Electronic – DIY Inductor: Solve for number of turns without knowing the length

Air core inductors can have small capacitance. You can get an idea how small by considering the capacitance of two wires parallel to each other of length \$\pi D\$ (where \$D\$ is the coil diameter). That's like the turn to turn capacitance in the coil, and it's a small number. Of course, the total capacitance will be more than just the series combination of these turn to turn capacitances since the windings are coupled and will add in kind of a series/parallel way, but still a small number. That might make it hard to just measure by resonance. Probe capacitance of ~15pF or 20pF could be close to the coil capacitance.

Medhurst did a lot of work during the 1940's on inter-winding capacitance of air cored single layer solenoids. An equation that is an extension of Medhursts work is:

\$C_{\ell }\$ = \$\frac{4 \ell \epsilon _o \epsilon _{\text{rx}} \left(\frac{1}{2} k_C \left(\frac{\epsilon _{\text{ri}}}{\epsilon _{\text{rx}}}+1\right)+1\right)}{\pi \text{Cos}^2 \psi }\$

where
\$k_C\$ = \$\frac{0.106 D^2}{\ell ^2}+\frac{0.717439 D}{\ell }+0.933048 \left(\frac{D}{\ell }\right)^{3/2}\$

and \$D\$ is coil diameter, \$ \ell\$ is the coil length (or height), \$\psi \$ = \$\tan ^{-1}\left(\frac{p}{\pi D}\right)\$, \$p\$ is the pitch or turn spacing, \$\epsilon _o\$ = 8.854pF/m, \$\epsilon _{\text{ri}}\$ is relative permittivity of the coil form (2.56 for polystyrene, for example), and \$\epsilon _{\text{rx}}\$ is relative permittivity of space external to the coil (probably just air). For your case of \$D\$ = 0.3048m, \$ \ell\$ = 1.1811m, and \$p\$ = 1.08712mm, \$C_ \ell\$ ~ 21pF.

Reference for this is "The self-resonance and self-capacitance of solenoid coils" by David Knight. The equation is 5.3 on p 25, but there is a lot of detail of its development.

You might also be interested in the work of Gary Johnson who has written a book about Tesla coils. Here is a sample "Tesla Coil Impedance", and the rest is on his web page.

Inductance

Inductance is proportional to turns squared, all things being equal.

L2 / L1 = N2^2 / N1^2 ... 1 so
N2^2 = N1^2 x (L2/l1) ... 2 so
N2 = N1 x sqrt (L2/L1) .... 3

Apply 3. above

For 15 uH:
L1 = 22 uH
L2 = 15 uH
N1 = 50 so
N2 = 50 x sqrt(15/22) = 50 x 0.826 =~ 41.3 turns

So 41 turns in theory and maybe 40 - 42 range.

Similarly for 7 uH N = 50 x sqrt(7/22) =~ 28turns

Current rating:

Core magnetisation depends on "ampere-turns" so current handling capacity will increase with the inverse of the turns ratios.
I2 = I1 x N1/N2

So 15 uH & 41 turns will handle
9A x 50/41 =~~~= 11A
...
Thicker wire may be required depending on actual currents used, as copper losses increase with I^2.

Treat all such results with care and test in practice.

Insulation:

You do not describe the core type or former used.
You say 500V rating - I don't know what voltage you are using in practice but, be sure that insulation is (more than) adequate for the task. As you intend to use existing windings, take care not to nick insulation or have wire rub against core edge. If a bobbin former or insulated core is used this should be easy enough to manage. Odds are a 500-V rated part does not use a bare core and depend on wire insulation alone :-)