therefore constant along each poloidal field line, soThe poloidal components of the same force-free equation then reduce to the Grad-Shafranov equation for a force-free magnetic fieldThe poloidal current density isAmpère's law around an azimuthal circle gives the enclosed poloidal electric current:up to the orientation sign. Thus $F$ is the enclosed-current function in units ${\mu }_0/(2\pi )$.