"The maximum permissible current density of copper in open air is approximately 3.1×106 A/m2 of cross-sectional area, above which it begins to heat excessively."

I came across the above statement when reading about copper in wikipedia.

Reference: - https://en.m.wikipedia.org/wiki/Copper

If the statement is true, a copper conductor with crossectional area of 1 square milimeter can only carry 3.1Amps of current in open air. If this rating is exceeded than the wire will heats up too much.

How true is this?  How do cable manufacturers observe this limitation when designing cable and insulating the coppper with materials with poor thermal conductivity.? Within an insulation copper conductor will heats up much more faster.


1 Answer 1


This is very true! The power dissipation in cables during current flow leads to Joule heating. The generated power is proportional to the resistance of the conductor and the square of the current (Joule-Lenz law) $$P \propto R I^2 $$ The conductor material will break down at a certain point when more heat is generated than the conductor can dissipate. But as you suspected correctly, if you combine the conductor with an insulation, heat dissipation is limited. Ampacity or current-carrying capacity is the quantity that tells you the maximum current you can apply to a conductor in certain conditions. It takes into account the conductor material, its insulation, the surrounding medium and its temperature, etc. The ampacity for many different cables is well established and there are standards and tables where you can find all the data. You can check the American wire gauge or the National Electrical Code but there are many more examples on the web.

If you were a cable manufacturer, you could probably just stick to the standards that hold for your country. Otherwise you can also measure surface temperature or carry out an ampacity measurement: Prepare the environmental conditions and the conductor you want to characterize and increase the current until the conductor breaks down.


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