Re: DIY electromagnet under £100: Disappointing results

On Oct 2, 4:52 pm, John Larkin
<jjlar...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
On Fri, 02 Oct 2009 15:23:09 -0500, j...@xxxxxxxxxxxxx wrote:
On Fri, 02 Oct 2009 15:06:00 -0500, amdx wrote:

<j...@xxxxxxxxxxxxx> wrote in message
news:mMednbsMZM371lvXnZ2dnUVZ_rWdnZ2d@xxxxxxxxxxxxxx
On Fri, 02 Oct 2009 11:34:27 -0700, John Larkin wrote:
On Fri, 02 Oct 2009 13:07:51 -0500, John Fields wrote:
snip
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.

Yes by mistake I left out a factor of four due to length
400 turns would be about 4 times as long so that factor and
the factor of 4 from half-diameter would give .64 ohms
resistance rather than .16 ohms so now the calculation is
12.5A*12.5A*0.64ohms = 100W and 5000 ampere turns which is
the same as with #10 wire.

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- Hide quoted text -

- Show quoted text -

Yup!

I used an old Varian electromagnet with water cooling. Closed loop
'clean' water supply, the same water cooled a hugh bank of pass
transitors on the back of the power supply. (circa 1980's)

George H.
.

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