Cylindrical magnetostatic pressure balance

For an axisymmetric vector field independent of the axial coordinate and regular on the axis, Gauss's law for magnetism gives $B_R=0$. The magnetostatic Ampère-Maxwell equation then gives $j_R=0$, $j_{\varphi }=-c{B}_z^{\prime }/(4\pi )$, and $j_z=c(R{B}_{\varphi }{)}^{\prime }/(4\pi R)$ in Gaussian units. Radial magnetostatic equilibrium isFor enclosed axial electric current $I(R)$, the magnetostatic Ampère-Maxwell equation gives $B_{\varphi }=2I/(cR)$, henceThe first term contains pressure and axial magnetic pressure; the second combines toroidal magnetic pressure with the hoop force from magnetic tension.

For constant axial magnetic field and $p=p_0(1+R^2/a^2{)}^{-k}$ with $p_0>0$ and $k>0$, cylindrical magnetostatic pressure balance gives, with $x=R^2/a^2$,The nontrivial electric current is finite at infinity exactly when $k>1$, in which case $I^2(\infty )=2\pi c^2{p}_0{a}^2/(k-1)$. Either sign of $I$ is possible because the pressure balance fixes its square.

With no axial electric current, cylindrical magnetostatic pressure balance reduces to constant $p+B_z^2/(8\pi )$. For $p=p_0{e}^{-R^2/a^2}$ and $B_z(0)=0$, its radial solutions areAs $R\downarrow 0$, $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ and $j_{\varphi }\to -\sigma c\sqrt{8\pi p_0}/(4\pi a)$. These finite scalar limits do not give a smooth function of Cartesian position: $R=\sqrt{x^2+y^2}$ has a cusp at the axis, and the azimuthal unit vector has no unique limit there. Thus these data describe radial equilibrium away from the axis, but cannot satisfy smooth vector-field regularity on the axis without modifying the data.