 Solution

For a static equilibrium, linearization of magnetic induction givesThe rotating local basis must be differentiated. In particular, ${\partial }_{\parallel }\hat{\mathbf{x}}={\mathbf{b}}_0/R$ and ${\partial }_{\parallel }{\mathbf{b}}_0=-\hat{\mathbf{x}}/R$. Their contributions proportional to ${\xi }_{\parallel }$ cancel in the radial component, yieldingWriting $h=\delta B/B_0$, the magnitude component iswhere $\nabla \cdot \mathbf{\xi }={\xi }_x/R+i{k}_y{\xi }_y+i{k}_{\parallel }{\xi }_{\parallel }$ when ${\partial }_x{\xi }_x=0$.

Set $M=B_0^2/(4\pi )={\rho }_0{v}_A^2$ and $\delta \Pi =\delta p_{\perp }+Mh$. The transverse $y$ component of the linear kinetic magnetohydrodynamic momentum equation isSince $k_y\gg k_{\parallel }\sim \omega /v_A$, perpendicular magnetic pressure enforces the leading pressure-balanced plasma perturbation condition $\delta \Pi =0$, givingIn the radial component, the curvature part of the perturbed magnetic force is $-2Mh/R$: one contribution comes from $({\mathbf{B}}_0\cdot \nabla )\delta \mathbf{B}$ and one from $(\delta \mathbf{B}\cdot \nabla ){\mathbf{B}}_0$. The perturbed pressure anisotropy contributes $-(\delta p_{\perp }-\delta p_{\parallel })/R$. After the leading total-pressure constraint has removed ${\partial }_x\delta \Pi$, the radial equation isSubstituting $Mh=-\delta p_{\perp }$ givesAlthough $h$ is small at low plasma beta, its contribution $Mh$ to the force balance is of the same order as $\delta p_{\perp }$ and must be kept.