# How to power an electromagnet?

Is it better run an electromagnet of a bigger voltage(12V) with a smaller current(150mA) or with a smaller voltage 5V and a bigger current? Which will heat it up less? Does it even matter? The electromagnet will have to work for several hours so heat is my main problem. Also should I use thicker or thiner wire? I'm trying to build a electromagnet for a ferrofluid project.

active| newest | oldestWhich one to do you use ?

The coil with the most turns will have the ability for the largest magnetic Flux.

Flux is a product of the number of turns and the current.

In fact you may find that you can reduce the current and still produce

the Flux necessary for your ible.

So the coil with the most turns has the best chance of operating cooler.

If you need more Flux you can add a NIB round ( same diameter as the bolt ) permanent magnet making the coil run cooler. Then for no Flux reverse the

coil current ( still runs cool ).

A

Add a NIB magnet and now you get more Flux for the same current !

And if you need to Zero the Flux, Just a little Negative current.

Please enjoy the Pic.

A

Thanks in Advance.

Now I see how I could have made a mistake by appending the image of the electromagnet made by winding wire around a bolt. I'm trying to make something like a door lock. So I'm trying to make it more of a rectangle(110mm X 15mm). Well actually I'm trying to make a 9-segment display that would use ferrofluids to display the time. So alltogether 4 displays, each 9 electromagnets. I'm still in the experimenting phase trying to figure out if it is feasible.

Anyway thanks again for all your answers. All of them have been wery helpfull but

icengs answer of the use of NIB magnets was a total revelation to me.With a fixed diameter, seandogue has pointed out that your field strength (or "flux") scales with I×N, where N is the number of turns. So that argues for higher current.

Heat dissipation from resistance scales with the

squareof the current: P = I×V = I^{2}R. That argues for lower current (which helps quadratically), or for lower resistance (which helps linearly).Note that since P = IV, if you raise the voltage

andlower the current, keeping the product constant, you will get exactly the same heat dissipation!The resistivity (resistance per unit length) of wire scales inversely with cross-section. That argues for larger diameter wires, but that in turn means fewer turns (and hence lower flux) for a give spool.

So what you need to do is figure out what your real system constraints are, and optimize after all those constraints have been applied. For example, you can estimate the total cross-sectional area of wire that will fit in the spool, and use that to balance the maximum possible N vs. gauge (diameter). Once you've chosen the wire gauge, you can look up the resistivity, and use the spool dimensions to determine the total length. and hence resistance. And so on.

Although I can't be sure, since the only frame of reference is the nut/bolt combination, it looks as though you've used something on the order of 30 or 40 gauge. As such, you'll need to be ~careful not to exceed its limits.

(For instance, you probably don't want to try to pump a few amps through it)The magnetic field is proportional to

I, where_{b}NIis the current through the solenoid, and_{b}Nis the number of turns.Measure the real resistance of the solenoid using a multimeter to get a baseline value, from which you can compute a ballpark voltage for a given input voltage

For future reference, Lower N, higher I will have a faster turn on time due to a lower inductive component to the solenoid.

If you don't have a variable voltage supply, consider getting hold of a rheostat, ie, a power potentiormeter, for varying the input voltage.

Your solenoid has a fixed real resistance, so the steady state current (the current after it has been energized for a short period of time) through it is dictated by the applied voltage. This is why I suggest measuring the resistance and calculating the necessary voltage for a given solenoid current

V = I x R

As to the efficiency you may just have to experiment and try both ways. In General more turns more magnetic flux.