Hi there,
how can an I specify the magnetic field of a radially magnetized ring magnet in 3D? I tried to do it similar to tutorial 8 (only in 3D): Model --> Material --> MgDyn --> Magnetization 1, 2, 3. When I'm right, Magnetization 1, 2, 3 corresponds to the axes x, y, z and I would need to do a coordinate transform but how do I have to define this in Elmer? Otherwise, I might set the coordinate system to "Cylindric Symmetric" but what's the meaning of the coordinate mapping 1, 2, 3 in this case?
Thanks
magnetic field of a radially magnetized ring magnet
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Re: magnetic field of a radially magnetized ring magnet
Hi
You can write small functions to make the coordinate transformations. You can use the fact that r=sqrt(x^2+y^2), sin(phi)=y/r, cos(phi)=x/r etc. So for example r*sin(phi) would be just y. MATC is usefull for testing, for example:
-Peter
You can write small functions to make the coordinate transformations. You can use the fact that r=sqrt(x^2+y^2), sin(phi)=y/r, cos(phi)=x/r etc. So for example r*sin(phi) would be just y. MATC is usefull for testing, for example:
Code: Select all
MySinPhi = Variable "Coordinate"
Real MATC "tx(1)/sqrt(tx(0)^2+tx(1)^2)"
Re: magnetic field of a radially magnetized ring magnet
Thanks for the explanation. I tried to implement it like this and I was able to perform a simulation. However, the resulting magnetic flux density doesn't seem to be rotationally symmetric (see attachment). Here is the code for the radially magnetized ring magnet, that I used:
Can you please tell me whether this implementation is correct?
Code: Select all
Material 1
Name = "N42 NdFeB Magnet"
Density = 7500
Magnetization 2 = Variable "Coordinate"
Real MATC "900.0e3*tx(1)/sqrt(tx(0)^2+tx(1)^2)"
Magnetization 1 = Variable "Coordinate"
Real MATC "900.0e3*tx(0)/sqrt(tx(0)^2+tx(1)^2)"
Relative Permeability = 1.05
End
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Re: magnetic field of a radially magnetized ring magnet
Hi,
Yes, the sif looks correct assuming that your case is centered origin. However, the picture does not look as if you would have rotational symmetry. Are only the magnets cylindircal but not the fully geometry.
You can also break symmetry by the mesh. Particularly for a coarse mesh it is quite impossible to have a perfectly symmetric results.
-Peter
Yes, the sif looks correct assuming that your case is centered origin. However, the picture does not look as if you would have rotational symmetry. Are only the magnets cylindircal but not the fully geometry.
You can also break symmetry by the mesh. Particularly for a coarse mesh it is quite impossible to have a perfectly symmetric results.
-Peter