Re: DIY electromagnet under £100: Disappointing results

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>>>and with 39 feet of wire in it, since #10 has a resistance of about 1 >>>milliohm per foot, it'll have a resistance of 39 milliohms. >>> >>>Now, let's say that this thing can dissipate 100 watts, continuously. >>> >>>Since: >>> >>> P = I²R >>> >>>then: >>> P 100W >>> I = sqrt --- = sqrt --------- = 50.64 ~ 50 amperes, >>> R 3.9e-2R >>> >>>the voltage across the coil would be, >>> >>> E = IR ~ 50A * 3.9e-2R ~ 2 volts >>> >>>and its IT would be 50 amperes * 100 turns = 5000 ampere - turns. >>> >>>Google "ampere-turns" for some rather more detailed info. >> >> If an electromagnet is thermally limited, and you plan to fill the >> available winding window, I think it doesn't matter what size wire you >> use. So size the wire to match whatever power supply you have, to get as >> many watts into the coil as it can stand. >> >> It's like a transformer. A 100 VA transformer is the same size whether >> it has a 120 volt primary, a 240 volt primary, or a dual primary. > > A persuasive argument but would it not be the case that if > you wound with wire half the diameter of #10 say #16 that > there would be 400 turns with .16 ohms resistance (both four > times as much as with #10) but half as much current for the same > power level since 25*25*.16 = 100 so there are 10000 ampere turns > which is twice as much as before.

Prove to me that 400 turns of #16 is the same length of wire as 100 turns of #10 wire. Assume a 1" diameter core and 2" length. I've looked for an online calculator, no luck so far. Mike

Reply to
amdx
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It all boils down to current density per unit cross-sectional area of the available window. A 1 cm square area, solid copper, might support

100 amps thermally. Slice it into four sections using thin insulators, 0.25 cm^2 each, and you've changed nothing, except that you have four times as many "turns". It's just as if you had four windings and could decide to run them in series or in parallel. Each winding doesn't know if it's being used in series or in parallel; it contributes the same flux for the same ohmic power loss.

Water cooling helps here.

John

Reply to
John Larkin

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