I'm trying to create a permanent magnet. I have a steel rod. I wrap AWG
6 around the rod twice (so I got a two layer coil) and directly connect it to a 12V DC battery for like a tractor or snowmobile (235 cold cranking amps).Very little happens. A max of 5A comes out and I can't even make an electromagnet.
Can anyone tell me what I can do to get an extremely high amperage pulse? Are there any power supplies that will generate a high amp current? I'm thinking like 80 amps at 12V.
Wire sufficient FGCs in parallel to give you 1 Farad.
Charge to 1kV
Discharge through 10 turn 6" diameter coil made from 1/4" copper rod.
Important .... just before you discharge, phone for the ambulance because the ignorance that you display is certain to kill you during the process.
(Wasn't his advice given here recently?)
--- Well, let's see...
#6 AWG wire has a diameter of about 0.167" and a resistance of about
0.395 ohms per 1000', so if we assume that you've got a 12" long steel rod with a diameter of 1", then you'll be able to get: 12" length n = ------------------ = 71.86 ~ 71 turns 0.167" diameter on it.The steel rod has a circumference of:
C = pi D = 3.14* 1" = 3.14 inches,
so the length of the first layer will be
L1 = C n = 3.14" * 71 turns = 222.94" ~ 223"
Since the wire has a diameter of 0.167", the second layer will be wound on the diameter of the rod plus twice the diameter of the wire:
D2 = 1" + 0.167" + 0.167" = 1.334"
and its length will be:
L2 = pi D2 n = 3.14 * 1.334" = 297.4" ~ 297"
So, the total length of the winding is:
L = L1 + L2 = 223" + 297" = 520"
Since the wire has a resistance of 0.395 ohms per 1000 feet, that's
0.000395 ohms per foot, and the resistance of the winding would be: 0.000395R R = ----------- * 43.33 = 0.017 ohms. 1'Assuming your battery's internal resistance is about equal to the winding's resistance means that you'll have 12V across 0.034 ohms,
and the current into the coil will be:
E 12V I = --- = -------- ~ 353 amperes R 0.034R
Since you're only getting 5A into the coil I'd suggest that your battery isn't fully charged, that it's bad, or that whatever's between the battery and the coil is limiting the current.
BTW, if you do get 353A into the coil, it'll be dissipating:
P = IE = 353A * 6V = 2118 watts
which is about 50 watts per foot, so it'll get very hot very quickly.
The battery will also be dissipating the same amount of power, so you'll need to be very, very careful when you're experimenting with it and if you ever get it working right.
JF
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.