Why do electromagnets have itty bitty wires?

Question

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?

Answer 1

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.

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.

Answer 2

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

Answer 3

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.

Answer 4

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.