# How are diamagnets repelled from high field regions?

1. Sep 6, 2014

### cf123

Imagine an electromagnet with two pole faces, one flat (N) and one pointed (S). If you put a diamagnet in between them, I know that it will be attracted to the high-field region (i.e. the S pole where the field lines are more concentrated).

On the other hand, they always say that diamagnets will have an induced moment that cancels the field inside of them. That would mean that the diamagnet in this example could be viewed as a dipole, with its N pole closer to the pointed S pole of the electromagnet, and with its S pole closer to the flat N pole of the external magnet.

I would, from this analysis, then expect my diamagnet to be attracted to both poles of the electromagnet, but more so to the pole with the higher field: the S pole. That is paramagnetic behavior, and obviously this is completely wrong. What am I missing?

2. Sep 6, 2014

### Meir Achuz

A diamagnet would have its South pole closer to the South Pole of the electromagnetic. I think you may be confusing the relation between B and H. B will not simply equal mu H near the end of the object.

Last edited: Sep 6, 2014
3. Sep 7, 2014

### cf123

I understand that the diamagnet would have to have its SP closer to the SP of the electromagnet, in order to make the forces work out correctly in this example. What I do not understand is how having an induced moment in that direction would cancel the field inside the object, and not make it stronger... (The field lines from the electromagnet would run S-->N and the field lines inside the dipole would run parallel, not antiparallel to that...)

4. Sep 8, 2014

### Meir Achuz

You have to distinguish between the 3 fields, B, H, M. When you say "field lines" which field do you mean?
In a rod with no magnetic properties, B and H are the same inside and out, and M=0. I am using Gaussian units so that we can compare B and H without mu0. If the rod has negative susceptibility (mu<1), with B = mu H,
B inside the rod will be less than it was before, and H inside the rod
will be greater. Since B-H=4pi M, M will be in the opposite direction.
Outside the rod, B equals H and both are weaker than they were with no susceptibility.
If you draw a picture of the rod with zero susceptibility, and then one for negative susceptibility, putting in all three fields on each I think the situation should be clear.
The B and H fields will always be in the direction of the external magnetic field in each case. The magnetization field, M, will always be in the opposite direction.
I hope this helps.

5. Sep 8, 2014

### cf123

Alright, my understanding is this:

For a diamagnetic material, u<1 and B=uH = H+4piM.

Therefore, M must point in the opposite direction of H, whereas B and H point in the same direction.

Now, I draw a picture:

eN S----->N eS

eN=electromagnet north pole,N=diamagnet north pole, etc. That is, we place our diamagnetic dipole in between two poles of an electromagnet, with its moment opposing the field.

If, then we say that due to the shape of the electromagnet, the region near eS is higher field than the region near eN, won't this diamagnet be attracted to the high field region?

6. Sep 9, 2014

### Meir Achuz

Your picture is wrong. It should be eN N<----------------<S eS.
There is effective magnetic surface charge at each end of the diamagnetic given by ${\bf M\cdot{\hat n}}$. This puts a S Pole at the right and a N pole at the left. It is M, not B or H that determines the effective magnetic poles.

Last edited: Sep 9, 2014
7. Sep 9, 2014

### cf123

I'm sorry for asking so many questions. How are B, H, and M not all pointing the same direction in your picture? Won't the vectors point in the same direction (S to N and eS to eN) in your picture?

8. Sep 9, 2014

### Meir Achuz

B and H are to the right, but M is to the left in that picture.
This makes B<H in the diamagnet.