# Discussion Forum

## Surface current density unit

 Topics: 4.3, 4.2, AC/DC Module
RSS feed   |   Turn on email notifications   |   9 Replies   Last post: October 11, 2012 2:23pm UTC

Guillaume Le Devehat

October 4, 2011 12:23pm UTC

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

Edgar Kaiser

October 4, 2011 1:25pm UTC in response to Guillaume Le Devehat

Re: Surface current density unit

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

Guillaume Le Devehat

October 4, 2011 2:06pm UTC in response to Edgar Kaiser

Re: Surface current density unit

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

Ivar Kjelberg

October 4, 2011 2:25pm UTC in response to Guillaume Le Devehat

Re: Surface current density unit

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

Guillaume Le Devehat

October 4, 2011 4:33pm UTC in response to Ivar Kjelberg

Re: Surface current density unit

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

Ivar Kjelberg

October 4, 2011 6:48pm UTC in response to Guillaume Le Devehat

Re: Surface current density unit

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

--
Good luck
Ivar

Jakub Zlamal

October 9, 2012 12:05pm UTC in response to Ivar Kjelberg

Re: Surface current density unit

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"

Ivar Kjelberg

October 9, 2012 7:49pm UTC in response to Jakub Zlamal

Re: Surface current density unit

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]

--
Good luck
Ivar

Jakub Zlamal

October 11, 2012 10:24am UTC in response to Ivar Kjelberg

Re: Surface current density unit

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.

Ivar Kjelberg

October 11, 2012 2:23pm UTC in response to Jakub Zlamal

Re: Surface current density unit

Hi

indeed on last image it looks OK ;)

--
Good luck
Ivar

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