Permanent magnet strength

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
[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.

eddybob123

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

clem

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

eddybob123

Okay, I looked it up on the Internet but I don't understand how it fits into magnetism.

eddybob123

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".

stephenbosley

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.

Hassan2

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