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Calculating electromagnet – magnetism vs heat

perka

New Member
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?
 

Pommie

Well-Known Member
Most Helpful Member
Sure, use any online calculater to work out the resistance of your coil and then the magnetic strength is turns x current and the heat is current squared times resistance.

Mike.
 

perka

New Member
Thanks for the answers. I did some research and did this calculation:
circumference of the spool: diameter of the screw * PI :: 1 * 3,14 = 3,14
Length of wire: Number of turns * circumference :: 500 * 3,14 = 1570 cm
The resistance of the wire / cm (ohm): 0,00032
Total resistance of the wire: resistance per cm * length :: 0,00032 * 1570 = 0,5024
Current:
I = V/R, I = 5 / 0,5024 = 9,95 A

Im using wikipedia to get information about the copperwire:
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes

Two 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.html
Solenoid Magnetic Field Calculation
here 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 magnet

So in my example above I will get:
heating about 70 degrees
79.64 gauss in the center of the magnet


In other words I will go for a thicker wire to get down the heating ...
Im also using a diode to handle the EMF

Would you agree on this calculation?

Thanks
 

Pommie

Well-Known Member
Most Helpful Member
Don't have time to go through all your calculations now but will just make a couple of suggestions.

The diameter of the coil is the average diameter. I assume there is limited length and so the coil will have multiple layers.
I think the 70C is for uncoiled wire. When wrapped tightly it will be much higher. The enamel will break down at around 200C.

I would suggest thinner wire with more turns. Supplying 10A at 5V is not as easy as 4A at 12V.

Mike.
 

Pommie

Well-Known Member
Most Helpful Member
But thinner wire means more turns and magnetic strength is turns * amps. Sometimes thinner wire will result in a stronger magnet.

Mike.
 

ClydeCrashKop

Well-Known Member
Most Helpful Member
I would suggest buying a relay or solenoid with the size & strength coil that you want. They will be professionally made and as efficiently as possible. If you can use it, a relay will have an iron core with better permeability than a steel screw.
 

unclejed613

Well-Known Member
Most Helpful Member
But I guess thinner wire will produce not as strong magnet.
the "strength" of the magnet is a function of ampere-turns. if you have 100 turns and 1A flowing, that's 100 ampere turns, and it will have the same field strength as a magnet running 10A through 10 turns, or 1000 turns with 0.1A.
 

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