In videos like this one by electroboom, people make electromagnets with itty bitty wires. It seems the intuition is more turns, greater magnetic field. But I'm not so sure about that.

Suppose we have some annealed copper wire at 20°C. (I used this site to calculate the resistances). With wire at 1/8th of an inch and 1 foot long, the resistance is 0.6583 mΩ. If we halve the diameter, then the cross sectional area is divided by 4, the length increases by 4 times (quadrupling the number of turns), and the resistance increases 4 times per foot and 16 times over all, and is now 10.533 mΩ.

The magnetic field near a wire is given by \$B=\mu_0I/2\pi R\$ (where R is the distance from the wire, not the resistance.) In the case with 1/8th inch wire, the current will be 16 times larger, but will have 4 times fewer turns. The end result is that the magnetic field is 4 times greater with the thicker wire.

So why do people use thin wire? Is it always best to maximize number of turns and minimize wire diameter (within other engineering constraints)? Is it always best to maximize wire diameter and minimize turns? Or is this a case where we're trying to match the resistance in the inductor to something else (such as the internal resistance of the power supply, or the resistance in the rest of the circuit), and if so, what? What design considerations dictate the wire diameter?

  • \$\begingroup\$ Concentrating it in a small a space as possible also matters. \$\endgroup\$
    – DKNguyen
    Mar 2 at 19:03
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    \$\begingroup\$ only low power or high voltage electromagnetic components have really thin wires. If you look at high power magnetic components designed for cars (12V), like starter motors, choke solenoids, alternators, etc., you'll find that their winding wires are relatively thick. \$\endgroup\$ Mar 4 at 0:29
  • \$\begingroup\$ It depends on the Wattage of the magnet (or rather, how many Amps it is expected to operate at). The more current you expect to flow through your magnet the thicker the wires need to be to avoid melting the wires. Otherwise you only need to use the bare minimum thickness that avoids melting so you chose the lightest wire you can for the amount of current you expect. If you look at high speed, high power brushles motors (ones operating at 10kW or more) you will find thicker wires (though still very thin) than what you'll see on smaller motors. \$\endgroup\$
    – slebetman
    Mar 4 at 3:02

5 Answers 5


Regardless of the voltage it is designed for, a solenoid of a given strength wound with a similar fill of the bobbin will tend to dissipate the same power all other things being equal.

So if the voltage is doubled, the resistance must be increased by a factor of four. With 1/2 the current, you will need 2x the number of turns to get the same flux so the wire must be half the cross-sectional area or 0.71 times the diameter to fit in the available space.

Here is a typical commercially available solenoid that we can assume was designed optimally.

enter image description here

That relationship may break down at very large wire diameters because of the wire geometry (rectangular wire cross sections can be used) or at very small wire diameters because of insulation. The so called 'fill factor' is a measure of this- how much of the area of the winding window is actually filled with conductor (and not insulation or space between conductors).

Given that you have to supply (in the above case) 1.3W to the solenoid, would you prefer 180V at 7.2mA, 12V at 110mA or 0.8V at 1.6A? Each might have a valid use case. The wire will be a lot thinner in the first case so it might cost more to buy the wire, to wind it and it might be less reliable especially in an environment that is less than benign. Such as the 50\$\mu\$m wire used to wind a coil in my car's key fob.

  • 8
    \$\begingroup\$ Such as the 50μm wire used to wind a coil in my car's key fob. I sense some exasperation here... \$\endgroup\$ Mar 2 at 18:34
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    \$\begingroup\$ Your senses are working well. 8-\ \$\endgroup\$ Mar 2 at 18:43
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    \$\begingroup\$ @AndrewMorton I learned a new word today, exasperation. \$\endgroup\$
    – winny
    Mar 2 at 19:21
  • \$\begingroup\$ So I see how to go from one optimally designed electromagnet at one voltage to another optimally designed electromagnet at a different voltage (double the voltage, quadruple the resistance). But what if I just want to know the optimal wire width for my application? Can I get that just knowing the voltage and internal resistance of my source? \$\endgroup\$ Mar 2 at 19:49
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    \$\begingroup\$ You can estimate the power capability of the physical arrangement (and duty cycle) you have in mind , and if you know the winding window you should be able to find a wire size that fills the window and dissipates the given power for your supply voltage. We don't normally try to match internal resistance of the source with a solenoid- we'd prefer the source stays at 12V (say) rather than being loaded down to 6V and half the power lost in the source- but there might be cases where that is appropriate (maybe gas valves). \$\endgroup\$ Mar 2 at 20:01

Funny: I actually designed the B12 unit above. The wire diameter/ turns/resistance is a function of geometry/heat dissipation/ voltage. With a given geometry, you only have so much coil space and so much heat dissipation ability, that determines the power, and then it is basically ohms law to get the resistance in the coil that will fit and only generate that amount of heat in watts. As the voltage goes up, the resistance and turns go up but the watts (and amp turns stay about the same. So you get the same force out/ watt

  • \$\begingroup\$ Terms like "above" and "below" don't work very well when not everyone sees the same order of answers. It's better to link to the specific post you mean (use the "share" link to get a permanent URL). \$\endgroup\$ Mar 17 at 8:16

It's only the weight, the mass of copper, or equivalently the volume, that's winding the electromagnet that matters, not how many turns it's implemented with.

Consider the magnet to be wound of two windings of the same number of turns N of the same cross section wire A. When energised with the same voltage and current, each winding will create the same H-field, and dissipate the same power.

Connect the windings in series. You now have 2N turns of A area wire.

Connect the windings in parallel. You now have N turns of effectively 2A area wire.

In either connection, you have the same volume of copper, same H-field, same power dissipation. But you have different numbers of turns of different section wire. All that changes at the terminals is the supply voltage and current changes, with the same power input. That's assuming you can freely change the voltage and current of your power supply.

Let the number of windings go to any number, and it's easy to see how you can get to any number of turns, of any cross section. Only the total amount of copper matters.

If you go to the limit and use one turn (which incidentally is a very simple number to do calculations with), then in practice your lead-in wires will have difficulty handling the high current and low voltage drop. But that's a practical, not a theoretical, issue. In practice, you use many turns to get your supply voltage up, your current down, and the resistance of the lead-in wires down to a tiny fraction of the total magnet resistance.


The magnetic field near a wire is given by \$B=\mu_0I/2\pi R\$ (where \$R\$ is the distance from the wire, not the resistance.) In the case with 1/8th inch wire, the current will be 16 times larger, but will have 4 times fewer turns. The end result is that the magnetic field is 4 times greater with the thicker wire.

You are assuming that you hold the voltage constant. If you hold the power into the coil constant (i.e., hold \$I^2 R_c\$ constant, where \$R_c\$ is the coil resistance), then two coils wound with the same volume of copper will, to a first-order approximation, generate the same magnetic field.

This comes about because the coil's magnetic field is proportional to the current times the number of turns, the number of turns is inversely proportional to the wire area, and the resistance (as you've figured out) is inversely proportional to wire area squared. So 20 turns at 1 amp should give you just about the same field as 200 turns at 0.1 amp.

Note that if you're designing a system, then the size of your actuators are roughly determined by the amount of power they can handle, while the size of any energy-storage elements (batteries, capacitors, coils) are roughly determined by the amount of energy that they need to hold. You can determine this for yourself with some catalog searches and a bit of linear regression, or by working things out from first principles.

So -- within reason -- you can change your main supply voltage without significantly changing the size of the components needed. Usually where it starts to matter is when the size, weight or expense of the interconnecting conductors gets too great (driving you to a higher supply voltage), because you have concerns with shock safety, insulation thickness, or spark or corona discharge (driving you to a lower supply voltage), or because you have an existing supply rail.

This is why appliances and industrial machines have the same sized light bulbs, power supplies, and motors in Europe and North America, even when the prevalent line voltages differ by a factor of 2.


When designing coils for applications like starters and alternators vs DC solenoid, it’s essential to consider the wire diameter (gauge). One often overlooked factor is inductance, which significantly impacts coil behavior. Let’s explore this:

  1. Inductance and Wire Diameter:
    • Thicker Wire (Lower Gauge):
      • Has fewer turns due to its larger cross-sectional area.
      • Results in lower inductance.
    • Thinner Wire (Higher Gauge):
      • Requires more turns for the same length of wire.
      • Leads to higher inductance.
  2. Inductance’s Role:
    • Inductance opposes changes in current.
    • Coils with higher inductance:
      • Struggle with rapid current changes (e.g., pole changes in BLDC motors).
      • Exhibit slower response times.
    • Coils with lower inductance:
      • Adapt better to alternating currents and rapid switching.
      • Enable faster transitions between magnetic poles.
  3. Design Considerations:
    • Engineers balance wire gauge for optimal performance:
      • Thinner Wire:
        • Lower current requirements (due to more turns). Smaller supporting circuit components requirements
        • Higer voltage required.
        • Slower switching (due to higher inductance).
      • Thicker Wire:
        • Faster switching (due to less turns).
        • Higher current required, larger supporting circuit components requirements
        • Lower voltage requirements

in summary, understanding the trade-offs between wire diameter, inductance, and coil behavior is crucial for optimizing motor designs.

  • \$\begingroup\$ I tried two different tools. "100% Chance that your text is generated by AI" and "77% of text is AI-generated". \$\endgroup\$
    – winny
    Apr 2 at 14:59

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