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Is it possible to produce over 1000 lb of magnetic force over a distance of 4 inches(0.101 meters), using an electromagnet?

How difficult would it be? Explaining the general spec's of such an electromagnet? Most importantly,the power input... how much power would this electromagnet need? A few kWatts?

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  • \$\begingroup\$ But I think you can find your answer in internet(by googling) and specially Wikipedia. en.wikipedia.org/wiki/Electromagnet \$\endgroup\$
    – Roh
    Commented Feb 27, 2014 at 6:14
  • \$\begingroup\$ Use an N53 (or close) Neodumium magnet as the item attracted and you'll greatly ease your task. Extending a "rule of thumb" from elsewhere - With an iron core you'll get approx field near pole faces out to about half core depth. / With a neodymium magnet this applies for ~~~ half magnet thickness. / There are various simulation packages (some free) that allow you to investigate this. / More data needed: What are you trying to do? \$\endgroup\$
    – Russell McMahon ♦
    Commented Feb 27, 2014 at 8:29
  • \$\begingroup\$ See daycounter.com/Calculators/Magnets/… and easycalculation.com/engineering/electrical/solenoid-force.php \$\endgroup\$
    – Russell McMahon ♦
    Commented Feb 27, 2014 at 8:40
  • \$\begingroup\$ Scrap yards pick up old cars with magnets that strong. \$\endgroup\$
    – Optionparty
    Commented Feb 27, 2014 at 14:36
  • \$\begingroup\$ @Key You need an electromagnet to produce a magnetizing force about 200,000 T/m in the required gap. So an electromagnet maden -let's say- by an iron core of 0,5m diameter and 0,5m length and 500 turns, can do the job. The current you need is about 38A. That means you can use a magnet wire AWG 10 in a current density of 800A/cm^2 to keep the temperature under 80 degr.Celsc. If your DC voltage supply is 48V then the required power will be about 1.85kW. This electromagnet will have a final holding force of about 2.7 tones! \$\endgroup\$
    – GR Tech
    Commented Nov 12, 2014 at 18:03

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Here is a formula that is useful

Force = \$(N\cdot I)^2\cdot 4\pi 10^{-7}\cdot \dfrac{A}{2g^2}\$

  • F = Force
  • I = Current
  • N = Number of turns
  • g = Length of the gap between the solenoid and the magnetizable metal
  • A = Area

With 100 amps through a 100 turn coil of area 1 sq m, the force on a magnetizable metal at 100mm is 6283 newtons (about 1400 pounds).

But what about the permeability of the electromagnet's core and the permeability of the "iron" it's attracting to - air forms the gap and it dictates the flux density.

The input power needed to do this is zero because no work is being done in creating this force. However, if you actually want to know the losses you need to think about how thick the copper wire is to pass this current and maybe trade off number of turns with amps.

Here's an online calculator and here is a page that explains the theory.

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  • \$\begingroup\$ Wait, how is the input power = 0 and why isn't work done? I'm confused. And can this be true, I thought it would very difficult to produced such forces over large distances! From this formula it seems doable. + Still trying to understand the 2nd link. \$\endgroup\$
    – Pupil
    Commented Mar 1, 2014 at 12:05
  • \$\begingroup\$ Work is measured in watts. No work has been done unless you move something hence no power but, you've got resistive losses that lose power as heat and there is energy in the magnetic field you have created that will get turned to heat when you remove excitation from the coil. \$\endgroup\$
    – Andy aka
    Commented Mar 1, 2014 at 14:11
  • \$\begingroup\$ Ah, I forgot to state this. But the main purpose of such an electromagnet is to pull a ferromagnetic object with that rate of force over a distance r. So X amount of Watts is used to do that work. I assume nothing would change correct? W = F(of magnetic attraction) x r(gap)? \$\endgroup\$
    – Pupil
    Commented Mar 2, 2014 at 1:17
  • \$\begingroup\$ In addition, in order to produce such large forces... work is done in watts to generate a magnetic field B? That is then dissipated into heat? \$\endgroup\$
    – Pupil
    Commented Mar 2, 2014 at 4:29
  • \$\begingroup\$ aka the energy stored in a magnetic field I.e. half L x I squared. \$\endgroup\$
    – Andy aka
    Commented Mar 2, 2014 at 11:20

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