= Solution
In <magnetostatics>, $-\nabla p-\rho\nabla\Phi+\mathbf f_L=0$. The plane-parallel <Newtonian gravitational potential> and all thermodynamic fields are independent of $y$, so the $y$ component of the <Lorentz force> must vanish:
$$
\nabla\psi\times\nabla B_y=0.
$$
Away from a null of $\nabla\psi$, this says that $B_y$ is constant along each connected contour of the <Cartesian magnetic flux function>. Thus, on a regular flux region,
$$
\boxed{B_y=b(\psi).}
$$
This statement is local; disconnected contours with the same numerical flux label need not share one global function without the usual flux-tube connectivity assumption.
Since $\nabla B_y=b'(\psi)\nabla\psi$, the transverse <magnetostatic equilibrium> equation becomes
$$
\boxed{\frac1{\mu_0}
\left[\nabla^2\psi+b(\psi)b'(\psi)\right]\nabla\psi
+\nabla p+\rho\nabla\Phi=0.}
$$
This is a <Cartesian magnetostatic flux-function equilibrium>. Projecting it along a transverse <magnetic field line> gives hydrostatic balance along that line; projecting across the line balances the gas pressure and weight against <magnetic pressure> and <magnetic tension>.
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