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Permanent magnet strength

by eddybob123
Tags: magnet, permanent, strength
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eddybob123
#1
Oct27-11, 10:19 PM
P: 115
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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clem
#2
Oct28-11, 07:32 AM
Sci Advisor
P: 1,256
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.
eddybob123
#3
Oct28-11, 05:37 PM
P: 115
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"

eddybob123
#4
Oct28-11, 06:54 PM
P: 115
Permanent magnet strength

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
Termotanque
#5
Oct28-11, 07:00 PM
P: 34
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.
clem
#6
Oct29-11, 07:16 AM
Sci Advisor
P: 1,256
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.
eddybob123
#7
Oct29-11, 01:59 PM
P: 115
But how do I calculate the force? How do I calculate it in dipoles?
clem
#8
Oct29-11, 02:49 PM
Sci Advisor
P: 1,256
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.
eddybob123
#9
Oct29-11, 05:08 PM
P: 115
Quote Quote by Termotanque View Post
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.
clem
#10
Oct30-11, 06:16 AM
Sci Advisor
P: 1,256
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.
eddybob123
#11
Oct30-11, 01:29 PM
P: 115
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.
clem
#12
Oct30-11, 05:22 PM
Sci Advisor
P: 1,256
One picture is worth 31 words.
eddybob123
#13
Oct30-11, 06:54 PM
P: 115
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.
Attached Images
File Type: bmp untitled.bmp (168.5 KB, 22 views)
clem
#14
Oct31-11, 10:45 AM
Sci Advisor
P: 1,256
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.
eddybob123
#15
Oct31-11, 06:47 PM
P: 115
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.
clem
#16
Nov1-11, 10:40 AM
Sci Advisor
P: 1,256
and the distance between the magnets
eddybob123
#17
Nov1-11, 05:38 PM
P: 115
Say, 1.5 centimetres.
clem
#18
Nov1-11, 08:11 PM
Sci Advisor
P: 1,256
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.


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