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rsr_life
Aug30-07, 12:00 PM
Hello,

I have a cylindrical Neodymium permanent magnet. I need to find the pull force of the magnet at various distances from the magnet. Is there an expression relating the pull force(or attractive force) to the magnetic flux density B and the distance from the cylindrical magnet? What is the expression that tells us how the B varies with distance?

The parameters I have for the magnet are here

http://www.kjmagnetics.com/proddetail.asp?prod=D2C .

Thanks,

rsr_life

berkeman
Aug30-07, 01:05 PM
I don't work with magnets much, but this may help:

http://en.wikipedia.org/wiki/Magnet




.

rsr_life
Aug30-07, 07:33 PM
The wiki doesn't seem to have an expression listed for calculating magnetic flux density at a distance for a cylindrical permanent magnet. There's no current involved, so I just want to be able to simulate and get field values at different points from the magnet.

Seems there should be a straightforward expression to do this, maybe involving partial derivatives.

Thanks.

mdelisio
Aug31-07, 11:33 AM
Far enough away, most magnets can be approximated as a magnetic dipole moment m, which is a vector. For a magnetic dipole m at the origin and oriented in the z-direction, you get

\vec{B} = {\mu_o m \over {4\pi r^3}} (2\cos\theta \hat{r} + \sin\theta\hat{\theta})

The trick for your problem will be to find out a good guess at m. As usual, Wikipedia has some good information.

rsr_life
Aug31-07, 02:39 PM
Thanks for that. What about points closer to the magnet - near the surface, for example, where the gradient is steeper?

dlgoff
Aug31-07, 06:34 PM
Maybe you can find an equation by experiment. Could you use some type of spring device (to messure weight) with a iron-mass on it and find the force at different points around the magnet? Say; leave the spring hanging and moving the magnet at different positions under the mass?