# Electromagnet Design

by I_am_learning
Tags: design, electromagnet
 P: 669 I am trying to design a very powerful electromagnet for experimental reasons. However, I am wondering at his absurd sounding conclusion I arrived at! The governing equation of the magnetic field generated by a solenoidal electromagnet is B = u0nI Provided that I keep these things constant 1. length of solenoid = 15 cm 2. diameter of solenoid = 2cm 3. Voltage to be applied = 24 Volts 4. Max power = 72 watts. (hence max current = 3A) then I can increase B, by simply increasing n. But to increase n, I need longer wire, hence more R (resistance); so less than 3A current may flow if I wound more wires to increase n. However, I can always reestablish 3A current by simply increasing the wire diameter. Thus it seems to me that I can attain B as high as I desire, by simply winding more wire and increasing the wire diameter, and my electromagnet will never consume more than 72 watts. My intuition tells me this must not be possible, but I don't see why?
 P: 477 Hi there, From what I remember, n represents the number of time you wind the wire per unit lenght. And, it seems to me to be logical that you will have a hard time building a very intense electromagnet with a 72W power output. How intense do you want you magnetic field??? I believe the solution to your question stands in the maximum power output. Your wire will definetely get longer, and the resistance will increase. Therefore, you will need more power to increase the current output. Cheers
 P: 141 you have forgotten the diameter of the coil in your equation. increasing wire diameter means increasing coil diameter.(wire becomes hard to bend) also, if you put a ferromagnetic core inside your coil it becomes more powerfull.(permeability increases)
P: 669
Electromagnet Design

 Quote by fatra2 Hi there, From what I remember, n represents the number of time you wind the wire per unit lenght.
Yes, you remember well.

 Quote by fatra2 How intense do you want you magnetic field???
As, strong as I can afford, given that the power consumption is within 72 watts.

 Quote by fatra2 I believe the solution to your question stands in the maximum power output. Your wire will definetely get longer, and the resistance will increase.
I already mentioned that in the OP. So, I said that I will increase the wire diameter to reduce the resistance.

 Quote by fatra2 Therefore, you will need more power to increase the current output. Cheers
I am not trying to increase the current output, so I won't require more power.

P: 669
 Quote by SirAskalot .(wire becomes hard to bend)
Hard but not impossible. I am seeking theoretical explanation here. Of course, I have practical limitations, I won't be able to even get wires as large as 1cm dia.
Nevetheless, I am here to know, the fault in my reasoning presented in OP.
 Quote by SirAskalot if you put a ferromagnetic core inside your coil it becomes more powerfull.(permeability increases)
Of course.
 P: 477 Hi there, Well one solution would be to cool down the electrical circuit. That way you would be playing on the resistivity of the wire, therefore decreasing its resistance. Cheers
 P: 141 Well, you don't actually "use(convert)" any energy in your coil/inductor. You just store it in a magnetic field. The only power used are the I^2R (resistive) losses. Until you use your coil to lift something, then the magnetic field changes due to the "re-action" field of the ferromagnetic substance. Then your 70W might not be enough. Increasing the turns changes the self-inductance (L), and you can store more energy. Tough there are a time aspect involved, B needs time to build up. Also your steady state current are given by ohms law: I=U/R. As far as im concerned, the only limitations are the practical. And you aint got any MRI machine using your setup.
P: 669
 Quote by SirAskalot Well, you don't actually "use(convert)" any energy in your coil/inductor. You just store it in a magnetic field. The only power used are the I^2R (resistive) losses. Until you use your coil to lift something, then the magnetic field changes due to the "re-action" field of the ferromagnetic substance. Then your 70W might not be enough. Increasing the turns changes the self-inductance (L), and you can store more energy. Tough there are a time aspect involved, B needs time to build up. Also your steady state current are given by ohms law: I=U/R. As far as im concerned, the only limitations are the practical.
First of all, this sort of explanation was what I was seeking. Thanks.
Its no problem for me that It will take time for B to setup.
I would like your further explanation at this point.
1. Ok, you seem to agree that by using wire of heavy diameter, I can attain High 'B' and yet within 70 watts. Now, what happens when I try to pull objects by that field? Does power consumption increase or the electromagnet fail to attract? Whats the theory behind that?
Any useful(in sense that it explains this point) links are welcomed.

 Quote by SirAskalot And you aint got any MRI machine using your setup.
I am aware of that. If it was possible people wouldn't have made such heavy power MRIs. :)
 P: 4,663 I have attached a thumbnail with the important equations for designing an electromagnet (solenoid) with (second figure) and without (first figure) an air gap. The air gap can be either inside or outside the coil, as long as the geometry of the iron (ferrite) is unchanged. Note that for the elecromagnet (solenoid) with an air gap, as the air gap gets shorter, the magnetic field field gets higher. Please review. If you provide a very detailed description of what you want to do, I can provide more guidance. Do you want the magnetic field in a magnetic material or in air? How big a volume of magnetic field do you need? Is this an ac (pulsed) or dc electromagnet? Bob S Attached Thumbnails
P: 669
 Quote by Bob S I have attached a thumbnail with the important equations for designing an electromagnet (solenoid) with (second figure) and without (first figure) an air gap. The air gap can be either inside or outside the coil, as long as the geometry of the iron (ferrite) is unchanged. Note that for the elecromagnet (solenoid) with an air gap, as the air gap gets shorter, the magnetic field field gets higher. Please review. If you provide a very detailed description of what you want to do, I can provide more guidance. Do you want the magnetic field in a magnetic material or in air? How big a volume of magnetic field do you need? Is this an ac (pulsed) or dc electromagnet? Bob S
What I am trying to do with this electromagnet is to levitate objects.

The hall sensor (which am still to learn) will sense the distance of the object, and accordingly send feed back signal so as to lower or raise the power of electromagent, so as to keep the object suspended.
So, the electromagnet I am going to need is
1. Powerful
2. Continuous operation
3. Air gap
< you may know more :)>
 P: 4,663 The on-axis magnetic field of a finite-length solenoid is given by http://www.netdenizen.com/emagnet/so...insolenoid.htm For 3 amps max, the best choice of enamel-coated copper wire is 16 Ga.; 3.7 amps current carrying capacity, and 4.09 ohms per 1000 ft. For 24 volts, 3 amps, the length of wire should be about 700 2000 feet (8 ohms). You can add a ferrite rod to the solenoid to get perhaps a factor of two in magnetic field. An iron rod is too slow (eddy currents). You will not be able to get 72 watts out of the coil if you run it continuously (it will run too hot); you may have to run it 1 minute on, 10 minutes off. Maybe using forced air cooling will be sufficient to cool 72 watts continuously. A Hall effect magnetic field pickup is probably too slow to respond to the movement of the ferromagnetic object. You might consider photodetectors to detect the elevation of the object. You will also need a fast amplifier to control the solenoid current (preferred to voltage control). I hope this helps. Bob S
P: 669
 Quote by Bob S The on-axis magnetic field of a finite-length solenoid is given by http://www.netdenizen.com/emagnet/so...insolenoid.htm
It don't have Iron core, right? Where in the formula should I include its u, if I add iron core.

 Quote by Bob S For 3 amps max, the best choice of enamel-coated copper wire is 16 Ga.; 3.7 amps current carrying capacity, and 4.09 ohms per 1000 ft. For 24 volts, 3 amps, the length of wire should be about 700 2000 feet (8 ohms).
What is the formula for predicting the wire diameter. Why shouldn't I use 10 AWG, instead (0.9989 Ohms/feet), hence 8000 feet (hence more N) for 3A ?
 Quote by Bob S You can add a ferrite rod to the solenoid to get perhaps a factor of two in magnetic field. An iron rod is too slow (eddy currents).
Ferrite rod and iron rod is different? What is a ferrite rod?

 Quote by Bob S You will not be able to get 72 watts out of the coil if you run it continuously (it will run too hot); you may have to run it 1 minute on, 10 minutes off. Maybe using forced air cooling will be sufficient to cool 72 watts continuously.
I will see it.

 Quote by Bob S A Hall effect magnetic field pickup is probably too slow to respond to the movement of the ferromagnetic object. You might consider photodetectors to detect the elevation of the object. You will also need a fast amplifier to control the solenoid current (preferred to voltage control).
I will see it, too :)
 Quote by Bob S I hope this helps. Bob S
Helping. :)
P: 4,663
You can add a ferrite rod to the solenoid to get perhaps a factor of two in magnetic field. An iron rod is too slow (eddy currents). The on-axis magnetic field of a finite-length solenoid is given by
http://www.netdenizen.com/emagnet/so...insolenoid.htm
 Quote by thecritic It don't have Iron core, right? Where in the formula should I include its u, if I add iron core.
Based on my earlier post with a thumbnail analysis of a coil with a gap, the formulas indicate a hi-mu core in the solenoid will add about a factor of two to the external magnetic field. Note that the on-axis field drops off very fast below the end of the solenoid. This means that as the ferrite object gets closer to the solenoid, the attraction force gets stronger very fast. So if the magnetic field is constant, the vertical position of the ferrite object is unstable. You can overcome this with fast negative feedback from your vertical position monitors

 What is the formula for predicting the wire diameter. Why shouldn't I use 10 AWG, instead (0.9989 Ohms/feet), hence 8000 feet (hence more N) for 3A ?
Here is a table of copper wire gauge, wire diameter, wire amp rating, and ohms per 1000 feet.

http://www.interfacebus.com/Copper_Wire_AWG_SIze.html

Based on your requirement of 72 watts and 24 volts, you need wire with a 3 amp rating (16 Ga. wire or larger) and a resistance of 24 volts/3 amps = 8 ohms, which is 2000 feet of 16 Ga. wire. Using 10 Ga wire for 3 amps is a waste of copper.

 Ferrite rod and iron rod is different? What is a ferrite rod?
A ferrite rod is a ferromagnetic rod with very high resistivity, so eddy currents will not be a problem. You will need fast control of the solenoid current to respond to motion of the ferromagnetic object. Large ferrite rods are on the web.
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You will not be able to get 72 watts out of the coil if you run it continuously (it will run too hot); you may have to run it 1 minute on, 10 minutes off. Maybe using forced air cooling will be sufficient to cool 72 watts continuously.

 I will see it.
I think your coil can dissipate about 10 watts w/o forced air cooling.

Bob S
P: 669
 Quote by Bob S Using 10 Ga wire for 3 amps is a waste of copper. Bob S
Since, 10 Ga wire has much lower resistance/length I might be able to make more number of turns with same resistance and current.
I am well ready to pay for the extra 'waste' copper, if I get extra magnetic field.
P: 4,663
 Quote by thecritic Since, 10 Ga wire has much lower resistance/length I might be able to make more number of turns with same resistance and current. I am well ready to pay for the extra 'waste' copper, if I get extra magnetic field.
Yes, you can get more turns by using 10 Ga. wire.

But consider the following. If you use 10 Ga. wire instead of 16 Ga. wire, the maximum length (because of resistance) increases to 8000 feet. For the 16 Ga. wire, 2000 feet is equivalent to 9700 turns (using the 2 cm diameter). Because you can get only about 115 turns per layer on a 15 cm long solenoid, there will be ~84 layers of wire (over 10 cm thick). This thickness is well beyond the useful thickness for the solenoid. If you switch to 8000 feet of 10 Ga. wire, the wire length is 4 times larger, there are 2 times less turns per layer (so twice as many layers), and each layer is twice as thick, leading to a coil with thickness 4 x 2 x 2 = 16 times larger. This is totally unrealistic.

The best thing you can do is switch to a 6 volt, 12 amp power supply (suitable for 10 Ga. wire). The coil design would be much more reasonable.

Bob S
P: 669
 Quote by Bob S For the 16 Ga. wire, 2000 feet is equivalent to 9700 turns (using the 2 cm diameter)
Just to mention, This is a little flawed. See this formula I derived (google doc document),
I just included the increasing radius, nothing much done. The actual number of turns comes around 1517 turns.

But these don't make any difference to the point you are trying to make, I see.

 Quote by Bob S This thickness is well beyond the useful thickness for the solenoid. If you switch to 8000 feet of 10 Ga. wire, the wire length is 4 times larger, there are 2 times less turns per layer (so twice as many layers), and each layer is twice as thick, leading to a coil with thickness 4 x 2 x 2 = 16 times larger. This is totally unrealistic.
I see the point, the solenoid would go drastically increasing in diameter. And as the site you linked to seem to suggest, there is a optimum diameter for which B will be maximum.
I am currently trying to gather every thing and find out a formula that will give the wire diameter required to reach the optimum Geometry. However, the first step, finding out the optimum geometry is still to go. I also, already started email conversation with the author of the site and he thinks I am heading in right direction.
Since there is no rush in my project, I am trying to go slowly, trying to learn everything that comes in path.