Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 1

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

PutFor an axisymmetric vector field depending only on the cylindrical radius $R$, the solenoidal vector field condition isHence $R{B}_R$ is constant. The hypothesis $B_R\to 0$ on the axis forcesThe azimuthal and axial components of $\nabla \times \mathbf{B}=q\mathbf{B}$ are thenEliminating $B_{\varphi }=-B_z^{\prime }/q$ gives the closed ordinary differential equationOnce $B_z$ is known, $B_{\varphi }=-B_z^{\prime }/q$ and $B_R=0$ determine the full cylindrical force-free magnetic field.

Here $q=\kappa /R$, so the equation from part (b) becomes the Euler-Cauchy equationThe power-law ansatz $B_z=R^s$ gives the indicial equationFor $0<\kappa <1/2$, the general field is thereforeThe polynomial discriminant changes sign atFor $\kappa >{\kappa }_c$, define $\mu =\sqrt{{\kappa }^2-1/4}$. A real form of the solution isThus the field has a log-periodic oscillation: its phase is periodic in $\ln R$, so it oscillates as the radius changes geometrically.

At $\kappa ={\kappa }_c=1/2$, the indicial equation has the multiple root $s=-1/2$. The general axial component isSince $B_{\varphi }=-2R{B}_z^{\prime }$, the azimuthal component isConsequentlyas $R\to \infty$; the same ratio is identically one when $D=0$. A magnetic field line has tangent parallel to $\mathbf{B}$, so its asymptotic angle $\vartheta$ with the $z$-axis satisfiesHence the field lines become helices making the constant anglewith the axis.