In the aligned frame, define the <Alfvén number>
$$
A_i^2=\frac{u_i^2}{v_{Ai}^2}
=\frac{\mu_0\rho_i u_{xi}^2}{B_x^2},
$$
where the second equality uses $\mathbf u_i\parallel\mathbf B_i$. Since $j=\rho_i u_{xi}$ and $B_x$ are common,
$$
A_i^2=\frac{\mu_0j u_{xi}}{B_x^2}.
$$
Therefore
$$
\boxed{\frac{\rho_2}{\rho_1}
=\frac{u_{x1}}{u_{x2}}
=\frac{A_1^2}{A_2^2}}.
$$

Using alignment in the tangential momentum flux gives
$$
\rho u_xu_y-\frac{B_xB_y}{\mu_0}
=\frac{B_xB_y}{\mu_0}(A^2-1).
$$
Its continuity then yields
$$
B_{y2}(A_2^2-1)=B_{y1}(A_1^2-1),
$$
or
$$
\boxed{\frac{B_{y2}}{B_{y1}}
=\frac{A_1^2-1}{A_2^2-1}}
$$
whenever the ratios are determinate.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-314/3/c/solution