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Results - Density current

Posted 11 set 2013, 08:12 GMT-4 Low-Frequency Electromagnetics Version 4.3 4 Replies

João Fernandes
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Hello,

I'm current working in a 2D simulation of a conceptual motor and I need some help interpreting the results.

I defined a copper conductor as the boundary of a cylinder of iron. The copper conductor has a surface density current [A/m]. After computing, when I use a "1D plot - line graph" to check the density current on that surface, the result is in [A/m^2] and not in [A/m]. Why?
I also was especting a Lorenz force contribution on that surface with the units [N/m^2] but the result is in [N/m^3].

Can someone explain me why is that?

Thanks for your atention,
Best reggards,
João Fernandes


4 Replies Last Post 12 set 2013, 04:50 GMT-4
Jens Krause
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Posted: 1 decade ago 11 set 2013, 08:28 GMT-4
Hallo,

which exact quantity are you plotting?

For me mf.Jsz is in A/m and mf.Jz is in A/m^2, as I expect.

Regards

Jens
Hallo, which exact quantity are you plotting? For me mf.Jsz is in A/m and mf.Jz is in A/m^2, as I expect. Regards Jens
João Fernandes
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Posted: 1 decade ago 11 set 2013, 08:41 GMT-4
Hi,

Now i realise that i was ploting mf.JZ so the units were in [A/m^2].

But i also have a conductive layer (surface) next to the current surface that i defined. I was expecting to induce a density current [A/m] on the conductive layer, but the only option to plot inductive currents is mf.JiZ.

I am trying to compare the amplitude of the density currents but one is [A/m] (source) and the other is [A/m^2] (induced). How can i have both in the same units?

Best reggards,
João

Hi, Now i realise that i was ploting mf.JZ so the units were in [A/m^2]. But i also have a conductive layer (surface) next to the current surface that i defined. I was expecting to induce a density current [A/m] on the conductive layer, but the only option to plot inductive currents is mf.JiZ. I am trying to compare the amplitude of the density currents but one is [A/m] (source) and the other is [A/m^2] (induced). How can i have both in the same units? Best reggards, João
Jens Krause
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Posted: 1 decade ago 12 set 2013, 04:35 GMT-4
Hi Joao,

If I understand you correctly, you have:

A: a circle (area=0) with surface current density in A/m
B: a ring (area>0) with volume current density in A/m^2

If you integrate Jz in B along the thickness, that would be
give a surface current density in A/m. Is that what you are looking
for?
You would use a general projection to perform such an integration.


Regards

jens
Hi Joao, If I understand you correctly, you have: A: a circle (area=0) with surface current density in A/m B: a ring (area>0) with volume current density in A/m^2 If you integrate Jz in B along the thickness, that would be give a surface current density in A/m. Is that what you are looking for? You would use a general projection to perform such an integration. Regards jens
João Fernandes
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Posted: 1 decade ago 12 set 2013, 04:50 GMT-4
Hello Jens,

The figure in annex better explains what I was saying. (Check attach file)

I realized that I can define materials as surfaces (without area). Now I ask if I can induce a current on that surface (on conductive material in figure). If I can, than how can I plot the induced current density in [A/m] to compare with the source density current?
The only plot I can do now is showing me the induced current as [A/m^2], but the conductive material is one surface (without area). How can I convert to [A/m]?

I hope now is easy to understand and thanks for your help.

The best regards,
João



Hello Jens, The figure in annex better explains what I was saying. (Check attach file) I realized that I can define materials as surfaces (without area). Now I ask if I can induce a current on that surface (on conductive material in figure). If I can, than how can I plot the induced current density in [A/m] to compare with the source density current? The only plot I can do now is showing me the induced current as [A/m^2], but the conductive material is one surface (without area). How can I convert to [A/m]? I hope now is easy to understand and thanks for your help. The best regards, João

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