Calculating electromagnet – magnetism vs heat - Page 1
Forum section: electronics / beginners

author: perik
author_rank: Newbie
timestamp: August 21, 2019, 12:57:48 pm
content: Im making an electromagnet that will be more or less permanently turned on so I need to avoid heating problems. I have studied this a bit and came to the conclusion that:the lower the gauge (AWG) (the thicker the wire) and the more turns around the core => the greater current it can handle and the lower the heat.with this in mind I would have taken something like 500 turns around a 10 mm diameter screw, with pretty thick copperwire awg 20 (0.5 mm) or something similliar would be okey?What do you say? Is there a way of calculating these heatingproblems before I decide the dimension and number of turns of my copperwire?

author: Kleinstein
author_rank: Super Contributor
timestamp: August 21, 2019, 01:41:19 pm
content: The thickness of the wire only changes from low voltage, high current to high voltage low current. The heat will be the same of the same amount of copper is used and the same field is created.To reduce the heating, one should use the maximum amount of copper that fits in place and try to keep the magnetic field concentrated to where it is needed. Using an iron core of the right shape can help a lot. It is not only the iron inside but also the shape and return path outside.

author: radiolistener
author_rank: Super Contributor
timestamp: August 21, 2019, 02:04:23 pm
content: Did you tried to use neodymium magnets? It doesn't eat electricity, no heating. And it has very strength magnetic field

author: TimFox
author_rank: Super Contributor
timestamp: August 21, 2019, 02:41:17 pm
content: One method to improve heat removal from an electromagnet is to wind insulated tape instead of wire, but insulated wire is much easier to find. There are textbooks on electromagnet design that cover this. Otherwise, Kleinstein is correct: for a given fraction of a constant volume filled by copper, the same excitation (in ampere-turns) will need the same power, with higher voltage in thinner wire.

author: soldar
author_rank: Super Contributor
timestamp: August 22, 2019, 07:22:51 am
content: An electromagnet needs more current to activate it than to hold it so it is common to reduce current when just holding.

author: perik
author_rank: Newbie
timestamp: August 22, 2019, 11:30:02 am
content: Thanks for the answers. I did some research and did this calculation:circumference of the spool:diameter of the screw * PI :: 1 * 3,14 = 3,14Length of wire:Number of turns * circumference :: 500 * 3,14 = 1570 cmThe resistance of the wire / cm (ohm):0,00032Total resistance of the wire:resistance per cm * length :: 0,00032 * 1570 = 0,5024Current:I = V/R, I = 5 / 0,5024 = 9,95 AIm using wikipedia to get information about the copperwire:https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizesTwo questions reagarding the cable:1) The Ampacity - is that what temperature the current will generate? In my case than if you read the table, since my current is 9,95 A that would generate about 70 degrees? Is that correct?2) Fusing Current - that is the current that will melt the wire? Preece is longer than 10 seconds and Onderdonk is over 1 second or 32 ms? Is that correct?So in my case I should go for a thicker wire (lower AWG) to be able to take down the heat .Then to calculate the magentism I use this page:http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.htmlSolenoid Magnetic Field Calculationhere I get with my values: 157 m long, 500 turns, 9,95 A in current and 200 in relative permeability = 79.64 gauss in the center of the magnetSo in my example above I will get:heating about 70 degrees79.64 gauss in the center of the magnetIn other words I will go for a thicker wire to get down the heating ...Im also using a diode to handle the EMFWould you agree on this calculation?Thanks

author: IanB
author_rank: Super Contributor
timestamp: August 22, 2019, 11:42:28 am
content: It was explained to you above that for a given size of coil (winding volume) and a given magnetic field the thickness of the wire does not change the heat produced. Therefore you should not decide on the wire thickness based on heat generation, you should base it on the power supply you will use to run the coil (voltage and current capability), and to some extent mechanical considerations (robustness, cost to wind, etc.).To see why the wire thickness doesn't change anything, consider that the magnetic field is a function of ampere-turns. If you double the cross-section of the wire you will halve the number of turns that fit into the space, which means you will have to double the current to obtain the same magnetic field. If you double the area and halve the length you will reduce the resistance by a factor of 4. But doubling the current will increase the power per unit of resistance by a factor of 4. The two 4's cancel out, so the heat generated will remain the same.

author: Kleinstein
author_rank: Super Contributor
timestamp: August 22, 2019, 11:52:31 am
content: The is no simple relation from current in a given wire to temperature rise. It very much depends on the environment. Especially as a coil the heat flow is pretty much restricted. It is more like on can treat the coil as a single block. 5 V and 10 A are 50 W - quite some heat and would probably result in more than 70C, maybe 150 C.For the wire one would likely want several wires on top of each other. If the inner diameter is fixed at 1 cm one may consider something like 3-5 cm a reasonable outer diameter. Ideally one would use thicker wires the further out. As it is kind of inconvenient to combine different wires it is not used very often. This is used when really looking at efficiency and highest fields.The formulas from the link looks flawed - they are OK without a core, but with a core they seem to have missed the demonizing factor. One can just multiply with the relative permeability if the core is closed or infinite long. A finite length core will cause a smaller field. AFAIR , for a simple cylinder the the amplification gets no larger than the length to diameter ratio, even of the relative permeability is higher.

author: perik
author_rank: Newbie
timestamp: August 22, 2019, 12:10:47 pm
content: hmm. ok thanks. So which way should I go to get a decent powerfull electromagnet with a lot less heat. It is going to stand on for long time so need to have the heat around 30 celcius rather than 150 celcius. :/

author: Kleinstein
author_rank: Super Contributor
timestamp: August 22, 2019, 03:20:13 pm
content: The first point is to get a good idea what is really needed. Does one need a large homogeneous field, a strong holding force or just a high field at a small area. Also the point of needing control is important, as permanent magnets can help quite a lot for a static field. So what is the intended use of the magnet ?The main idea is to use a rather large amount of copper and concentrate die field to where it is actually needed.There can be a kind of trade of between more material used and higher electric power, if a iron core can be used. I high field may need high power and active cooling (e.g water). Especially without a core it just needs power, possibly lot's of it.

author: radiolistener
author_rank: Super Contributor
timestamp: August 22, 2019, 03:21:48 pm
content: Quote from: perik on August 22, 2019, 11:30:02 amHere is my draft calc for your case:Wire material = CopperWire diameter = 0.5 Wire turns = 500Solenoid diameter = 0.01 Wire length = 15.707 Wire resistance at DC = 1.3423 Voltage = 5 Current = 5 / 1.3423 = 3.725 Solenoid length = 0.25 Solenoid H = 7450 [A/m]Solenoid B = 0.009354466  (= 93.54466 )Solenoid L = 97.0397 Heating = 5^2 / 1.3423 = 18.6 Magnetic flux depends on coil winding density and current. Heating depends on resistance, which depends on wire thickness. If you will use fat wire (for less heating), you will get smaller winding density and smaller magnetic flux. And vice versa.For comparison, Neodymium magnet N52 has about B = 1.43 - 1.48  ( = 14300 - 14800 )https://www.aliexpress.com/item/32978000774.htmlAnd it don't consume electricity and no heating at allIn order to get the same magnetic flux from the coil calculated above, you will need to apply 590 Amps / 792 Volts.