00:01
Hello students, in this question we are given an ellipsoid.
00:04
So, the ellipsoid is in the x, y and z axis.
00:11
And it looks something like this.
00:16
Okay, so it has an x component which is equal to 2a and the y component is 2a, again 2a and the z component is actually 2b.
00:34
Okay, so with those in place, now we need to determine the magnetization of, we need to find out the magnetic pole densities for this geometry.
00:47
So first of all, let us take this in cartesian coordinates.
00:52
So it looks like something, equation looks like x square by a square plus y square by a square because the length is 2a, right? so this is 2a.
01:04
So the half of that will be the x component plus y square z square by b square will be equal to 1, right? so x can be written in terms of phi and theta.
01:22
We can determine that it is as 2a times sine phi cos theta and y will be equal to 2a times sine phi sine theta and z will be equal to 2b times cos theta.
01:42
So the surface is divided into small regions of differential area da.
01:47
So for each region, the magnetic pole density can be determined by dp by da will be equal to m0 by 4 pi times cos theta.
01:59
So here m0 is the uniform magnetization in ellipsoid...