# Permanent magnet strength

1. Oct 27, 2011

### eddybob123

I've recently attempted to calculate the force of repulsion between two neodymium magnets. The problems arise fairly early when I don't know how. I know about ampere's force law and all that, but I couldn't find an equation concerning permanent magnets.

Obviously, two magnets with rectangular faces and side 5 and 6 have more force on each other than two magnets with half that size, because more magnetic field lines interact with each other.

I need an equation

Thanks

2. Oct 28, 2011

### clem

In Gaussian units, the force between two magnets, touching or very close together, is
F=2pi MM'A, where M and M' are the magnetization of each and A is the are of contact.
In terms of pole strength, it is F=2pi gg'/A.

3. Oct 28, 2011

### eddybob123

Does it work with different shaped magnets, such as a cylindrical magnet acting on an elliptical disc magnet? And what do you exactly mean by "area of contact"

4. Oct 28, 2011

### eddybob123

What happens when they are not close together? What units do you measure F in? I know I have a lot of questions but I' am just a beginner in physics

5. Oct 28, 2011

### Termotanque

If the magnets are not close, the magnetic field is far from uniform and you'll have a rough time calculating the force.

One thing that you can do is to suppose that they are far apart, and calculate the magnetic force on one magnet (thought as a dipole), due to the magnetic field of the other magnet, also thought as a dipole.

6. Oct 29, 2011

### clem

That formula is for two flat faces in contact, like the ends of bar magnets touching.
A is the area of contact. In Gaussian units, F is in dynes. Other configurations have different results.

7. Oct 29, 2011

### eddybob123

But how do I calculate the force? How do I calculate it in dipoles?

8. Oct 29, 2011

### clem

If the magnets are far apart compared to their size, the force is that of two dipoles.
If the are flush together, then post #2 applies.

9. Oct 29, 2011

### eddybob123

How exactly do I calculate the force? I googled it but I couldn't find a simple formula.

10. Oct 30, 2011

### clem

After 5 posts, you still haven't let us know the configuration of the two magnets. There are a number of formulas for different circumstances. Some configurations need complicated integrals. Post #2 gves one such formula. You "exactly ... calculate the force" by putting numbers into it.

11. Oct 30, 2011

### eddybob123

THe cylindrical magnet is steadily placed at a certain point, and the elliptical disc magnet is attached to a wooden bar on the width, which is attached to a rotating pivot.

12. Oct 30, 2011

### clem

One picture is worth 31 words.

13. Oct 30, 2011

### eddybob123

Sorry.
I uploaded this image from the program Paint. I need an equation to calculate the force of magnetism on the cylindrical magnet to wherever on the pivot the elliptical disc magnet is.

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14. Oct 31, 2011

### clem

The radius of the cylindrical magnet, the thickness of the elliptical magnet, its two axes, and the distance betwen the magnets are all important to decide what approximation to use.

15. Oct 31, 2011

### eddybob123

Suppose the radius of the cylindrical magnet is 1 and the thickness of the elliptical magnet is 0.3. Suppose the two axes of the elliptical magnet is 2.1 for its height, and 0.9 for its width. All units are in centimeters.

Just so you know, I don't need an exact answer. All I'm looking for is a formula to help calculate these things.

16. Nov 1, 2011

### clem

and the distance between the magnets

17. Nov 1, 2011

### eddybob123

Say, 1.5 centimetres.

18. Nov 1, 2011

### clem

For the distances you mention, a reasonable approximation would be to consider the elliptical magnet to be a dipole m, and the face of the cylindrical magnet to be like a uniformly charged disk. A formula for the force would be
$$F=\frac{2\pi R^2 Mm}{(d^2+R^2)^{3/2}}$$,
where R is the radius (1 cm} of the cylindrical magnet, M is its magnetization, and d is the distance (1.5+1.1/2) from the face of the cylinder to the middle of the elliptical magnet. This is all in Gaussian-cgs units. You could measure M by the force to separate two identical cylindrical magnets given in post #2. You could measure the magnetic moment m by the torque in a known B field (in gauss)
by torque=m B cos\theta.
This approximation should be reasonable until you get too close together or too far apart, when more complicated formulas would be needed.

19. Nov 1, 2011

### eddybob123

What about all the other positions on the rotating pivot? What will happen to the equation?

20. Nov 2, 2011

### clem

It gets much more complicated, requiring a Legendre polynomial expansion.
The formula would be simpler if d>>R.