I'm not an expert in the electromagnetic field and need to design a solenoid actuator with 1 kg force approx. I am wondering if I change the wire diameter of the coil, what happens in solenoid's parameters? for example, the current passing the wire, voltage or force of the solenoid would be changed if the wire diameter goes up or down?

  • \$\begingroup\$ What have you found in terms of research? Look on here, there are some relevant questions with answers... \$\endgroup\$ – Solar Mike May 23 '19 at 14:56

The wire diameter directly affects how much current the solenoid can handle (resistance changes with diameter). The pulling force is determined by the ampere-turns (among other factors such as core type, winding diameter, winding length, etc.). The more turns, the more pulling force. As you add turns, however, the resistance of the solenoid as a whole will go up.

If you have a solenoid and change only the wire gauge of its winding, the following would have to change:

Larger gauge wire:

  • Overall resistance decreases
  • Current increases
  • Pulling force increases
  • Outside diameter of solenoid would increase due to extra mass in wire

Smaller gauge wire:

  • Overall resistance increases
  • Current decreases
  • Pulling force decreases
  • Outside diameter of solenoid would decrease due to less mass in wire

Note that you should consider the current capacity of the wire gauge you are using, along with the duty cycle. You can increase pulling force by increasing voltage (and therefore current). However, if you don't limit current, the wire will heat up and can eventually fail. If you only actuate the solenoid briefly, you can allow higher power. Some people refer to this as a "kick" when supplying higher voltages to solenoids than what would be considered 100% duty-cycle appropriate.

There are some helpful calculators for solenoids, such as this one at CalcTool.


Flux density in a solenoid is proportional to the number of turns and current.

enter image description here

As you can see, wire size and/or resistance is not in the equation, the size of the wire only indirectly affects the result. Too large and you can't fit the turns into the space. Too small and the wire will overheat, or the resistance is so large that you need a higher voltage than desired to get the same current.

Also, since most people want to drive the solenoid with a constant voltage, you want the desired current to result from the rated voltage.


The two most important parameters you have for the strength of an electromagnet are

The Power you have available and
The weight of Copper you have available

Although this doesn't seem like enough, all the other terms like volts, current etc cancel out. The more you can spend on copper to build it, and power to run it, the stronger the field you will get.

Now, do you make that copper (for instance) 1000 turns of 2mm2 wire, or 2000 turns of 1mm2 wire? It depends on what voltage and current power supply you have available. Consider a magnet with two 1000 turn 1mm2 windings. Put them in series, you have 2x the voltage. Put them in parallel, you have 2x the current. But in both cases, the same magnetic field, and the same power dissipation.

So it doesn't matter what diameter wire you use, if you haven't chosen your power supply yet.

  • \$\begingroup\$ I will fit my design in 24 volts power supply with 400 mA, what is your advisement about wire size? is 34 AWG reasonable? (wind up to 60 ohms) \$\endgroup\$ – milad Chalipa May 23 '19 at 15:47
  • \$\begingroup\$ @miladChalipa What's the winding volume of the magnet? What's the duty cycle of the magnet? \$\endgroup\$ – Neil_UK May 23 '19 at 15:49
  • \$\begingroup\$ I can redesign the spool size according to wire volume, so it is not a problem. for the duty cycle, in the beginning, will be 100% but after 10~20 ms, 25% I think. \$\endgroup\$ – milad Chalipa May 23 '19 at 15:53

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