Electromagnet, and unanswered questions

  • Thread starter nikunter
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  • #1
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Hello All, My name is Nik. I am very glad I found this forum. For the past couple months i have been researching (YES, I do use the search function) information on electromagnets. I have done a lot of reading on wiki so I am familiar with its E.M. page.

I am building an electromagnet. Like others I am curious and have questions/concerns.
I am trying not to repost old questions but some answers just didnt do me justice.

I am trying to build an E.M. to spec. The only goal I have is it needs to lift 500 - 1000 lbs.
My power source will either be 12v car batteries, or eventually LiPo / LiIon batteries.
I will run them in series to maximize battery life.

The only constraints I have are the size and height of the iron core. It can have a diameter no larger than 4 - 5", and a height no larger than 4"

The problem I am having is what wire to choose. I know that a copper magnet wire has a resistance, and so do the batteries. I know it can only let through so much current.



Are there any exmaples of E.M.'s I can actually learn from? I am tired of searching and finding the same, grade 2 level, iron nail and wire with a tiny battery. I understand the theory but now i need some variables to plug in.

so to sum this up:
- What gauge wire (I know if you add length you add R, add area and you reduce the R)
- what will it take for a power supply

Also, Where is the EMF the strongest? is it along the sides? or at the base?

any help is appreciated. Some of you can understand when you're passionate about learning something you spend many nights awake. Thank-you very much. Helping me would reduce the amount of years the stress is taking from my life haha.
 
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Answers and Replies

  • #2
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ok, ok.. I'll read some more.

If anything, does anyone know of a good resource site for magnet wire.
 
  • #3
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Enamelled copper wire is used by motor rewinders and transformer manufacturers, and is in plentiful supply although it is certainly not cheap.
As for models from which you could learn, any industrial steel foundry which uses scrap steel will probably use electromagnets like the one you have in mind to move their steel around. They'll often put such a magnet on a crane, in place of a hook, and pick up tons of steel at a time.
 
  • #4
uart
Science Advisor
2,797
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The problem I am having is what wire to choose. I know that a copper magnet wire has a resistance, and so do the batteries. I know it can only let through so much current.

Hi nikunter. typically with coil design you have some maximum amount of cross-section available for copper and the goal is to obtain a given number of ampere-turns (NI). Your overall lifting power (force) will in fact be proportional to (NI)^2 though the exact proportionality factor will depend on the core and load geometry and permeability, so to some extent you may need to experiment to find just what NI you require.

Now here's the interesting thing about developing a certain NI for a given voltage source and available copper area. If you double the number of turns then you double the length of copper but you also halve the cross-sectional area of each turn (due to constrained overall copper cross section). So taken together this implies a 4 times increase in resistance (and decrease in current) for a 2 times increase in turns. So (perhaps counter intuitively) you actually get less ampere-turns the more turns of copper that you use (assuming that you still use all available copper area). You do however draw much less current from the battery if you use more turns.

Perhaps you could start by choosing a current level that is suitable for your battery and an overall winding cross section that is appropriate for your operating conditions and then work out the maximum number of ampere turns that you can fit into that space @ the given battery current drain.

The relevant formula is : [itex]R = \rho L/A[/itex], where "L" is the wire length (meters), "A" is the wire cross section (m^2) and [itex]\rho[/itex] is approx 2E-8 for copper.
 
  • #5
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so, I have to ask now,
How does this apply to a coil that you would pass the magnet across for power generation?
Does lower resistance equal more current out for the same magnet/speed?
(and my computer does not see the formulas in the black )

dr
 
  • #6
3
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Hi nikunter. typically with coil design you have some maximum amount of cross-section available for copper and the goal is to obtain a given number of ampere-turns (NI). Your overall lifting power (force) will in fact be proportional to (NI)^2 though the exact proportionality factor will depend on the core and load geometry and permeability, so to some extent you may need to experiment to find just what NI you require.

Now here's the interesting thing about developing a certain NI for a given voltage source and available copper area. If you double the number of turns then you double the length of copper but you also halve the cross-sectional area of each turn (due to constrained overall copper cross section). So taken together this implies a 4 times increase in resistance (and decrease in current) for a 2 times increase in turns. So (perhaps counter intuitively) you actually get less ampere-turns the more turns of copper that you use (assuming that you still use all available copper area). You do however draw much less current from the battery if you use more turns.

Perhaps you could start by choosing a current level that is suitable for your battery and an overall winding cross section that is appropriate for your operating conditions and then work out the maximum number of ampere turns that you can fit into that space @ the given battery current drain.

The relevant formula is : [itex]R = \rho L/A[/itex], where "L" is the wire length (meters), "A" is the wire cross section (m^2) and [itex]\rho[/itex] is approx 2E-8 for copper.


Thank-you very much!
You've basically answered everything I've been looking for. But of coarse learning new things means more questions. I'll just ask one though,

[itex]\rho[/itex] is approx 2E-8 for copper.

2E-8, what is that? E=? and why -8
and if its the certain resistance per meter or something, doesnt that change with each gauge?
I can't find any info on gauge resistance online. The page I did find was for regular copper wire where as I am looking for the magnetic stuff.
 
  • #7
4,662
5
What is the shape of the 500 pound object you have to pick up? Does it have a flat surface bigger than the size of the magnet? Do you plan to machine the magnet iron core out of low carbon steel, or build it out of stock cold-roll? I suggest you think about using ~50 amps for 10 seconds-this will help determine the copper conductor size. It looks like you want ~400 amp-turns per meter. See thumbnail.

The simplest shape is a horeshoe electromagnet, , but it may not be the best.

Bob S
 

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  • #8
uart
Science Advisor
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2E-8 is shorthand for 2 times 10^(-8) or 0.00000002.

If you prefer then use mm^2 (mm times mm) as your units for "A" (still use meters for L) and then you can use just 0.02 for [itex]\rho[/itex]. BTW. It's called resistivity and is a function of the material, this is the approx value for typical copper wire.
 
  • #9
uart
Science Advisor
2,797
21
and if its the certain resistance per meter or something, doesnt that change with each gauge?

No, that's the beauty of multiplying by "L" and dividing by "A", it works for any size wire. :)
 
  • #10
4,662
5
Thank-you very much!
You've basically answered everything I've been looking for. But of coarse learning new things means more questions. I'll just ask one though,



2E-8, what is that? E=? and why -8
and if its the certain resistance per meter or something, doesnt that change with each gauge?
I can't find any info on gauge resistance online. The page I did find was for regular copper wire where as I am looking for the magnetic stuff.
The copper wire resistance (ohms per 1000 feet) is about 0.1 ohms for 0 (ought) gauge, 1 ohm for 10 Ga, 10 for 20 Ga, 100 for 30 Ga, 1000 for 40 Ga. It is a logarithmic scale, easy to remember.
Bob S
 

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