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Magnet Coil Turns and wire calculation 
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#1
Sep2613, 05:26 AM

P: 5

Hello Friends
I am working on home made electromagnet and I have only volt & Ohms details like 12Volt DC, 24Volt DC & 240V AC and ohms is 80Ω, 100Ω & 120Ω for all for example 12VDC 80 ohms coil, 12VDC 100ohms,24VDC 80Ohms coil...etc. now i want to know which copper gauge we have to choose to create that magnet coil and how many turns i have to take for proper magnet.. is there any equation to calculate to choose...wire and turns I want to also know the ampere capacity of that magnet coil. Thanks you very much.. 


#2
Sep2613, 09:07 AM

Sci Advisor
Thanks
P: 1,924

Do you know the cross section area and length of the winding ?
Take a look at a wire table. See; http://en.wikipedia.org/wiki/America...AWG_wire_sizes The diameter, resistance per length and current in amps need to be juggled to limit the current while maximising the number of ampere*turns for the voltage you select. 


#3
Sep2613, 01:46 PM

P: 1,212

I think a warning needs to be said here, because from the sounds of your post you're considering attaching a copper wire to the mains power without a good understanding of the basic electronic circuit elements...
be careful! 


#4
Sep2713, 03:37 AM

P: 5

Magnet Coil Turns and wire calculation
Dear thank you for your replay i don't have much knowledge for the magnet coil
I want to know how to calculate if you tell me in detail with example ... Thanks to All, I also got some example on below...may be it will help us dear For DC, we may be able to do some simplified experiments to give us some interesting results. So lets say you have a 10 ohm, 600 turn coil. For DC, we'll simplify the current as a resistive calculation of I=V/R = 50/10 = 5 amps. Call this COIL 1: i=5,N=600,A=1,g=.5 (A and g explained below) Now remove 100 turns, lets assume the resistance is reduced linearly. 500/600 * 10 Ohms = 8 1/3 Ohms, and I = 50/8.33 = 6 amps So COIL 2: i=6,N=500,A=1,g=.5 Here's where it gets interesting. Go to this site for a simplified estimation of the coil force, and plug in both sets of numbers, and compare the resulting coil force for each coil. The A and g terms are related to the geometry of the coil, and you can use appropriate values for a pinball coil. http://www.daycounter.com/Calculator...lculator.phtml You'll find that the force is identical for both coils. So by removing winds, we're increasing the current needed to produce the *same* force. CURRENT GOES UP, WINDINGS GO DOWN, FORCE STAYS THE SAME! Essentially, we are making the coil LESS EFFICIENT by removing winds, and achieving no gain in force. The real case may differ from this simplified version somewhat, but this experiment shows that removing windings is a bad idea. We're only increasing the amount of current the coil is going to use, increasing the stress on the driver circuit, and getting no gain in force. 


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