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The specs:

  • The electromagnet has 7,2 cm of height, 1 cm of radius.
  • A hole with 2,5 mm of radius in the middle.
  • An aluminium foil has 0,024 mm of height so I assumed the insulator layer has the same height (0,024 mm) and there is some leftover space between the foils (since there is a little bit of air between them) with 0,024 mm for the sake of simplicity, rounding the number to an arbitrary 0,072 mm of layer height.
  • If we divide 7,2 centimetres by 0,072 mm, it could be said this has 100 turns/layers (I think.)

Maybe if the insulation layer is just enamel, the height could be reduced to just 0,049 layer height and give something around 150 layers and double the strength (I think.)

The problem:

I tried some online solenoid strength calculators, but every electromagnet force calculator gives me a different number, so I just want to be sure I'm not just calculating things incorrectly.

Also, most electromagnet calculators base themselves on conventional electromagnets with wires, unlike the Bitter configuration as I'm trying to build. Not to mention that this kind of electromagnet allows for a higher amperage because of the cooling holes.

A Bitter electromagnet is a "simple", but very powerful configuration of electromagnet arrangement. Basically, it is "just" a "hamburger" cylinder made of sheets of conductive metal one above other separated by an insulator with a "hole" on it so the current can flow through the layers.

illustration showing the current flow and arrangement in a Bitter electromagnet

For the sake of simplicity, let's just imagine it is a continuous flat spiral.

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    \$\begingroup\$ The Bitter configuration has current density varying with radius, most magnet calculators won't take that into account. You may be able to kludge that by calculating for several concentric hollow solenoids. Your insulation thickness is far too generous, fill it with as much metal as you can, both for physical strength and adiabatic heat capacity. \$\endgroup\$
    – Neil_UK
    Commented Nov 15, 2022 at 12:34
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    \$\begingroup\$ Aluminum is considerably less conductive than copper... not a wise choice for high-current applications (too much resistive heating.) For high-current applications, choose a silver-plated copper foil for highest conductivity (lowest resistive loss.) \$\endgroup\$
    – rdtsc
    Commented Nov 15, 2022 at 12:46

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