Axisymmetry makes the azimuthal direction geometrically distinct, so the field splits into a meridional poloidal part and an azimuthal toroidal part:
$$
\mathbf B=\mathbf B_p+B_\phi\mathbf e_\phi.
$$
The divergence-free condition on the poloidal field permits the <poloidal magnetic flux function>
$$
\boxed{\mathbf B_p=\nabla\psi\times\nabla\phi},
$$
or in cylindrical coordinates
$$
B_r=-\frac1r\frac{\partial\psi}{\partial z},
\qquad
B_z=\frac1r\frac{\partial\psi}{\partial r}.
$$
The magnetic flux through a circle of radius $r$ at fixed $z$ is
$$
\Phi_p(r,z)=2\pi[\psi(r,z)-\psi(0,z)].
$$
Thus, after choosing $\psi=0$ on the axis, $2\pi\psi$ is the enclosed poloidal flux and surfaces of constant $\psi$ are magnetic surfaces.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-314/4/c/solution