Assume you have a rectangular bar of copper with finite length L. You apply a voltage difference at two points on the top surface of the bar; (x1,y1) and (x2,y2). Because the bar is finite in length, I believe that the current will not be uniformly distributed. How would you go about calculating the voltage and current density at an arbitrary point within the bar?

I have tried to think through a few models that would approximate this, but I keep falling short. I think there must be a better way. Ideally, there would be an actual equation that I could derive that would give me the voltage and current density at any point within the bar. I have also wondered if a finite element method approach is best here.

If the bar is relatively thin in the z direction, I wonder if the problem can be simplified by instead considering a copper sheet with no thickness.

If anyone could point me in the right direction, I would really appreciate it.

  • \$\begingroup\$ you can deal with finite boundaries by the method of images. No current will flow across the bar edge, so you can 'reflect' a current contact in the boundary so that it creates a zero current sheet there. Rinse and repeat for all boundaries. Now you have an infinte grid of current sources and sinks distributed across an infinte half-plane. Solve for one, then sum as many of the decreasing contributions as you need. \$\endgroup\$ – Neil_UK Feb 10 at 16:17
  • \$\begingroup\$ > WHY do you think this way? ", I believe that the current will not be uniformly distributed." This only depends on termination and Kelvin effects but it is uniform otherwise. Go read about Volume Resistivity conversion to linear resistance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 10 at 16:59
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 because the bar is finite. Consider the example where the two contact points are very close to each other at one end of the bar. Would the majority of the current not flow in the area close to the points? Why would the same amount of current be flowing at the opposite end of the bar? The resistance of this copper bar is nonzero. I’m thinking in terms of a current divider where the resistance of the direct path between the two points is far lower than a path that loops to the opposite end of the bar, so proportionally more current would flow near the two points. \$\endgroup\$ – mblem22 Feb 10 at 17:40
  • \$\begingroup\$ Yes those are all Kelvin contact errors. I assume you want to avoid these \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 10 at 17:42
  • \$\begingroup\$ When you say "at two points" do you literally mean a geometric point...something with zero size? or is the current applied uniformly across the width of the bar? \$\endgroup\$ – Elliot Alderson Feb 10 at 17:49

I would use a finite element analysis. Break the bar into small cubes of material and then model each cube as six resistors, all connected together at the center and one connected to each face. Assume that your two points are at the exact centers of two cubes. Use SPICE, or construct the resulting resistor network and calculate the current through each resistor.

This is a numerical approximation, of course, but it gets better as you increase the number of cubes (i.e. make them smaller).

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  • \$\begingroup\$ This approach makes sense to me. I think I can do this in 2d since the bar is relatively thin in my application. I would definitely like to make the mesh as fine as possible, but I’m not entirely sure how to go about it. I’ll look into generating a spice netlist programmatically. Thanks for your feedback. \$\endgroup\$ – mblem22 Feb 10 at 19:56

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