Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 314 / 4

For an axisymmetric vector field, write $\mathbf{B}=B_R{\mathbf{e}}_R+B_{\varphi }{\mathbf{e}}_{\varphi }+B_z{\mathbf{e}}_z$. Pure differential rotation givesThis vector has no azimuthal component and is independent of $\varphi$, so its curl has zero radial and vertical components. Its azimuthal component isThe final braces are $R\nabla \cdot \mathbf{B}=0$ by Gauss's law for magnetism. The ideal magnetohydrodynamic induction equation therefore yields axisymmetric magnetic winding:Differential rotation creates a toroidal magnetic field from the fixed poloidal magnetic field. A stationary field under the same purely rotational assumptions requires Ferraro's law of isorotation, ${\mathbf{B}}_p\cdot \nabla \Omega =0$.

Let $M$ be the central stellar mass and take positive Keplerian rotation, $\Omega (R)=\sqrt{GM/R^3}$. The initial poloidal magnetic field stays fixed by part (a), while $R{\Omega }^{\prime }(R)=-(3/2)\Omega (R)$. Integrating the toroidal induction equation with $B_{\varphi }(R,0)=0$ givesThe initial field is a solenoidal vector field on the disk region $R>0$, since ${\partial }_R(R{B}_R)=0$. Assuming $B_0\ne 0$, the amplitude ratio isIt reaches one atThus the equality time is the same fraction $1/(3\pi )\approx 0.106$ of the orbital period at every radius. If “orbital time” instead means ${\Omega }^{-1}$, the corresponding fraction is $2/3$. These are kinematic solutions with prescribed rotation and the neglected magnetic back-reaction.

In the Gaussian units of the paper, the total magnetic pressure is $p_B=(B_R^2+B_{\varphi }^2)/(8\pi )$. Define the initial plasma beta ${\beta }_p=8\pi p_0/B_0^2\gg 1$, assuming a nonzero initial field. The solution from part (b) givesFor any $t>0$ this ratio decreases strictly with $R$, tends to infinity as $R\downarrow 0$ and tends to $1/{\beta }_p<1$ as $R\to \infty$. The magnetic-pressure equality radius under Keplerian winding is therefore unique in the formal profile on $R>0$:The same expression proves $p_B>p$ for $R<R_{\mathrm{eq}}$ and $p_B<p$ for $R>R_{\mathrm{eq}}$. At equality, $|B_{\varphi }|/|B_R|=\sqrt{{\beta }_p-1}\gg 1$, justifying the toroidal-dominated approximationThe source's existence claim refers to the indefinitely extended radial profile and the stated neglect of back-reaction. For a disk with a finite inner and outer edge, an equality radius lies in the disk only while the displayed value lies between those edges. If the initial field were zero, no magnetic winding or equality radius would occur.