 Solution

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.