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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.

For the homologous spherical flow $\mathbf{u}=q\mathbf{r}$, the ideal magnetohydrodynamic induction equation and $\nabla \cdot \mathbf{B}=0$ give the material derivativeLet $x=r/R(t)$ and use the self-similar ansatzPart (b) gives $q=3\dot{R}/(4R)$, and thereforeThe radial induction equation becomes $x{f}^{\prime }=6f$. The boundary value $f(1)=1$ gives $f=x^6$. The solenoidal vector field condition requireswhich also matches the tangential shock value in part (c). Hence the interior field is

Outside the shock, the uniform-field lines obey $r\sin \theta =\text{constant}$. Inside, the magnetic-field-line equation givessoA sketch therefore shows straight exterior lines refracting at the spherical shock into north-south symmetric curves that bow toward the equatorial interior before leaving through the opposite hemisphere. The field strength falls as $r^6$ toward the centre.