Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 4

The representationgivesHence $\mathbf{B}\cdot \nabla \psi =0$: the level sets of the poloidal magnetic flux function are magnetic surfaces. Moreover, the magnetic flux through a circular disc isThus $2\pi \psi$ is the enclosed poloidal magnetic flux up to an additive axial reference.

The flux-function form solves $\nabla \cdot \mathbf{B}=0$ identically. With $k=k(\psi )$, the poloidal relation $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and axisymmetry solve steady mass conservation. The ideal induction equation is solved by the velocity representation with the field-line angular velocity $\omega =\omega (\psi )$. The azimuthal momentum equation integrates once to the magnetohydrodynamic angular-momentum invariant $\ell =\ell (\psi )$.

What remains is the pressure or entropy equation and the two poloidal components of momentum. Projecting poloidal momentum along a field line produces the Bernoulli invariant in part (c); projecting across magnetic surfaces produces the transfield or Grad-Shafranov equation that determines their shape. An equation of state and boundary conditions complete the problem. The stellar gravitational potential is prescribed, so no self-gravity Poisson equation remains to solve.

Although the Lorentz force can exchange energy between poloidal and toroidal motion, it does no net work in the frame rotating with a magnetic surface. Dotting steady momentum with the poloidal field and using the mass-loading, isorotation, and angular-momentum invariants combines the magnetic work with the toroidal kinetic term to giveThereforeis constant on each magnetic surface. Hereis the gravitational potential plus the centrifugal potential in the frame corotating at the field-line angular velocity. This is the modified Bernoulli function of the wind.

Define the poloidal Alfvén number bywhere $u_p=k{B}_p/\rho$. Thus, along a given field line,density decreases as the Alfvén number grows.

The azimuthal velocity representation givesSubstitution into $\ell =r{u}_{\varphi }-r{B}_{\varphi }/({\mu }_{0}k)$ yieldsAt the Alfvén surface, $A^2=1$. Smoothness requires the numerator of the solution for $B_{\varphi }$ to vanish there, soSolving away from the surface then gives

Ideal flux freezing anchors a field line in the highly conducting disc, so its field-line angular velocity naturally equals the angular velocity of its footpoint. A circular orbit in the central potential is Keplerian:

Put $r=r_0+\delta r$ and expandThe linear radial term vanishes because $GM/r_0^2={\Omega }_0^2{r}_0$. The radial and vertical second derivatives are respectively $-3{\Omega }_0^2$ and ${\Omega }_0^2$, so

If the field line makes angle $\theta$ from the vertical, then locally $\delta r=z\tan \theta$. Its quadratic potential change iswhich is negative exactly whenThis is the local magnetocentrifugal acceleration launching criterion.

The velocity representation givesThereforeIn the launching region, $r\simeq r_0\ll r_a$ and $A\ll 1$, so part (d) givesConsequently

The modified Bernoulli kinetic energy isThus a decrease in ${\Phi }_{cg}$ is shared between poloidal acceleration and relative toroidal motion. If $R$ is approximately constant, the same potential drop produces a poloidal kinetic increase smaller by the factor $1/(1+R^2)$ than the estimate that neglects the toroidal term. If $R$ varies strongly, that term can absorb or return energy, so monotonic decrease of ${\Phi }_{cg}$ alone no longer proves an equally direct increase of $u_p^2/2$.