How do I calculate repulsion force between two neodymium magnets?

In summary: Gravity, in that they are both governed by an inverse square law. Friction is proportional to the inverse of the distance between two objects.In summary, the force between two magnets is proportional to the product of their magnetizations and the distance between them.
  • #1
eddybob123
178
0
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
 
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  • #2
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
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
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
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
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
But how do I calculate the force? How do I calculate it in dipoles?
 
  • #8
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
Termotanque said:
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.

How exactly do I calculate the force? I googled it but I couldn't find a simple formula.
 
  • #10
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
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
One picture is worth 31 words.
 
  • #13
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
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
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
and the distance between the magnets
 
  • #17
Say, 1.5 centimetres.
 
  • #18
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
[tex]F=\frac{2\pi R^2 Mm}{(d^2+R^2)^{3/2}}[/tex],
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
What about all the other positions on the rotating pivot? What will happen to the equation?
 
  • #20
It gets much more complicated, requiring a Legendre polynomial expansion.
The formula would be simpler if d>>R.
 
  • #21
Okay, I looked it up on the Internet but I don't understand how it fits into magnetism.
 
  • #22
Shouldn't it fit into an inverse square law of some sort, like gravity? Magnetism is similar to gravity in several ways except with gravity, you don't care what the object looks like, as long as it has a designated mass. And with magnetism, the area of the faces closest to the other magnet varies the result. Obviously, two cylindrical magnets each weighing 100 grams one meter apart, both with a base area of ten square centimeters, has less force pulling on them than two cylindrical magnets each weighing 100 grams one meter apart, both with a base area of twenty square centimeters. So it is kind of like the inverse square law, except with an extra bit added concerning the area of the face of the magnet. I need to know that "extra bit".
 
  • #23
I guess the question we are asking ourselves is "can you calculate the flux of a permanent magnet and how does one go about do that?" Yes it is possible with a coil or inductor.
 
  • #24
This post is quite old and you might have found the solution. If not, you need to use Maxwell Stress Tensor. It makes the calculation very convenient. http://www.fieldp.com/documents/stresstensor.pdf
 

What is the formula for calculating repulsion force between two neodymium magnets?

The formula for calculating repulsion force between two neodymium magnets is F = (m1 * m2) / r^2, where F is the force in Newtons, m1 and m2 are the magnetic strengths of the two magnets in Tesla, and r is the distance between the two magnets in meters.

What is the unit of measurement for repulsion force between two neodymium magnets?

The unit of measurement for repulsion force between two neodymium magnets is Newtons (N). This is the standard unit for force in the International System of Units (SI).

How does the distance between two neodymium magnets affect the repulsion force?

The repulsion force between two neodymium magnets is inversely proportional to the square of the distance between them. This means that the force decreases as the distance increases.

What factors can affect the accuracy of calculated repulsion force between two neodymium magnets?

The accuracy of calculated repulsion force between two neodymium magnets can be affected by factors such as the strength and shape of the magnets, the distance between them, and any external magnetic fields present.

Can the repulsion force between two neodymium magnets be stronger than the individual magnetic strengths of the magnets?

Yes, the repulsion force between two neodymium magnets can be stronger than the individual magnetic strengths of the magnets. This is because the force is dependent on the product of the two magnetic strengths, not just their individual values.

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