:

The inductance is not changing. What you are measuring is not pure
inductance but the coil has a stray capacitance. That is what is making
the
coil seof resonate.

__________________________________________________ _______

I am well aware of that, but you are tap dancing around the relevance of
the formula X=2*pi*F*L.

Just answer this: If I have a coil of very high Q (no appreciable
resistance), and I apply 100 volts of 100 MHz AC to it, and measure a
current of 2 milliamps through it, then:

1. What is its reactance?
2. What is its inductance?

--
Bill, W6WRT

I am not tap dancing around that formula. To be a self-resonance , the coil
must have some capacitance somewhere. You are leaving that out by just
using one formula. You also have to use the Xc=/(2pi*f*C) to get the
capacitance and then to find the final results you have to use Xo=Xl-Xc.

As to what you are asking for , it could be almost anything for the
inductance. It all depends on the stray capacitance. ?Remember you into
this self resonance so there has to be some capacitance. There is an
infinante number of solutions to the inductance of this problem.

The simple circuits that are most often found to measuer inductance do not
take the capacitance that may be there into concideration.

Just as if you are using a simple SWR meter and have a SWR of 2:1. You do
not know if you have a 25 ohm resistor or a 100 ohm resistor hooked to it.
Either will show 2:1. Also if there is any reactance , either inductive or
capacitive the simple SWR meter will not show that either. Again there is
an infiniante number of capacitors and inductor and resistors that will show
up as a 2:1 swr. You need a beter instrument.