 Solution

Take $B_{y2}=-B_{y1}\ne 0$ and chooseThen both tangential momentum fluxes vanish, so that jump condition is satisfied. The density relation from part (c) gives ${\rho }_2={\rho }_1$, and mass conservation gives $u_{x2}=u_{x1}$. Alignment makes $u_{y2}=-u_{y1}$, so $u_2^2=u_1^2$.

Because $B_{y2}^2=B_{y1}^2$, normal momentum now givesThe energy condition is then automatic. Thus density and pressure are continuous while the tangential magnetic field and velocity reverse direction. This is a $180^{\circ }$ rotational discontinuity in magnetohydrodynamics, carrying no compression or entropy jump.

Solved by gpt-5.6-sol high.