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## Surface current density unit

Dear All,

I am currently doing a simulation on Comsol 4.2 on a 2D magnetic field representation.

My question is regarding the input value of current density within magnetic fields module.

I would have expected that the current density unit would be A/m2 but it is written A/m which is basically a unit for magnetic fields.(see attached screen shot)

Does anybody has a clue?

Thanks a lot,

Guillaume

10 Replies Last Post Oct 8, 2015, 3:03 AM EDT
Posted: 6 years ago
Guillaume,

on your image it looks like you define a surface current on a boundary. This is why you have the unit A/m. The total current in this case is this linear current density times the length of the boundary in r-direction (and/or z-direction).

If you define azimuthal current density on a domain (cross section) you will get the unit A/m2.

Cheers
Edgar

Posted: 6 years ago
Dear Edgar,

Thanks for explanation. The boundary on which I applied the current is actually a very thin conductive surface(1um thick); that is why I took it as a boundary.

In the reality, I have a cross-sectional area of 50um2 which is very tiny.

The biggest value allowed in this crosssection is 5e6 A/m2. Shall I put 5e6 A/m in my simulation?

Cheers,
Guillaume

Posted: 6 years ago
Hi

you should respect units, but in 2D-axi a surface is in fact represented by a loop 2*pi*r*dr where dr is a small element of the boundary(=edge) of length L. Do not forget the 2*pi*r when needed ;)

Furthermore, a "surface current" in 2D-axi is in fact a current on a boundary = edge so you must normalise = divide the current by your thickness to get the surface density.

You say it looks "like a line" but your selection is also a line and not a domain = surface (in 2D-axi) so for me something is not understood (by me or you ? ;)

--
Good luck
Ivar

Posted: 6 years ago
Hi,

I know about the fact that a boundary in 2D-axi is actually a loop in 3D

I assumed that the thickness of my coil was almost equal to 0, that's why I took a boundary to model it.

The only thing I missed I guess is to divide the value of 5e6 A/m2 that I have by the thickness of the layer.

Guillaume

Posted: 6 years ago
Hi

carefull, if you have a current density [A/m^2], you must multiply it by the thicknes [m] to get the linear current density [A/m]. I was assuming you started from the current [A], that you need to divide by the thickness [m] to get to the line density [A/m] ;)

As I said, with COMSOl it's easy just check your units, systematically

Note today [rad] and Lagrange multipliers "lm's" do not have (always) units defined just as functions are untiless and require unitless operands, and Parameters used for parametric sweep or continuation sweeps are to be defined as scalars without units

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Good luck
Ivar

Posted: 5 years ago
I have prepared two samples in 3D. One with surface current density (10turns I=1A) coil of the length 9mm so the surface current density is 10*1/9e-3 [A/m] and another case with real coil.

Resulting Bz in the coil is "the same"

Posted: 5 years ago
Hi

your case 2 is strange (add a streamline and look at the magnetic flux lines)

to generate a magnet use the 2nd capture, or your test file, or even the test 2, but add a central cylindrical coordinate system and select the coil + add a domain external current on this cylinder with a cylindrical phi current density of the appropriate value 10*1[A]/Section[m^2]

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Good luck
Ivar

Posted: 5 years ago
I created test cases to check that I understand surface current density definition.

I do not see any problem with test2.mph, stream lines look good.

Posted: 5 years ago
Hi

indeed on last image it looks OK ;)

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Good luck
Ivar

Posted: 2 years ago
Hello

I have created a 3d metallic nanostructure and incident a range of wavelengths and tried to study the reflectance. I was curious to know the surface current density. In result model builder, I found an option for current and charge in electromagnetic definitions. there are many currents. Jx, Jy, Jz, Jnorm, and many more. It's difficult to understand which one I should opt? I would be very thankful if anyone could explain.

Hi

indeed on last image it looks OK ;)

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Good luck
Ivar

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