I've been wondering and testing selfmade magnets by wrapping iron with wires and applying a voltage to it which makes it a bit magnetic. I've been looking around the internet and can't seem to find formula's or calculation examples on this subject. If I have wires of example 0.22mm² and a voltage of 3.3V, 2Amps, 25 turns, how big must the iron core be to lift a weight of example 5 grams?


How much voltage and current do I have to apply when I have a constant value of the thickness of the copper wire, windings and iron core to lift 5gr.

Also, if I halve the amount of windings but double the thickness of the copperwire, is that the same?

Apologies in advance for possible duplicates of this matter, I've tried to use this http://onlinecalculators.brainmeasures.com/Electric/SolenoidCoil.aspx

When I fill in:

Current = 2A
Area = 0.00439m2 = Cilinder with a diameter of 20mm and a length of 60mm
Number of turns = 25 turns
length of the gap = 0.001m = 1mm gap (I intent to touch the magnet if the intented object)
Results = 0.069N = 7 grams.

However, the calculator doesn't seem to care about the coil thickness/wire diameter which I doubt the results of it capabilty of lifting that amount of weight.

  • \$\begingroup\$ Very related: electronics.stackexchange.com/questions/101127/…. Wire thickness has nothing to do with it providing a certain current can be passed reliably. \$\endgroup\$
    – Andy aka
    Commented Oct 5, 2018 at 14:01
  • \$\begingroup\$ The search term for this sort of magnet is "solenoid" or "solenoid magnet". If you can figure out how to refine your search so you're not getting a flood of car parts, you may find useful information. \$\endgroup\$
    – TimWescott
    Commented Oct 5, 2018 at 14:25

2 Answers 2


The the pull force depends only on the amp-turn product and the relative permeability of the core.

Having thicker wire only allows you to put more current in the wire.

Putting more current in the wire will create more force, but only to a point.

After the core saturates then you won't get much more force by adding current or turns.

As for coil voltage, it depends on the coil resistance.

EXAMPLE 1) For example, suppose you had a core that was 10mm diameter and 30mm long.
Lets say you have some 30AWG magent wire that is 0.3mm diameter.

30mm / 0.3mm = 100 turns

So you could fit 100 turns of 30 AWG wire in that space.

Packed in a coil you might run 100mA through the wire.

So your total Amp-Turns is 100mA * 100 = 10 Amp-Turns

The length of the wire is (pi * 10mm)/turn * 100 turns = 3140 mm.

30 AWG wire has a resistance of 338 milli-ohms per meter.

So the total coil resistance is 3140mm * 338 milli-ohms per meter = 1.22 ohms.

So the coil voltage would be 1.22 ohms * 100mA = 0.122V.

EXAMPLE 2) Suppose you had the same core, but were using 40 AWG magnet wire that has a diameter of 0.08 mm.

30mm / 0.08mm = 375 turns

Packed in a coil you might run 26mA through the wire.

So your total Amp-Turns is 26mA * 375 = 9.75 Amp-Turns.

The length of the wire is (pi * 10mm)/turn * 375 turns = 11,780 mm.

30 AWG wire has a resistance of 3.44 ohms per meter.

So the total coil resistance is 3140mm * 3.44 ohms per meter = 40.5 ohms.

So the coil voltage would be 40.5 ohms * 26mA = 1.053V.

Note that in both examples size of the core was the same and the Amp-Turns about the same (10.0 vs 9.75), so you would get about the same pulling force. But the coil voltage was different bcause one coil resistance was different.

As for the exact pulling force it is highly dependent on geometry and your core material. You may just need to experiment on that one.

Note that in this case the 30AWG wire used 100mA * 100mA * 1.22 ohms = 12.2mW of power.

The 40 AWG wire used 26mA * 26mA * 40.5 ohms = 27mW.

So in this case the 40AWG wire was less efficnet. The reason is that there was less total volume of copper (since the layer of 40AWG wire was thinner).
If we used four layers of 40AWG wire the height of the four layers would total to 0.32mm (30AWG was 0.3mm).

Making an approximation (since the wire diameter is small compared to core diameter)
For four layers of 40 AWG wire we have 375 Amp-Turns * 4 = 1500 Amp-Turns.

The coil resistance would be 40.5 ohms * 4 = 162 ohms.

To get 10 Amp-Turns we would need 10 Amp-Turns / 1500 Turns = 6.7mA.

The coil voltage would then be 6.7mA * 162 ohms = 1.085V

So clearly for a given wire size and Amp-Turns the coil voltage did not change when we increased the number of turns.

6.7mA * 6.7mA * 162 = 7.3mW.

So now the 40AWG wire is more efficient than the 30 AWG wire because we increased the volume of copper in the 40 AWG case.


All that matters for the pull performance of the solenoid is the amp-turn product of the winding and that the coil is well coupled to the core. The wire diameter is important only in that it determines the resistance of the coil, and hence the power dissipated by passing the current. If you use very heavy wire, you may be able to generate the current needed at a fraction of a volt dropped across it. If you're using a constant voltage supply, you'll want to size the wire to have a resistance that gives an appropriate current. If you use a constant current supply it won't matter.

PS Recheck you numbers you put into that calculator. Your area calculation is wrong.


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