Resolve the upstream uniform field $\mathbf B_0=B_0\mathbf e_z$ into components normal and tangential to the spherical shock:
$$
B_{r,1}=B_0\cos\theta,
\qquad
B_{\theta,1}=-B_0\sin\theta,
\qquad B_{\phi,1}=0.
$$
The <ideal magnetohydrodynamic shock conditions> preserve the normal magnetic component and multiply the tangential component of a weak passive field by the gas compression ratio. Part (b) therefore gives
$$
\boxed{B_r(R,\theta,t)=B_0\cos\theta,
\qquad
B_\theta(R,\theta,t)=-4B_0\sin\theta,
\qquad
B_\phi=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 4 / c / Solution

{title2=ideal MHD shock conditions}

The ideal magnetohydrodynamic shock conditions are the Rankine-Hugoniot jump conditions for mass, momentum, energy, and magnetic flux. For a dynamically weak magnetic field, the normal field is continuous while the tangential field is multiplied by the gas compression ratio.


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