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Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.