Electromagnetism, calculation of B

[usn7564]

Homework Statement

A very big iron plate with the thickness 2h have, far from the plates edges, had a hole with the radius a drilled through it. With respect to the holes axis of symmetry the plate has a magnetization

\bar{M} = M_0 \frac{a}{s} \hat{s}

where s is the distance from the axis of symmetry.

In the hole, centered around the axis there is a small circular conductor with the current I with radius b << a.

Find the force on the conductor.

The Attempt at a Solution

I went ahead and found the bound charges which proved to be only on the top and bottom surfaces which proved to be

\bar{K} = M_0 \frac{a}{s} -\hat{\phi} for the upper surface and the same in the reversed direction for the lower.

Now with simply the right hand rule I concluded that the only B which will give a net force is the one in the s direction (giving a force in the z direction). The force from phi disappears in the cross product of biot savart, the force from z disappears due to symmetry. Then using with the knowledge that the only force that matters is the z component of the force:

\bar{F} = \Delta(\bar{m} \cdot \bar{B}) = \bar{m} \cdot \frac {d \bar{B}}{dz}\hat{z}

So I get that due to m being dotted with B only the z component survives of B, but I also concluded that the z component of B would not give a net force. Yet it ends up being the only one that gives a net force? There's something I'm misunderstanding 100% here.

and seems my tophats aren't working, but they're the unit vectors

 

[dauto]

For hats use \hat instead of \^.

 

[usn7564]

for hats use \hat instead of \^.

That worked, thanks.