Force per unit area on conducting sheet in magnetic field Question from I.E. Irodov (3.260)




I’m having trouble with the (c) part because the external magnetic field should be towards the right to give such a result,hence the force on unit area should be downwards,I’m also unable to find the magnitude in (c) part.Any help would be appreciated.
 A: The case c) as depicted (assuming field lines are in the plane of the paper) implies two things:


*

*the current flows perpendicularly into the plane of the paper;

*the total field magnitude has left-to-right component on both sides, but field is weaker on the right-hand side.
This situation will occur if the external field has non-zero horizontal component $\mathbf B_{h}$ (left-to-right) and vertical component $\mathbf B_v$(bottom-to-top).
The horizontal component will contribute with force directed downwards (according to formula $\mathbf F = \int \mathbf j\times\mathbf B\,dV$). The vertical component will contribute with force directed to the right.
It is not clear why the solution in Irodov claims there is only force to the right. Perhaps it would be true if the plate was infinite and with zero ohmic resistance; then the sum of Lorentz forces acting on the current-forming mobile charge carriers would not translate into ponderomotive force on the plate, but would accelerate only the mobile charge carriers themselves (and change the current in the plate).
