The toroidal component of $(\nabla\times\mathbf B)\times\mathbf B=0$ gives
$$
\mathbf B_p\mathbin\cdot\nabla(rB_\phi)=0.
$$
Both $\psi$ and $rB_\phi$ are therefore constant along each poloidal field line, so
$$
\boxed{rB_\phi=F(\psi)}.
$$
The poloidal components of the same force-free equation then reduce to the <Grad-Shafranov equation for a force-free magnetic field>
$$
\boxed{-r\frac{\partial}{\partial r}
\left(\frac1r\frac{\partial\psi}{\partial r}\right)
-\frac{\partial^2\psi}{\partial z^2}
=F(\psi)\frac{dF}{d\psi}}.
$$
The poloidal current density is
$$
\mathbf j_p=\frac{F'(\psi)}{\mu_0}\mathbf B_p.
$$
Ampère's law around an azimuthal circle gives the enclosed poloidal electric current:
$$
\boxed{I_p(\psi)=\frac{2\pi rB_\phi}{\mu_0}
=\frac{2\pi F(\psi)}{\mu_0}},
$$
up to the orientation sign. Thus $F$ is the enclosed-current function in units $\mu_0/(2\pi)$.

Solution

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

{c}
{wiki=Grad–Shafranov_equation}

For an axisymmetric force-free field with $\mathbf B_p=\nabla\psi\times\nabla\phi$ and $rB_\phi=F(\psi)$, the flux function obeys
$$
-r\frac{\partial}{\partial r}\left(\frac1r\frac{\partial\psi}{\partial r}\right)
-\frac{\partial^2\psi}{\partial z^2}
=F\frac{dF}{d\psi}.
$$


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