# Electromagnet Design Questions

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1. Dec 19, 2011

### bschwartz

Hey guys, so I have a couple questions about electromagnets that I was hoping to get some insight on here. Basically, I've been tasked with designing an electromagnet that is low-heat (<150F or so) yet high pull force. I'm improving on a past design where the electromagnet got way too hot (> 250F) too quickly (<15min). The past design consisted of a core with diameter 2" and length 6" with about 500 windings of copper wire.

I've been messing around with a spreadsheet found on this website http://www.coolmagnetman.com/magelect.htm and it seems that a shorter, thinner core 1"x4" with 1500 turns will be my best bet, because it draws much less power. However, I can't verify the spreadsheet with my own calculations so I don't know if it's really all that accurate.

Before I order the 1000' and start tediously winding the magnet, is there any way that I can be sure that this design will generate less heat than the previous model? I've tried modeling it in SolidWorks, but I'm not sure if thats accurate either since I'm getting temperatures of like 600F for both designs.

Is my design even on the right track? There's just a lot of uncertainty with this project and I would like to be somewhat confident in the design before I start ordering all the stuff lol.

Thanks for any help!

2. Dec 25, 2011

### Staff: Mentor

Are you powering this with smooth DC, or rectified mains AC? How much current have you got to play with? Is there a size constraint on the final product? What is the application?

It seems you are using a solid cylindrical core, so you end up with basically a bar magnet. If you could wind around a former of the shape of a horse-shoe, the poles would be closely spaced and the field between them would be stronger. A crane picking up scrapped vehicles uses a giant horse-shoe electromagnet lowered onto the vehicle's roof.

The heat energy produced is entirely determined by the resistance of the winding. The faster the heat can be dissipated, the cooler the coil will be at thermal equilibrium.

Does the magnet need to be powered continuously? If not, it may be feasible to turn the current off or reduce it to fractional power for some of the time, so it would have some time to cool down.

3. Dec 26, 2011

### bschwartz

I'm fairly certain its a DC 19V power supply that we will be using. We can control the voltage by using a motor controller with potentiometer to vary the voltage from 0-19V. It's a high current psu so it can supply 43A, but we are working in the range of 7-10A or so. The application is a ferrofluid exhibit and we are using the electromagnet to "pull" the ferrofluid up a sculpture, similar to this

It won't need to be powered on continuously, but ideally we would like it to be able to withstand being on for as long as the user(s) is using it. The current design heats up way too fast so that one user may interact with it for a few minutes but then a cool down period of 15min or so has to happen before it can be powered on again. Not very good for an exhibit that will probably be seen by 1000+ ppl every day.

We are hoping to design a magnet that offers the same electromagnetic force as the older magnet but doesnt draw as much power and therefore, a combination of fans/heatsinks will hopefully be enough to keep it relatively cool or at least allow it to be used regularly without a long cool down period.

Last edited by a moderator: Sep 25, 2014
4. Dec 27, 2011

### Staff: Mentor

You have access to the ends of the winding, so can measure the coil's resistance? How many ohms?

You'll be able to predict whether a different winding will get just as hot because if it has the same resistance then for the same current it will get just as hot.

5. Dec 27, 2011

### bschwartz

Well no, not yet. We haven't wound the new magnet nor have we bought the supplies to wind the magnet. We want to make sure that the new magnet design won't heat up as much as the older one before we order anything.

The older magnet has about ~2.3 Ohms of resistance so it draws around 8 A of current. However, in the spreadsheet above it says that the old magnet should have a resistance of 0.7 Ohms and therefore it would draw ~25 A of current. We obviously know this isn't true due to measuring the current while the magnet is operating. This and a couple other reasons make me believe that the spreadsheet isn't as accurate as we would hope. However, even if the numbers aren't exactly right I want to believe that the trends are accurate. From the spreadsheet, the designed magnet will have a resistance of 2.7 Ohms and draw 7 A.

I think that we are just going to go off the theory that for an electromagnet, you want as many turns as possible. So, the thought is to use 14 gauge wire, wind it about 1500 times around a 1" x 4" mild steel core, and that should generate the same magnetic force as the old one for a fraction of the power. We believe this because we are increasing the number of turns (and length of wire), which should increase the resistance, which lowers the current, and therefore decrease the power draw.

So basically we would just have a massive 5" x 4" magnet, including all the wire, which we hope won't heat up as much because there will be so much more resistance. This sounds right in theory, doesn't it? Haha

6. Dec 28, 2011

### Staff: Mentor

If the cores have similar properties. the magnet strength is dependent on ampere-turns. If you use that slimmer core, then for the same number of turns you look like using roughly 20% less wire by weight, so resistance will drop by that amount.

Heat produced = Power losses = I2.R

so that smaller R would permit you to increase current slightly, for the same amount of heat produced in the winding. However, if you can increase I, then you can decrease the number of turns and still maintain the original magnet strength.

You can predict whether a new wire gauge will get just as hot. Measure its diameter and use the wire tables to determine resistance. http://en.wikipedia.org/wiki/American_wire_gauge#Table_of_AWG_wire_sizes

What is the gauge of your current wire? Measure the width of a couple of tightly packed strands, and divide by that number to find diameter accurately. 14 AWG gives 6 turns/cm, and has 2.5 ohms per 1000ft.

I don't understand why you would want the coil's resistance to limit the current. The ideal would be a coil with practically no resistance, this will allow you to reduce the voltage from your DC supply while keeping the current at its present level. With lower resistance, and a lowered voltage, there will be less heat produced in the winding.

That demonstration video you linked to---is there a simple bar-magnet type of electromagnet moving up and down inside the inverted cone?

7. Dec 28, 2011

### Staff: Mentor

If you double the number of turns, you can halve the current yet still maintain the same magnet strength. But doubling the turns has doubled the length, hence doubled the resistance. The result being a halving of the amount of heat produced in the winding, providing you don't change the wire gauge.

8. Dec 28, 2011

### bschwartz

I don't know for sure about the sculptures in the video below, but from testing our own sculptures and set up, all it is is an electromagnet directly beneath the sculpture (which is completely solid). The fluid will "climb" up the sculpture when the magnet is powered up (supplied more voltage). The sculptures we are designing around are about 4 inches high with the base diameter of the cone being 2in. To get a good interaction with the sculptures, the electromagnet needs to be pretty powerful so that's why we need an obscene amount of wire, which undoubtedly will increase resistance.

We won't be using the same length of wire for the new magnet, it will be roughly 5x the amount used in the old magnet. This is so we can get a lot more turns and hopefully be able to lower the current while still being strong enough to give a nice interaction between the fluid and sculpture. From the documentation we received, the wire of the old magnet is 16 gauge, however we haven't verified that by our own measurements.

The reason we need to limit the current by the resistance of the coil is because we need to keep the strength of the magnetic field relatively constant when compared to the last model. The equation for a closed circuit magnetic field is given below (this isn't the case for my situation since it is NOT a closed circuit, but the trends should be the same).

$B=NI\mu/L$

Say the model last year produced 1.5T of magnetic field. This amount of magnetic field was strong enough to "pull" the fluid all the way to the top of the sculpture so we want to try and obtain the same strength of field, but decrease the current. This is done by increasing N, the number of turns. To increase N, we went with a thinner core and also a lot more wire. This will increase the resistance of the circuit no matter what, but that will also decrease the current and since P=I2R, this should lower the overall power draw.

9. Dec 28, 2011

### Staff: Mentor

The sculpture is of a non-magnetic material, is it?

The windings on your old coil have bulked it out from a core of 2" to a final diameter of 6", is that correct? And at max it runs 8A at about 2.3 ohms? You can easily verify that resistance by measuring the voltage at that current. Are the windings all higgledy-piggledy, or are they carefully layered, and with each turn tightly packed against its neighbour like on a cotton reel?

A narrower diameter core will allow you to fit additional turns, thereby increasing the field off the end. But I have my doubts about a shorter core length doing anything for you. I think that equation you quote is being mis-applied.

10. Dec 30, 2011

### bschwartz

Sorry, I feel like I never clarified what the old magnet looked like. It is a 2" (diameter) by 6" (long) solid steel core. There are only about 600 windings on the core so its final diameter is ~2.25".

And those sculptures are magnetic. I'm not sure, but they appear to be the same material as the core of the electromagnet (1018 mild steel).

And yeah, there's definitely the chance that the equation is being misapplied since we are not dealing with a closed-circuit electromagnet. However, I haven't really been able to find any equations that will help determine the strength of the magnetic field. But the strength isn't necessarily our main problem. It's just keeping the magnet cool enough to run for extended periods of time without overheating.

11. Dec 31, 2011

### Staff: Mentor

If you were to wind another identical winding on top of the original one on your old magnet, then they could share the current equally. Overall heat produced would drop by 50%, but magnetic strength would stay as it is.

You wouldn't wind exactly the same number of turns, but wind it so that each winding had the same resistance so they'd divide the current equally. Maybe wind even fewer turns than this, so 15% more current would flow through the outer coil where the heat is better dissipated (I presume). Are you blowing air around the outside of the winding to cool it?

Or consider two more windings, if there is no restriction on bulkiness, then current in each would be 1/3 and heat produced would be 1/3 present level. (You'll find you have to reduce the voltage to 1/3 present setting, of course, to maintain current.)