In Reply to: You want to vertify your data on self-inductance? posted by cheap-Jack on January 23, 2007 at 12:47:22:
As such, it depends heavily on the actual wire spacing. The best one can hope for is zero thickness insulation and about 160 nH, or .16 uH per foot, for a wire pair. Nobody uses zero thickness insulation of course.The terman equation is best for calculation of a wire pair inductance, but that also has a component of the equations which has the 15 nH per foot internal (or self) inductance number I quoted..
Here is the self (or internal) inductance for a cylindrical wire:Albert Shadowitz, "the electromagnetic field", Dover, 1988, pages 216-217..
Note, I can't put the integral equations here, suffice it to say he completed the derivation, the formula I'll give is the end result..
The total inductance for a single cylindrical conductor is:
Lint = μ0 / 8 π.
μ0 = 4 π times 10-7 henries/meter.
This becomes (4 π /8 π)times 10-7 henries/meter
or, 1/2 times 10-7 henries/meter
this is 5 times 10-8 henries/meter
or, 50 times 10-9 henries/meter
50 nanohenries per meter.
12/39.4 = .304 feet/meter.
50 times .304 = 15.2 nanohenries per meter.
(Ok, so I rounded the .2 off)
Cheers, John
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Follow Ups
- that is for a wire pair..not a single conductor, and it's kinda high. - jneutron 13:10:36 01/23/07 (2)
- (yah, I know) - jneutron 13:35:28 01/23/07 (0)
- ah, forgot to add the second part.. - jneutron 13:29:54 01/23/07 (0)