= Solution
On one <poloidal flux surface>, the <field-line angular velocity> $\omega(\psi)$ is constant. In a frame rotating at that rate the relative <velocity> is $\mathbf v=\mathbf u-R\omega\mathbf e_\phi=k\mathbf B/\rho$, parallel to the <magnetic field>. The <Lorentz force> is perpendicular to this motion, and the <Coriolis force> also does no work. Centrifugal acceleration comes from the effective potential $-R^2\omega^2/2$. Since <entropy> is constant along the surface, <pressure> work integrates to <specific enthalpy> $w$, with $dw=dp/\rho$ along the flow.
Therefore <kinetic energy> relative to the rotating <poloidal flux surface>, <enthalpy> and effective gravitational-plus-centrifugal potential have a constant sum:
$$
\boxed{\frac12|\mathbf u-R\omega\mathbf e_\phi|^2+w+\Phi-\frac12R^2\omega^2=\varepsilon(\psi)}.
$$
This is a corotating <energy> integral, not the laboratory <energy> alone: <Maxwell stress tensor> stresses can transfer <energy> and <angular momentum> in the laboratory frame. In fact the expression equals $|\mathbf u|^2/2+w+\Phi-\omega Ru_\phi$, namely the total <magnetohydrodynamic Bernoulli invariant> minus $\omega\ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen <angular velocity> is fixed along the particular surface being followed.
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