 Solution

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.