I’d have trouble faking this with a straight face:
That’s measured with the 56 turn winding connected directly to a bench power supply, cranking up the current, taking the reading, and turning the current back down again, so as to avoid cooking the poor thing inside its PLA armor:
The “49E” sensor came from one of the bags of eBay fallout. They saturate around 4.25 V; the outputs above 4 V lose their linearity due to the sensor, not ferrite saturation.
calculations guesstimates suggested 25 turns would produce full scale at 5 A, so 56 turns should top out at 2.2 A. Frankly, given all the imponderables in this lashup, a factor of two seems pretty close.
Offsetting the output by -1 A would yield a 2 A range that’s just about exactly right. Unfortunately, some fiddling about with neodymium magnets suggests that you (well, I) can’t stuff enough opposing field into the slit without saturating (some part of) the ferrite core, reducing the permeability, and blowing all the assumptions.
So that suggests a buck winding, obviously with more turns to allow less current for the same magnetizing force. Wrapping 110 turns reduces the buck current to 500 mA and assuming a bit over an inch/turn requires 10 feet, which is nearly 1 Ω of 30 AWG wire: the buck current dumps another 250 mW into (a somewhat larger version of) that PLA armor.
Or just throw away half of the Hall effect sensor range and use an op amp along the lines of the LED current sensor.