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Magnetostatic cur-curl formulation in 2D
- creativeworker
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4 years 5 months ago - 4 years 5 months ago #2729
by creativeworker
Replied by creativeworker on topic Magnetostatic curl-curl formulation in 2D
Thank you very much for the clarification. Is it then correct to say that for 2D Magnetostatics it does depent on how my source structure is?
For example for a simple coil around an iron core:
* I can cut the the coil saggital such that three rectangles are created. The current (load) is then a scalar and so is the solution (vector potential A). In this case using H1 (Lagrange nodal elements) is correct and there is no problem with the "jumping" field.
* I can cut the system transversal such that a circle and a ring are created. Then the current is in plane and so will the vector field be. To bring that correct to finite elements, I would have to use HCurl-Space (Nedelec edge elements)?
Thanks again for your effort!
For example for a simple coil around an iron core:
* I can cut the the coil saggital such that three rectangles are created. The current (load) is then a scalar and so is the solution (vector potential A). In this case using H1 (Lagrange nodal elements) is correct and there is no problem with the "jumping" field.
* I can cut the system transversal such that a circle and a ring are created. Then the current is in plane and so will the vector field be. To bring that correct to finite elements, I would have to use HCurl-Space (Nedelec edge elements)?
Thanks again for your effort!
Last edit: 4 years 5 months ago by creativeworker.
- mneunteufel
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4 years 5 months ago #2747
by mneunteufel
Replied by mneunteufel on topic Magnetostatic curl-curl formulation in 2D
Yes, you are right.
In the first case H1 is the correct choice and the jumping coefficient doesn't make problems. It is like a Poisson problem with e.g. a discontinuous heat conductivity coefficient
[tex]div(\lambda \nabla u)=f.[/tex]
Just a remark: If you want to use a Zienkiewicz-Zhu (ZZ) type error estimator you have to interpolate the "flux" into HCurl, not VectorH1, like is done here in HDiv.
In the second case HCurl will produce better results than VectorH1.
Best,
Michael
In the first case H1 is the correct choice and the jumping coefficient doesn't make problems. It is like a Poisson problem with e.g. a discontinuous heat conductivity coefficient
[tex]div(\lambda \nabla u)=f.[/tex]
Just a remark: If you want to use a Zienkiewicz-Zhu (ZZ) type error estimator you have to interpolate the "flux" into HCurl, not VectorH1, like is done here in HDiv.
In the second case HCurl will produce better results than VectorH1.
Best,
Michael
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