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Use the central-field identity and substitute the prescribed relation between the toroidal magnetic field and axial magnetic field:The flux-tube axial field virial identity bounds the last average by $B_p^2$, while $p(0)\ge 0$. Therefore the uniform-twist central axial field bound isThe stronger exact expression is $B_z(0{)}^2=(1+{\gamma }^2)B_p^2-2{\mu }_{0}p(0)-(4{\mu }_0{\gamma }^2/s_0^2){\int }_0^{s_0}sp(s)ds$.

A toroidal magnetic field introduces an inward hoop force. In cylindrical coordinates, the radial magnetic tension is $-B_{\varphi }^2/({\mu }_{0}s)$, so cylindrical magnetostatic pressure balance in SI units becomesAt the field-free boundary $P(s_0)=p_e$. HenceRegularity on the axis gives $B_{\varphi }=O(s)$, which makes the integral finite and gives $B_{\varphi }(0)=0$. Evaluating at the axis proves the central-field identity:The positive hoop-tension contribution allows the central axial field to exceed $B_p$ without negative pressure. For an explicit regular example, put $x=s^2/s_0^2$, choose a constant $a>0$ and $0<\epsilon <1$, and take inside the flux tubeSubstitution verifies the radial balance, with continuous zero field at $s=s_0$ and positive gas pressure. For $a=3$, $\epsilon =1/4$, the central value is $B_z(0{)}^2=3B_p^2/2$.

To obtain the cross-sectional average, multiply the radial balance by $s^2$ and integrate by parts:The toroidal contributions cancel. The resulting flux-tube axial field virial identity isThis is at most $B_p^2$, and is strictly smaller whenever the weighted pressure integral is positive. In particular, continuous matching with $p_e>0$ guarantees strictness. If only $p\ge 0$ is assumed without a nondegenerate gas or continuous positive boundary pressure, the justified conclusion is the non-strict bound; the identity specifies exactly when equality occurs.

For a flux tube with $B_{\varphi }=\gamma (s/s_0)B_z$, regularity at the axis and cylindrical magnetostatic pressure balance give $B_z(0{)}^2=B_p^2-2{\mu }_{0}p(0)+{\gamma }^2⟨B_z^2⟩$. The flux-tube axial field virial identity and $p\ge 0$ give the bound. A twisted toroidal magnetic field can support a central axial field above the untwisted magnetic pressure limit even though its cross-sectional mean obeys that limit.