- #1

- 8

- 0

I've just been learning about surface magnetization currents circulating around hypothetical square loops. Since the magnetization is uniform the circulation currents cancel where the square loops are adjacent to one another and it can therefore be said that the current circulates around the edge of the magnet where the current is a bound current. I understand this and the derivation.

I am struggling a bit with the derivation of the same situation but now without uniform magnetization. I'll explain where I can get up to.

First looking at two loops, A and B. Loop A lies perpendicular to the z-direction and is centred on the point [itex](x,y,z)[/itex]. It represents part of a slice of thickness [itex]\delta z[/itex], and has sides of length [itex]\delta x[/itex] and [itex]\delta y[/itex]. The z-component of the magnetic moment of this volume element is [itex]M_x (x,y,z) \delta x \delta y \delta z[/itex], so the current that flows around the loop to generate this moment is

[tex]I(x,y,z)=(magnetic moment)/area=M_z (x,y,z) \delta z[/tex].

Loop B has the same dimensions but is centred at [itex](x+ \delta x, y, z)[/itex]. We can write the current in this loop as

[tex]I(x+ \delta x, y, z)=M_z (x + \delta x , y, z) \delta z [/tex].

The net current in the y-direction at the boundary between the two loops is

[tex]I(x,y,z)-I(x+ \delta x, y, z)=[M_z (x,y,z)-M_z(x + \delta x,y,z)] \delta z=-(\partial M_z/ \partial x) \delta x \delta z[/tex]

This is the bit that I'm getting stuck on now. Can someone please explain how we get the minus partial derivative and where the [itex]\delta x[/itex] comes from please? If you could break it down into smaller steps it would be greatly appreciated. Thanks in advance for any help.

Regards

Brian