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Re: [femm] ring magnet



In a message dated 11/28/00 10:04:47 AM Eastern Standard Time, 
ac319@xxxxxxxxxxxxx writes:


> Dear femm experts,
> I am trying use femm to model a problem which I am not successful so
> far: I have two disk shaped ring magnets, both of them are magnetized
> along the symmetry axis. The outer diameter of small magnet is less than
> the inner diameter of big magnet (about 1-2 mm) so that the small magnet
> can fit inside the big magnet.
> Question: Is it possible to model this configuration with femm. For
> example I want to fix the position of small magnet and position the big
> magnet in a way that both magnet's rotation axis coincides. If I move
> outer magnet along to rotation axis and parallel to rotation axis is it
> possible to measure relative vertical and lateral forces? Can femm do 3d
> configuration, or can the existing one be modified to do 3d ?
> best regards
> 
> Ahmet Cansiz
> Cambridge University
> IRC in Superconductivity
> 

It is possible to do part of what you are interested in with femm. As long 
as your magnet configuration is axisymmetric, you can do it with femm. I've 
attached a quick axisymmetric example with two concentric rings. When you 
are interested in computing forces, it is good to have a fine mesh in the 
region of the object upon which the force is to be measured--in this case, 
the magnets. In the example, I have defined a small region around the 
magnets with a pretty high mesh density. You'd then evaluate force in the 
postprocessor by drawing a contour around the magnet that you are interested 
in. Best results when the contour is through the air around the object 
rather than directly on the surface for numerical reasons.

For lateral displacements of the magnets, however, you really need a 3-D 
code. This isn't axisymmetric any more. However, if you are only interested 
in small perturbations about the axisymmetric case, you can infer axial 
forces from the axisymmetric force results. A fun result for magnets that 
have a unit permeability acting upon one another (and NdFeB, SmCo, and most 
types of ceramic magnets are closely approximated as having a unit 
permeability) is that the stiffnesses all have to add up to zero (i.e. Kx + 
Ky + Kz = 0). For an axisymmetric system, if Kz is the axial stiffness and 
Kr is the radial stiffness about a certain point, they must satisfy: Kz + 2 
Kr = 0. Therefore, if you can evaluate the axial stiffness, the force for a 
perturbation of some distance dr from the centered position is:
F = -(1/2) Kz dr
However, this result is accurate only for relatively small displacements.

Dave Meeker
--
http://members.aol.com/dcm3c

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<HTML><FONT FACE=arial,helvetica><FONT SIZE=2>In a message dated 11/28/00 10:04:47 AM Eastern Standard Time, <BR>ac319@xxxxxxxxxxxxx writes:
<BR>
<BR>
<BR><BLOCKQUOTE TYPE=CITE style="BORDER-LEFT: #0000ff 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px; PADDING-LEFT: 5px">Dear femm experts,
<BR>I am trying use femm to model a problem which I am not successful so
<BR>far: I have two disk shaped ring magnets, both of them are magnetized
<BR>along the symmetry axis. The outer diameter of small magnet is less than
<BR>the inner diameter of big magnet (about 1-2 mm) so that the small magnet
<BR>can fit inside the big magnet.
<BR>Question: Is it possible to model this configuration with femm. For
<BR>example I want to fix the position of small magnet and position the big
<BR>magnet in a way that both magnet's rotation axis coincides. If I move
<BR>outer magnet along to rotation axis and parallel to rotation axis is it
<BR>possible to measure relative vertical and lateral forces? Can femm do 3d
<BR>configuration, or can the existing one be modified to do 3d ?
<BR>best regards
<BR>
<BR>Ahmet Cansiz
<BR>Cambridge University
<BR>IRC in Superconductivity
<BR></BLOCKQUOTE></FONT><FONT COLOR="#000000" SIZE=3 FAMILY="SANSSERIF" FACE="Arial" LANG="0">
<BR>
<BR>It is possible to do part of what you are interested in with femm. &nbsp;As long <BR>as your magnet configuration is axisymmetric, you can do it with femm. &nbsp;I've <BR>attached a quick axisymmetric example with two concentric rings. &nbsp;When you <BR>are interested in computing forces, it is good to have a fine mesh in the <BR>region of the object upon which the force is to be measured--in this case, <BR>the magnets. &nbsp;In the example, I have defined a small region around the <BR>magnets with a pretty high mesh density. &nbsp;You'd then evaluate force in the <BR>postprocessor by drawing a contour around the magnet that you are interested <BR>in. &nbsp;Best results when the contour is through the air around the object <BR>rather than directly on the surface for numerical reasons.
<BR>
<BR>For lateral displacements of the magnets, however, you really need a 3-D <BR>code. &nbsp;This isn't axisymmetric any more. &nbsp;However, if you are only interested <BR>in small perturbations about the axisymmetric case, you can infer axial <BR>forces from the axisymmetric force results. &nbsp;A fun result for magnets that <BR>have a unit permeability acting upon one another (and NdFeB, SmCo, and most <BR>types of ceramic magnets are closely approximated as having a unit <BR>permeability) is that the stiffnesses all have to add up to zero (i.e. Kx + <BR>Ky + Kz = 0). &nbsp;For an axisymmetric system, if Kz is the axial stiffness and <BR>Kr is the radial stiffness about a certain point, they must satisfy: Kz + 2 <BR>Kr = 0. &nbsp;Therefore, if you can evaluate the axial stiffness, the force for a <BR>perturbation of some distance dr from the centered position is:
<BR>F = -(1/2) Kz dr
<BR>However, this result is accurate only for relatively small displacements.
<BR>
<BR>Dave Meeker
<BR>--
<BR>http://members.aol.com/dcm3c</FONT></HTML>

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