Question: Suppose that magnetic wire with resistivity similar to copper but with non-negligible permeability existed. What effect would magnetic wire have on the coil inductance versus a copper coil of the same size/construction?

This is of course a purely academic exercise, since such a material does not exist, though it could perhaps exist for plated wire at high frequencies, due to the skin effect.

EDIT:

I guess it's not entirely academic. isdi pointed out that "high permeability wire" actually exists.

According to [this](http://www.hongertech.com/high-quality-soft-magnetic-mu-metal-wire_p35.html): 

> 
> Mu Metal Wire is used in the production of inductors and electrical transformers. It has one of the highest magnetic permeability ratings of any metal or alloy, making it an industry standard in applications where metals with high magnetic permeability are needed.
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Answer: 
The [IR2110](https://www.infineon.com/dgdl/ir2110.pdf?fileId=5546d462533600a4015355c80333167e) is designed to supply +/-2A to drive the gate of your output devices.

You have an asymmetric drive caused by the 47 Ohm and diode across it. 

You have not specified what 2VM1 is, but assuming it's no more than a 15V supply your maximum gate drive current is about 0.3A during turn on. During turn off you may be exceeding the gate drive capability of the IR2110 since you then have 47 Ohm with whatever the effective Vf slope resistance of the diode is (along with a 0.8-0.9V offset). 

I'd suggest you could remove D2, 3, 5, 6 and replace the 47 Ohm with 10 Ohm resistors. This will make the turn on/off delays **the same** which is important. With a 15V VCC that will give you 1.5A charge/discharge current.
You have problem with charging the bootstrap capacitor cause its charging thru 2l at very high frequency he will never charge for needed value.