Along the open line, $B_p\sim F/r^2$ and the mass-loading relation with $u_p\to u_\infty$ gives
$$
\rho\sim\frac{kF}{u_\infty r^2},
\qquad
M_A^2\sim\frac{\mu_0ku_\infty}{F}r^2\longrightarrow\infty.
$$
The solutions of part b consequently have
$$
u_\phi\sim\frac\ell r,
\qquad
B_\phi\sim-\frac{\omega F}{u_\infty r}.
$$
The azimuthal <Alfvén speed> therefore approaches the nonzero constant
$$
\boxed{|v_{A\phi}|=\frac{|B_\phi|}{\sqrt{\mu_0\rho}}
\longrightarrow|\omega|\sqrt{\frac{F}{\mu_0ku_\infty}}}.
$$
For an unconfined outflow it is natural to take $\Phi\to0$ and $h\to0$ at infinity; also $u_\phi\to0$. The magnetic term in the <magnetohydrodynamic Bernoulli invariant> tends to
$$
-\frac{r\omega B_\phi}{\mu_0k}
\longrightarrow\frac{\omega^2F}{\mu_0ku_\infty}.
$$
Hence the <asymptotic energy of a radial magnetohydrodynamic wind> is
$$
\boxed{\epsilon=\frac12u_\infty^2
+\frac{\omega^2F}{\mu_0ku_\infty}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 4 / c / Solution

If $r^2|B_p|\to F$ and $u_p\to u_\infty$ along an open magnetic surface, then
$$
B_\phi\sim-\frac{\omega F}{u_\infty r},
\qquad
|v_{A\phi}|\to\omega\sqrt{\frac{F}{\mu_0ku_\infty}}.
$$
When enthalpy and gravity vanish asymptotically, the Bernoulli invariant becomes
$$
\epsilon=\frac12u_\infty^2+\frac{\omega^2F}{\mu_0ku_\infty}.
$$


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