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A cylindrical magnetostatic equilibrium is an axisymmetric vector field and pressure configuration independent of the axial coordinate. Its radial Lorentz force density balances the pressure gradient, as expressed by 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.

Decompose $\mathbf{u}={\mathbf{u}}_p+u_{\varphi }{\mathbf{e}}_{\varphi }$ and $\mathbf{B}={\mathbf{B}}_p+B_{\varphi }{\mathbf{e}}_{\varphi }$. In a steady axisymmetric vector field, $\nabla \times \mathbf{E}=0$ and $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$. The radial and vertical components of its curl areTherefore $R{E}_{\varphi }=C$ is a constant. Regularity at the symmetry axis sets $C=0$; equivalently, one imposes no toroidal loop electromotive force, $\oint \mathbf{E}\cdot d\mathbf{l}=2\pi C=0$. Under this usual condition,which implieswhere ${\mathbf{B}}_p\ne 0$. This is the regular-axis alignment of steady poloidal ideal flow. It allows nonzero toroidal components as well.

The axis or zero-loop-voltage condition is necessary for the unqualified conclusion. On an annular domain, $\mathbf{B}=(B_0/R){\mathbf{e}}_R$ and $\mathbf{u}=U_0{\mathbf{e}}_z$ give $\mathbf{E}=-(U_0{B}_0/R){\mathbf{e}}_{\varphi }$ with zero curl, yet the poloidal vectors are perpendicular. This can satisfy the remaining steady equations as well: choose constant mass density ${\rho }_0$, $\Phi =\pi G{\rho }_0{R}^2$ and $p=p_0-\pi G{\rho }_0^2{R}^2>0$ on the annulus. The field has zero curl, radial pressure balances gravity, the flow advects no equilibrium quantity, and ${\nabla }^2\Phi =4\pi G{\rho }_0$. It carries a nonzero loop voltage. Steady axisymmetry alone on an annulus therefore does not exclude such cross-field flow.

For an axisymmetric vector field, write $\mathbf{B}=B_R{\mathbf{e}}_R+B_{\varphi }{\mathbf{e}}_{\varphi }+B_z{\mathbf{e}}_z$. Pure differential rotation givesThis vector has no azimuthal component and is independent of $\varphi$, so its curl has zero radial and vertical components. Its azimuthal component isThe final braces are $R\nabla \cdot \mathbf{B}=0$ by Gauss's law for magnetism. The ideal magnetohydrodynamic induction equation therefore yields axisymmetric magnetic winding:Differential rotation creates a toroidal magnetic field from the fixed poloidal magnetic field. A stationary field under the same purely rotational assumptions requires Ferraro's law of isorotation, ${\mathbf{B}}_p\cdot \nabla \Omega =0$.

PutFor an axisymmetric vector field depending only on the cylindrical radius $R$, the solenoidal vector field condition isHence $R{B}_R$ is constant. The hypothesis $B_R\to 0$ on the axis forcesThe azimuthal and axial components of $\nabla \times \mathbf{B}=q\mathbf{B}$ are thenEliminating $B_{\varphi }=-B_z^{\prime }/q$ gives the closed ordinary differential equationOnce $B_z$ is known, $B_{\varphi }=-B_z^{\prime }/q$ and $B_R=0$ determine the full cylindrical force-free magnetic field.

A poloidal magnetic field lies in meridional planes through the symmetry axis, in contrast to a toroidal magnetic field. For an axisymmetric vector field, it is represented by a poloidal magnetic flux function as ${\mathbf{B}}_p=R^{-1}\nabla \Psi \times {\mathbf{e}}_{\varphi }$.