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Symmetry in Stationary Magnetic Fields

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I am looking to utilise symmetry with a magnetic fields simulation in order to reduce model complexity. The attached models are much simplified version of the actual design I am simulating, but are adequate to show my problem.

I have a simple coil, above which is positioned a permanent magnet assembly, comprising of 2 magnets in antisymmetry (one North up, one South up).

Attached are 3 models representing this simple setup. The first has no symmetry, the second has a single plane of symmetry and the third has symmetry in 2 planes.

The first 2 simulations are easily done using the coil geometry analysis node (and ensuing the length factor is given as 2 for the single plane symmetry version, so that the coil resistance is calculated correctly). The plots from the first 2 models also make perfect sense and are in agreement.

However, when I introduce a second symmetry plane, the results start to not make sense.

I have added a Perfect Magnetic Conductor boundary condition on the plane where the crossing field must be normal, and have introduced a mirror 3D dataset in order to get plots of the full cross section of the model. This mirror 3D dataset has the vector transformation set to antisymmetric, to account for the permanent magnet and coil antisymmetry in this plane).

The plots look OK, but I am a little concerned that the Fz in the coil has increased (having been very close to zero, basically a model rounding error) to about 0.21N. I am aware that force calculations are mesh dependant, but doubling the mesh densoty produces the same result, which makes me think there is something that needs adjusting somewhere... This force should be zero, theoretically.

Could this be because of an incorrect setup or boundary condition somewhere?

The attached Word doc shows images of the 3 models.

Any feedback greatly appreciated!

Mark



5 Replies Last Post 2020年7月9日 GMT-4 09:45
Edgar J. Kaiser Certified Consultant

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Posted: 4 years ago 2020年7月6日 GMT-4 13:04

Mark,

just a little thought. You apply a symmetry plane and a perpendicular antisymmetry plane. This cannot be correct for the coil.

Cheers Edgar

-------------------
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Mark, just a little thought. You apply a symmetry plane and a perpendicular antisymmetry plane. This cannot be correct for the coil. Cheers Edgar

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Posted: 4 years ago 2020年7月7日 GMT-4 05:32

Hi Edgar,

Of course, I was thinking in terms of the permanent magnet. I believe the coil (as you state) requires 2 antisymmetric planes, but the permanent magnet structure requires one plane (the xz plane) to be symmetric, yet the other (yz) plane to be antisymmetric.

Can 1/4 symmetry still be used? Is there some workaround that can be utilised? the actual geometry I am using is hitting the RAM limit of my machine, hence my desire to use a second symmetry plane if possible.

Thanks in advance for any feedback...

Mark

Hi Edgar, Of course, I was thinking in terms of the permanent magnet. I believe the coil (as you state) requires 2 antisymmetric planes, but the permanent magnet structure requires one plane (the xz plane) to be symmetric, yet the other (yz) plane to be antisymmetric. Can 1/4 symmetry still be used? Is there some workaround that can be utilised? the actual geometry I am using is hitting the RAM limit of my machine, hence my desire to use a second symmetry plane if possible. Thanks in advance for any feedback... Mark

Edgar J. Kaiser Certified Consultant

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Posted: 4 years ago 2020年7月7日 GMT-4 06:25

Hi Mark,

any symmetry you want to use must be respected by the whole model, the complete geometry, all fields and all boundary conditions. So I think you can only use the first plane.

Cheers Edgar

-------------------
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Hi Mark, any symmetry you want to use must be respected by the whole model, the complete geometry, all fields and all boundary conditions. So I think you can only use the first plane. Cheers Edgar

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Posted: 4 years ago 2020年7月7日 GMT-4 06:29

Hi Edgar,

I thought that was going to be the case... I think I was asking more out of hope than expecatation!

Thanks again for your confirmation... time to buy some more RAM!

Hi Edgar, I thought that was going to be the case... I think I was asking more out of hope than expecatation! Thanks again for your confirmation... time to buy some more RAM!

Durk de Vries COMSOL Employee

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Posted: 4 years ago 2020年7月9日 GMT-4 09:45
Updated: 4 years ago 2020年7月9日 GMT-4 09:48

Hi Mark,

The forces are computed by integrating the Maxwell surface stress tensor. Please consider adding fillets to your magnet, before investigating forces. Without the fillets, your Maxwell surface stress tensor with develop singularities at the (infinitely) sharp edges of your magnet. This will greatly reduce the accuracy of your force calculation. For the nonmagnetic copper coil, you could consider a domain integral of the Lorentz force instead.

More on fillets can be seen here: https://www.comsol.com/blogs/fillet-away-your-electromagnetic-field-singularities/

For the Lorentz force: https://www.comsol.com/model/electromagnetic-forces-on-parallel-current-carrying-wires-131

When it comes to forces normal to the symmetry plane: The integral is done over the geometry you have in your model, without considering symmetry. In reality, one half of your coil will pull in one direction, the other half will pull in the other direction, and if there is symmetry, the sum of both will amount to zero. If you integrate only over half your coil because the other half is not in your model, you will get a nonzero value.

Once you are aware of this behavior, you can compensate for it: If you are integrating over half a magnet or half a coil and you know there should be an equal and opposite-pulling counterpart on the other side of your symmetry plane, you can consider the force normal to your symmetry plane to be zero.

However, if your magnet does not cross the symmetry plane, but is oriented along the symmetry plane, it means you have an identical magnet on the other side and both magnets will repel each other (in case of a magnetic insulation boundary condition). If the symmetry plane is a perfect magnetic conductor, they will attract each other. In that case, there is actually a total net. force on each individual magnet. But the sum of the forces of both magnets is still zero.

You can even reason the other way around: If you could hypothetically cut a magnet in half, creating two monopoles (which has not yet been proven possible by science, but you CAN model it if you like), you will see the two poles attract each other (like positive and negative electrical charges).

So the poles of a magnet pull on each other. The main reason you would expect a value of zero, is because you typically consider the force on the entire magnet, not just on one of its poles.

We have a resource combing symmetry conditions, magnets, force calculations and fillets. It is located here: https://www.comsol.com/model/electromagnetic-force-verification-series-55871

For further questions related to your modeling, feel free to contact our Support team.

Online Support Center: https://www.comsol.com/support

Email: support@comsol.com

Hi Mark, The forces are computed by integrating the Maxwell surface stress tensor. Please consider adding fillets to your magnet, before investigating forces. Without the fillets, your Maxwell surface stress tensor with develop singularities at the (infinitely) sharp edges of your magnet. This will greatly reduce the accuracy of your force calculation. For the nonmagnetic copper coil, you could consider a domain integral of the Lorentz force instead. More on fillets can be seen here: [https://www.comsol.com/blogs/fillet-away-your-electromagnetic-field-singularities/](https://www.comsol.com/blogs/fillet-away-your-electromagnetic-field-singularities/) For the Lorentz force: [https://www.comsol.com/model/electromagnetic-forces-on-parallel-current-carrying-wires-131](https://www.comsol.com/model/electromagnetic-forces-on-parallel-current-carrying-wires-131) When it comes to forces normal to the symmetry plane: The integral is done over the geometry you have in your model, without considering symmetry. In reality, one half of your coil will pull in one direction, the other half will pull in the other direction, and if there is symmetry, the sum of both will amount to zero. If you integrate only over half your coil because the other half is not in your model, you will get a nonzero value. Once you are aware of this behavior, you can compensate for it: If you are integrating over half a magnet or half a coil and you know there should be an equal and opposite-pulling counterpart on the other side of your symmetry plane, you can consider the force normal to your symmetry plane to be zero. However, if your magnet does not cross the symmetry plane, but is oriented along the symmetry plane, it means you have an identical magnet on the other side and both magnets will repel each other (in case of a magnetic insulation boundary condition). If the symmetry plane is a perfect magnetic conductor, they will attract each other. In that case, there is actually a total net. force on each individual magnet. But the sum of the forces of both magnets is still zero. You can even reason the other way around: If you could hypothetically cut a magnet in half, creating two monopoles (*which has not yet been proven possible by science, but you CAN model it if you like*), you will see the two poles attract each other (like positive and negative electrical charges). So the poles of a magnet pull on each other. The main reason you would expect a value of zero, is because you typically consider the force on the entire magnet, not just on one of its poles. We have a resource combing symmetry conditions, magnets, force calculations and fillets. It is located here: [https://www.comsol.com/model/electromagnetic-force-verification-series-55871](https://www.comsol.com/model/electromagnetic-force-verification-series-55871) For further questions related to your modeling, feel free to contact our Support team. Online Support Center: [https://www.comsol.com/support](https://www.comsol.com/support) Email: [support@comsol.com](mailto:support@comsol.com)

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.