Resistance of aluminium rectangular wire I am working on upgrading an electromagnet design program. There are currently two types of wire used for the coil, circular and strip (read as: rectangular).
The current algorithms in this program are empirically derived with lots of undocumented constants and so one of the requirements is to move it more into the realm of accepted physics.
The formula I am using to calculate resistance is:
$$
R = \frac{\rho L}{A}
$$
These are aluminum wires so the value of $\rho$ boils down to $2.82\times10^{-8}$ at 20°C.
For circular wires this works perfectly, giving near enough the same results (enough to put it down to the increased accuracy).
For strip wires this doesn't work quite so well. The problem comes from a difference in the cross-sectional area of the wire.
The original program is doing $h^2$ as the area, where $h$ is the height of the wire. Now is that not wrong? The cross-sectional area should be $wh$ where $w$ is the width of the wire.
For example,
If the strip wire is 8.4mm by 2.8mm, the original program would do:
$$
2.8 ^ 2 = 7.84
$$
but should it not be doing:
$$
8.4\times2.8 = 23.52
$$
The entire original algorithm is:
$$
R = \frac{84.09L}{h^2}
$$
Is there a reason the original engineer may have done $h^2$ as apposed to the correct cross-sectional area?
I don't want to just declare his calculations as wrong as the results from this program have been used for the last 30 years and none of the magnets have gone bust or underperformed. Infact if anything they would have been over performing if this calculation is wrong.
 A: The difference way come from the fact the electrons repel themselves and thus tend to travel on the far edges of a conductor. In a circular section the electrons are evenly distributed around the perimeter giving you the nice predictable results.
With a rectangular section the electrons are crowding the corners changing the overall resistivity as the electron density changes around the perimeter.
PS. Consider using FEMM to simulate electron flow through a shape.
I did a test to see the results and I got the effect I expected. The field is concentrated around the edges, and the current density around the perimeter has spikes in the corners.
 
Update
Based on the comments I want to add that my point is that the resistivity calculation may depend on uniform current density (of sorts) and the rectangular cross section lends to non-uniform current density and thus might through the calculation off. 
The overall resistance on a wire is like a weighted average of the resistance of each fiber along the wire, with the current density the weighing factors. So if for any reason (I suspect) where the current density is high, so is the resistance (a non-linear effect) then the overall resistance of the conductor is not going be summarized simply by $R=\frac{\rho}{A} L$.
A: The original engineer may have been assuming a square crossection.  Yes, for DC anyway, the resistance of a wire is inversely proportional to the area of its corssection.
By the way, your resistivity of 2.82 x 10-8 can't possibly be right, as should be obvious from a dimensional analisys alone.
