Electronics Forum / Electronics / June 2008

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For the current:

#20 AWG weighs about 3lb. per 1000', so 5lb. would be:

    3lb       5lb
  ------- = ------- = 1667 feet long
   1000'       x'

A 12" diameter loop has a circumference of

   C = pi * D ~ 37.7"

So the loop has:

         1667' * 12"
    N = ------------ ~ 530 turns
           37.7"  

on it.

Now, go to:

http://www.technick.net/public/code/cp_dpage.php?aiocp_dp=util_inductance_circle

and fill in the boxes:

N  =  530      turns
R  =  0.152    meter (radius of circle)
a  =  0.000081 meter (radius of wire)
µr =  1              (relative permeability of air)

Click on "calculate" and 0.3216537332062735 will come up, so your coil
has an inductance of about 322 millihenries.


At 60 Hz it'll have a reactance of:

    Xl = 2 pi f L = 6.28 * 60Hz * 0.322H ~ 121.3 ohms

and, since it has a resistance of 25 ohms, it'll have an impedance of:

    Z = sqrt (R² + Zl²) = sqrt (625 + 14714) ~ 124 ohms.

The current through it then, will be:

         E     120V
    I = --- = ------ = 0.967 amperes.
         Z     124R

Since the reactive part of the impedance is lossless, the slick part
is that the coil will only be dissipating power in its resistance, and
with 967 milliamperes going through its 25 ohms, that comes out to:

    P = I²R = 0.967A² * 25R ~ 23 watts.

So, while it'll warm up somewhat, I'd bet that it wouldn't get
can't-keep-my-finger-on-it hot no matter how long you left it
connected to the mains.
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Seems like a shame... ;)

JF