Disclaimer: Hobbyist "maker" with limited experience and no relevant education. Please considerate and answer in detail. Feel free to skip to "Question".


I have an intention to build a simple solenoid for no other reason but my own interest. A basic review of theory brought me to the equation B = μIN/L whereby B = Field Strength,μ = Magnetic Permeability, I = Current, N = Coil Turns and L = Coil Length. For what ever reason my first observation was that there was surely an optimisation to be solved in the selection of wire gauge since an increased wire cross-section would increase your available Current Capacity (I) but decrease the Coil Turns (N) along a constant length. I then stopped short of attempting any calculation since I being proportional to the cross section area of a wire (in terms of Ampacity) and N being a first order relation, it seemed logic that the "bigger the wire the better, turns don't matter that much"... this can't be right. What further confused me was to investigate the current draw of commercial BLDC motors, of whatever size, and note that their amperage was surely beyond what would usually be carried on the wire gauges used for stators. Something to do with the AC driving? I sit confused.


  • What is the optimal design approach for selecting wire gauge for electromagnetic coils of a given length and diameter?
  • What factors are most import to consider?
  • Are there conventions or standards I should know about?
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    \$\begingroup\$ Limited f- range of response T=L/R , Bsat core, eddy current losses, loop area, magnetic coupling (leakage inductance) crosstalk, Litz Wire, air core vs high mu , high voltage insulation magnet wire, self-capacitance (resonance) winding patterns. , there's a lot of topics \$\endgroup\$ – Tony Stewart EE75 Apr 22 at 12:52
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    \$\begingroup\$ Why would you assume a constant length? More practically there might be a constant volume available for the copper. The drive voltage may be fixed (something like a relay). You may notice that DC relays of a given design tend to have the same coil size and power regardless of coil voltage. \$\endgroup\$ – Spehro Pefhany Apr 22 at 13:03
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    \$\begingroup\$ Realize that you are making an electromagnet, it will be much more powerful if you close the loop. en.wikipedia.org/wiki/… Same with a plunger solenoid, you want a path outside the coil. \$\endgroup\$ – Mattman944 Apr 22 at 13:31
  • \$\begingroup\$ @TonyStewartEE75 I'd already assumed I was scratching the surface of something much more complex. Do you perhaps have any recommended resources? Books? \$\endgroup\$ – George Kerwood Apr 22 at 17:09
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    \$\begingroup\$ It turns out that the volume for the copper is something that is determined fairly early on in the design process. The volume is always filled with turns of wire. If the wire is thick, there will be fewer turns. If the wire is thin there will be more turns. This holds true for transformers and motors. Note that solenoids are motors. If you want more inductance (higher voltage lower current) you use fine wire. If you want less inductance (lower voltage higher current) you use fatter wire. But you always fill the volume with copper. \$\endgroup\$ – mkeith Apr 23 at 3:13

Interestingly, the choice of wire gauge is not so important. You can pretty much design the whole rest of the system, and then pick the wire gauge according to how you want to trade off voltage and current at any given power level (P=VI).

Let's say you start with some electromagnetic system that has a coil made of AWG 21 wire...

You might measure I amps at V volts around N turns, and you might measure the DC resistance of the coil as R. The power lost to heat is I2R. The flux is proportional to IN.

Keep all dimensions the same, but switch to AWG 24 or so (pretend it's exactly half the cross-sectional area), Now:

  • The number of turns in the coil can double to 2N, because the cross sectional area of the wire is halved.
  • The DC resistance increases to 4R, because it's proportional to turns/area.
  • The current required to achieve the same flux is halved to I/2, because I/2 * 2N = IN
  • The voltage required to achieve that flux is doubled to 2V, because magnetic EMF is proportional to turns, and (I/2)*(4R) = 2IR.
  • The input power stays exactly the same: 2V*I/2 = VI
  • The power lost to heat stays exactly the same: (I/2)2*4R = I2R

So after changing the wire gauge, the capabilities and efficiency of the system are unchanged. We've just changed its operating voltage and its operating current in inverse proportions.

The capabilities of the system depend on the shape of the coil and other components, the current density in the coil and the material that the coil is made of. (heat losses are K * current_density * coil_volume, for some constant K that depends on the material)

  • \$\begingroup\$ Fantastic! This really clarified for me exactly my query so thank you very much for your time and detail. \$\endgroup\$ – George Kerwood Apr 23 at 8:59

Number of turns and current are variables that depend upon design goals. However, there is an optimization to be had in terms of geometry. A "Brooks coil" is a coil with a square cross section and an inside radius of the same dimension as a side of the square cross section. enter image description here (from www.circuits.dk)

The Brooks coil has the property of maximizing the magnetic flux for a given length of wire.


There is no ‘optimum’ but rather a compromise. A smaller cross-section wire will allow more turns but have greater resistance - this means more temperature rise. Ultimately the limiting factor is the temperature the wire insulation fails at - you want to stay away from this as far as possible.

How long do you want this solenoid energized for? For short term low duty cycle use you could tolerate the temperature rise. For example I’m working with some 12V solenoids at the moment. They draw 20A and will get extremely hot if left energised for over a couple of minutes. They are intended to unlock a mechanism so this limitation is not an issue.

For the joy of experimentation, I’d suggest you chose 2 or three different wire diameters and wind three solenoids. See what the outcome is and use this information to guide your next iteration.

  • \$\begingroup\$ All very useful constraints to consider, thank you for your time and explanation. \$\endgroup\$ – George Kerwood Apr 22 at 17:17

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