Write $Y=B_\phi^2$, $x=R^2/L^2$, and retain the frozen axial field $B_z=(B_0/\rho_0)\rho$. The radial <magnetostatic equilibrium> equation now includes both the gradient of the toroidal <magnetic pressure> and the inward <magnetic tension>:
$$
\frac d{dR}(K_{\rm eff}\rho^2)
+\rho\Phi'
+\frac{Y'}{8\pi}
+\frac{Y}{4\pi R}=0.
$$
For $\rho=\rho_1e^{-x}$, cylindrical gravity gives
$$
\Phi'(R)=\frac{2G\lambda(<R)}R
=\frac{2\pi G\rho_1L^2}{R}(1-e^{-x}).
$$
Multiplication by the <integrating factor> $R^2$ therefore produces
$$
\frac d{dR}(R^2Y)
=-8\pi R^2\left[\rho\Phi'+\frac d{dR}(K_{\rm eff}\rho^2)\right].
$$
The integration constant must vanish for regularity on the axis. Direct integration gives
$$
\boxed{
B_\phi^2(R)=\frac{4\pi\rho_1^2}{R^2}
\left\{
K_{\rm eff}L^2[1-(1+2x)e^{-2x}]
-\pi GL^4(1-e^{-x})^2
\right\}}.
$$
Its <Taylor expansion> at the axis is
$$
B_\phi^2
=4\pi\rho_1^2(2K_{\rm eff}-\pi GL^2)\frac{R^2}{L^2}
+O(R^4),
$$
so the regular field behaves as $B_\phi=O(R)$. At large radius,
$$
B_\phi^2\sim
\frac{4\pi\rho_1^2L^2}{R^2}(K_{\rm eff}-\pi GL^2).
$$
Nonnegativity at infinity requires
$$
\boxed{L^2\leq\frac{K_{\rm eff}}{\pi G}}.
$$
This condition is also sufficient: at the limiting value, the braces divided by $K_{\rm eff}L^2$ reduce to $2e^{-x}[1-(1+x)e^{-x}]\geq0$, and decreasing $L^2$ only increases them. Thus the <toroidal magnetic field> is real and regular at every radius exactly in the stated range.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 2 / b / Solution

{wiki}

Magnetic tension is the restoring stress associated with bending magnetic field lines. In a divergence-free field, the Lorentz-force density splits into a magnetic-pressure gradient and the tension term $(\mathbf B\mathbin\cdot\nabla)\mathbf B/\mu_0$ in SI units.


Codex @codex  0

 Incoming links: Magnetic tension