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Okay, so in Griffiths' Introduction to Electrodynamics, Griffiths clearly defines surface current density as follows:

when charge flows over a surface, we describe it by the surface current density, $K$. Consider a 'ribbon' of infinitesimal width dL running parallel to the current flow. If the current in this ribbon is $dI$, surface current density is $K=dI/dL$.

Now, I searched Google and some websites which are clearly telling me that surface current density is current per unit AREA, not length! Is there something wrong with my understanding of this concept or are both the definitions equivalent?

Griffiths similarly defined volume current density as current per unit area perpendicular to the current flow, while in my opinion, it should be current per unit VOLUME...

This is really very confusing to me, please clarify.

Thanks!

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  • $\begingroup$ can you provide references to the sources that say current density is one length dimension less than what griffith gives? $\endgroup$
    – Jim
    Mar 15, 2017 at 12:55
  • $\begingroup$ @Jim is there any way to add screenshots or do i paste the whole thing? actually griffith later wrote the biot savart law for SURFACE current density, K with the multiplication of the differential area element 'da'. $\endgroup$ Mar 15, 2017 at 13:07
  • $\begingroup$ the main problem that is arising here is that i've studied, in electrostatics, that surface charge density is charge per unit area and i'm wondering if that could be extended to electromagnetism and said that surface current density is current per unit area? $\endgroup$ Mar 15, 2017 at 13:09

2 Answers 2

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Let's start with charge density $\rho$ which is the charge per unit volume. To get the amount of charge on some object, we'd integrate over the volume. Current is defined as charge per unit time crossing some surface. So to describe a charge density moving, we get a current density $J$ which is amount of charge per area per time... dimensionally it is one less "per length" than $\rho$ and an additional "per time" .

Now look at 2D: $\sigma$ is charge per unit area, the surface current $K$ is dimensionally one less "per length" (the charge is now crossing through a 'line' one the surface instead of an area).

Now look at 1D: $\lambda$ (I think that is what Griffiths used) is the charge per length on a wire, $I$ is dimensionally one less "per length" (the charge is now crossing through a 'point' on the wire).

If that doesn't help and you are still confused about the da in the Biot-Savart like law for surface current, let me know and I'll add more about that.

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  • $\begingroup$ thanks a lot! your answer combined with NIGTKWAHI's answer totally cleared it :) $\endgroup$ Mar 15, 2017 at 15:40
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You are wrong that google results say that surface current density is a current per unit area. I did my own google search, and that was not said in any of the results.

My guess is that you have no problem with an infinitely concentrated charge being a zero-dimensional point, and a current happening along a line. Notice that a current occupies the region that a single charge moves through. So if you think about when happens when zero dimensional point charges move, you get a current that lives on a one-dimensional curve.

Now if the point charge is smeared out onto a one-dimensional line segment, then the corresponding current would be on a 2D ribbon, as described in griffiths. The same way that the charge on a line has dimensions of charge per length, the current density on the ribbon has units of current per length.

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  • $\begingroup$ my bad. wikipedia showed the definition of 'current density' when i searched 'surface current density' ---It is defined as a vector whose magnitude is the electric current per cross-sectional area at a given point in space.... i mistook that definition for surface current density $\endgroup$ Mar 15, 2017 at 15:38
  • $\begingroup$ "My guess is that you have no problem with an infinitely concentrated charge being a zero-dimensional point, and a current happening along a line. " ---- exactly where the problem was! $\endgroup$ Mar 15, 2017 at 15:40

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