 Solution

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.