For $\mathbf B=B\mathbf e_x$, the vertical magnetic perturbation is
$$
\delta B_z=ik_xB\,\widetilde\xi_z.
$$
The $z$ component of the Fourier-transformed equation of motion is therefore
$$
-\rho\omega^2\widetilde\xi_z
=-ik_z\widetilde{\delta\Pi}
-\rho v_A^2k_x^2\widetilde\xi_z.
$$
Using $\omega=vk_x$ gives
$$
\boxed{\frac{\widetilde{\delta\Pi}}{\widetilde\xi_z}
=-i\rho\,\frac{k_x^2}{k_z}(v^2-v_A^2)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-314/3/d/solution