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A stationary axisymmetric magnetic field in ideal purely rotational flow requires constant angular velocity along the magnetic field lines of its poloidal magnetic field. Otherwise axisymmetric magnetic winding generates a time-dependent toroidal component.

Use the poloidal magnetic flux function convention ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$, so $2\pi \Psi$ is magnetic flux up to a reference constant. Its poloidal magnetic field components areThe prescribed surface flux therefore givesThe constant offset ${\Psi }_1$ does not affect either field component. If instead $\Psi$ denotes the full physical flux rather than flux divided by $2\pi$, both component formulas acquire a common $1/(2\pi )$ factor; the subsequent logarithmic derivative relation is unchanged.

The Gauss's law for magnetism constraint is $R^{-1}{\partial }_R(R{B}_R)+{\partial }_z{B}_z=0$. Since $\alpha$ is constant with radius on the surface, it gives the power-law poloidal field near a disk surface relationFor a smooth one-sided field above the surface, $B_z(R,z)=B_z(R,0)[1-(\delta -1)\tan \alpha z/R]+O(z^2)$. Thus the field initially decreases with height if $\delta >1$, increases if $0<\delta <1$, and has zero first height derivative if $\delta =1$.

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.