Let $c_s^2\sim P_g/\rho$. The assumed magnetic pressure gives
$$
\frac{P_m}{P_g}\sim\frac{\rho r\Omega c_s}{\rho c_s^2}
=\frac{r\Omega}{c_s}=\mathrm{Ma},
$$
so the <plasma beta> obeys $\boxed{\beta=P_g/P_m\sim\mathrm{Ma}^{-1}}$. In the magnetically dominated regime, vertical <hydrostatic equilibrium> gives $P_m/\rho\sim\Omega^2H^2$. Combining this with $P_m/\rho\sim r\Omega c_s$ yields
$$
\boxed{\frac Hr\sim\left(\frac{c_s}{r\Omega}\right)^{1/2}
\sim\mathrm{Ma}^{-1/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 321 / 1 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-321/1/c/i/solution