Electromagnet that can generate 100 lbs at small size

  • #1
member 662135

Summary:

I wondering if it is possible to create an electromagnet that has a diameter of 0.5 cm that can generate 75 - 100 lbs of pull force.
I wondering if it is possible to create an electromagnet that has a diameter of 0.5 cm that can generate 75 - 100 lbs of pull force. I took physics in college so I have a basic understanding of how magnets work, but I don't have any practical knowledge. I don't know what is feasible.

I found an equation that describes what properties of a magnet need to be adjusted to accommodate for a desired strength and size:

F = (n x I)2 x mu x A/2g2
  • F = force = 430 N
  • I = current = 25 amps
  • g = length of the gap between the solenoid and a piece of metal = 2 cm
  • A = area = 7 mm2
  • n = number of turns = 250
  • mu = vacuum permeability = 4 x pi x 10-7
I understand that increasing the strength of the magnet is about increasing the number of turns and the current. I know 25 amps is a lot of current, but as far as I know it is reasonable. A thicker wire would be needed to handle that current, which might make 250 turns difficult. Is it feasible to do? Does this equation carry over well into reality, or do other things need to be taken into account?

Thanks
 

Answers and Replies

  • #2
davenn
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...........A thicker wire would be needed to handle that current, which might make 250 turns difficult.

impossible for anything under 20 cm maybe bigger

.......... Is it feasible to do?
not in the 0.5cm you want


....................Does this equation carry over well into reality,

yes, but not is the dimensions you want
 
  • #3
jrmichler
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I understand that increasing the strength of the magnet is about increasing the number of turns and the current. I know 25 amps is a lot of current, but as far as I know it is reasonable. A thicker wire would be needed to handle that current, which might make 250 turns difficult.
You can get your 25 amps through wire of any size. Just choose the largest wire that fits the space available, calculate the total length of the wire, and calculate the resistance. Multiply the resistance by your 25 amps to get the total heat generated, then divide by the thermal mass of the copper to get the rate of temperature rise. From the rate of temperature rise, and the maximum temperature rating of the wire insulation, calculate the time to reach that maximum temperature from the starting temperature. That will also tell you the necessary voltage to drive the electromagnet.

Given the small size of your electromagnet, do not be surprised if the time is less than one second. Because the time will be short, you can ignore the thermal mass of any core or supporting structure, and ignore heat transfer to surroundings.

When driving small coils with more than a few tens of volts, you need to consider voltage between adjacent turns. If the first turn is laying on the last turn, the full voltage is across the wire insulation. At some point, the voltage will arc through the wire insulation and short circuit the entire coil.

Do the calculations. I'd like to see the results.
 
  • #4
member 662135
You can get your 25 amps through wire of any size. Just choose the largest wire that fits the space available, calculate the total length of the wire, and calculate the resistance. Multiply the resistance by your 25 amps to get the total heat generated, then divide by the thermal mass of the copper to get the rate of temperature rise. From the rate of temperature rise, and the maximum temperature rating of the wire insulation, calculate the time to reach that maximum temperature from the starting temperature. That will also tell you the necessary voltage to drive the electromagnet.
I started the calculations but I don't have the time to continue looking for the properties of copper and wire insulation. I understand what you illustrating to me though.

Given the small size of your electromagnet, do not be surprised if the time is less than one second. Because the time will be short, you can ignore the thermal mass of any core or supporting structure, and ignore heat transfer to surroundings.
So basically, it's possible to make a magnet with these properties, but it will only last maybe one second? Let's drop the force of the magnet to 40 pounds and increase the diameter of the magnet's face to 1 cm. Given access to industrial materials and resources, do you think a magnet like this could be created that could run without any risk overheating (or any other technical issues i.e. a functional, long-lasting magnet)?
 
  • #5
member 662135
Let's drop the force of the magnet to 40 pounds and increase the diameter of the magnet's face to 1 cm. Given access to industrial materials and resources, do you think a magnet like this could be created that could run without any risk overheating (or any other technical issues i.e. a functional, long-lasting magnet)?
 
  • #6
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Are you talking 'pole size' or overall size ? And what length limit ??

Your project sounds like a variation of the old 'turning a long, strong nail into a simple electromagnet'...
 
  • #7
member 662135
Are you talking 'pole size' or overall size ?
What is pole size?

And what length limit ??
Do you mean how long is the magnet? It doesn't matter, I suppose 3 feet sounds reasonable.


Your project sounds like a variation of the old 'turning a long, strong nail into a simple electromagnet'...
Yea, kind of, except let's assume that we have access to whatever resources or materials we need. I'm just wondering if it's possible.
 
  • #8
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What is pole size?
The diameter of the core that contacts the job...

A tapered pole may be able to concentrate flux to the size you need.
 
  • #9
berkeman
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increase the diameter of the magnet's face to 1 cm.
I didn't read the whole thread in detail, but you understand that you need both the N and S pole faces of the electromagnet to be in contact with the flat face of the ferrous object you want to lift, right? Otherwise the lifting force will be greatly reduced.

So you need to think about how you are going to have both the N and S pole faces of the magnet be coplanar. The usual solutions are to use a Horseshoe style magnet, or a Cylindrical magnet...

https://int.frederiksen.eu/admin/pu...Files/Images/ecom/products/_dummy-product.png
243747


https://images-na.ssl-images-amazon.com/images/I/412HbOdxvDL._SX342_.jpg
243748
 
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  • #10
jrmichler
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It was pointed out in a PM that post #3 has an error:
****************
Greetings, I think you mean to say "multiply the resistance by the square of 25 amps" to get the power dissipated?
****************
He's right.
 
  • #11
SiD
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Hello...
I Had use the same formula to calculate required current.
Specifications:-
F=250 N
N=2500
A=0.00283 m^2 (The magnet acts as a close cylinder A=2πrl+2π r^20 where r=10 mm core dia. and l=85 mm length of winding)
g= 0.02 m
mu = vacuum permeability = 4 x pi x 10-7

After putting all the values in above formula i got the value of current = 3 A.(value of current use here is 5 A).
The resistance of winding= 1 kg ohm.

But when i made the practical with magnet the result were quit different.It was hardly capable to lift 9.81 N.Can you make suggestion what changes i have to make to get required power of magnet. I want to extend the tip of magnet minimum 1.5 cm from the winding.
 
  • #12
SiD
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Can any one suggest value of voltage to be supply
 
  • #13
sophiecentaur
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an electromagnet that has a diameter of 0.5 cm that can generate 75 - 100 lbs of pull force.
What separation distance do you need? The effect of only a small gap or unevenness can be drastic. The force that will 'hold' two faces together will be much much more than the force needed to attract the faces together in the first place. Take the example of a gas safety valve which, with a tiny voltage from a thermocouple in the pilot flame, can hold the valve open but will not pull a closed valve open - you need to provide the initial force.
 
  • #14
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Does this equation carry over well into reality
The equation is correct, but you need to consider the applicable conditions behind it.
 
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