> I get a descent boost curve at 30hz with a 53H inductor
At what impedance?
Bandwidth means Q.
Q is ratio of reactance to resistance.
> thats a reactance of 10'000ohms.
If "decent" means octave-wide or Q=1, then yes a perfect 50H inductor in a 10K resistor network would give "decent" Q=1 curve. And 10K is a customary nominal impedance for audio EQ systems.
If you change the resistors down 10X, to 1K, a perfect 50H inductor could give Q=10, a VERY narrow bandwidth (see Bob's room-node EQ problem), or a good 5H inductor could give the same Q=1. At 10 ohms (loudspeaker amp output), you only need 0.05H; but at typical levels it may need to carry high current.
Perfect inductors don't exist, and below 1KHz we normally must use many-many-many-turns on iron-cores, which makes a very im-perfect inductor.
In general, while you can add more and more turns to a core to increase H, you also add R. In filter apps, Q=10 at 30Hz is pretty good. (We use 50/60Hz power systems so that small iron can give Q>10 and a good Power Factor.)
And 53H or 160H on small iron means VERY small wire, very costly to buy and wind, and at some point there is more insulation (enamel) than copper. Q can be improved with bigger iron. But Q improvement is very slow for increasing size. Power-station 50/60Hz iron normally has Q>50 without any effort, but a 100KW core will crush your rack. And there is a balance between copper-loss and iron-loss, so that a really huge iron is not best for an audio inductor.
You are not the first to face this problem. When WWII radar researchers needed narrow low-audio filters, they turned to R-C Tee networks wrapped around amplifiers. Even with vacuum-tubes, 3 tubes and some R and C was a lot cheaper and lighter than L-C filtering.