Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 3

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$, the vector field has cylindrical componentsThe divergence in cylindrical coordinates is thereforeDirect use of the curl in cylindrical coordinates givesandThus, definingwe obtain

For the stated magnetic vector potential, part a givesApplying the same identity once more and using the Ampère-Maxwell equation in the magnetostatic limit givesExpanding the Lorentz force density ${\mathbf{F}}_m=\mathbf{J}\times \mathbf{B}$, using $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$ and the vector triple-product identity, separates its poloidal and azimuthal parts:where

In an axisymmetric magnetostatic state, pressure and gravitational forces are poloidal, so the azimuthal component of the Lorentz force density must vanish. Hence $f=0$ andThe two gradients are locally parallel, so $L\alpha$ is constant on each regular level surface of $\beta$. Thereforefor an arbitrary flux function $F$.

For a barotropic fluid, $\nabla p=\rho \nabla h$. Magnetostatic force balance and $L\alpha =F(\beta )$ giveThe left side is a gradient multiplied by $\rho$, so taking the curl shows that $(L\beta -F{F}^{\prime })/(r^2\rho )$ is constant on each $\beta$ surface. Absorbing the fixed factor ${\mu }_0$ into an arbitrary function $G$ gives the Axisymmetric magnetostatic Grad-Shafranov system