EMC Question of the Week: January 10, 2022
At 10 MHz, the internal inductance of a #24 copper wire is
- more than 100 nH/m
- about 50 nH/m
- negligible
- undefined
Answer
The best answer is “c.” At MHz frequencies and higher, the internal inductance of a wire is virtually always negligible. 50 nH/m would be the internal inductance of a round copper wire carrying a DC current. At 10 MHz however, the skin depth in copper is about 20 μm. The diameter of a #24 solid copper wire is nominally 511 μm. Since most of the current is concentrated on the outer edge of the wire, the internal inductance is considerably less than 50 nH/m. [In this case, less than 8 nH/m.]
Any reasonable application of this wire, would have an external inductance much higher than 8 nH/m. For example, the inductance per unit length of two #24 wires twisted together is about 500 nH/m. A 2-cm diameter loop of #24 wire has an external inductance of about 50 nH (or about 800 nH/m). Generally, the contribution of the internal inductance will always be less than the uncertainty in the calculation of the external inductance.
It would be incorrect to say that the internal inductance is undefined. It is well-defined and can be calculated at high frequencies. But since the internal inductance is negligible compared to the external inductance in virtually any application, this calculation is rarely worth the effort.
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