# DC electromagnet

1. May 16, 2008

### esander

I am trying to create an electromagnet that will repel another magnet when activated. Currently, I am using a 12v battery with 7400 mAh. For the coil I am using copper wire of AWG 9 and 12 (I am experimenting with both sizes). I have also tried air and iron cores. However, I have run into a snag. The permanent magnet I am using is a disc NdFeB magnet. With the air core, the magnet has an effect on the coil but it is not strong enough. When I add the iron core, the NdFeB magnet is strongly attracted to the core regardless of whether or not the current is flowing and in what direction it is flowing. Is this because the current is too low (or the NdFeB magnet is too strong) that the effects of the electromagnet are null?? Also, The iron core is a solid piece of iron and i have the coil wrapped tightly around it. Will the electromagnet become stronger if I loosen the coil and leave an air gap between the coil and core?? Will an iron ring work better as a core compared to a solid piece? Any information is helpful. Thanks.

Eddie

2. May 16, 2008

### marcusl

Good grief, get rid of the welding wire. You want a zillion turns of small gauge enameled magnet wire (maybe 24AWG, but you don't mention the size of your core.) Make sure you have enough turns that the DC resistance results in reasonable power dissipation. That is, V^2/R should be kept modest so you don't melt your wires. I suggest you get online and try to learn something about electromagnets before proceeding.

There are dozens of elementary sites. Here are a couple more advanced:

http://en.allexperts.com/q/Electrical-Engineering-1356/Electromagnet-Design.htm" [Broken]

http://www.geocities.com/CapeCanaveral/2404/design2.html"

Last edited by a moderator: May 3, 2017
3. May 16, 2008

### esander

The size of the core is approximately 1" in diameter (I am limited by space so the core must be small). You mention getting rid of the welding wire. The wire I have is enamel coated magnet wire, but it is the larger diameter.

4. May 16, 2008

### NoTime

Your problem involves the concept of magnetic circuits.
Some info here: http://en.wikipedia.org/wiki/Magnetic_circuit

Translated, you need more turns, more current or both.
It's going to be difficult to get enough turns using 12awg or even worse 9awg wire.
As Marcus noted you are probably better off with 24awg or perhaps 18awg wire.

5. May 17, 2008

### Phrak

Far out. The permanent magnet is overwhelming the influence of the windings on the core material.

This brings up an interesting question. I wonder, if at sufficient distance where the influence of the permanent magnet is small, the core is repelled from the permanent magnet rather than attracted. So tried it with a frig magnet and a rare earth without any confirming results. They always attracted in any orientation. They may be far too different in strength for a fair test, but that's all I've got in the way of magnets.

Last edited: May 17, 2008
6. May 19, 2008

### esander

Thanks for the responses guys. I figured that's what was wrong. I did a few calculations, but the attractive force of the electromagnet seems way too high (it would be great if it was correct, but i dont think it is). The equations I found are as follows: B=[mu(air)*mu(rel.)*I*N]/L and F=[(B^2)*S]/[2*mu(air)].

The values i've used are listed below. If someone could check my work, I'd really appreciate it.

mu(air)=4(PI)x10^-7
mu(rel)=5000 for soft iron
I=.533 amps
N=78 turns
L=.0508 m
The above yields B=5.1421

S=.00051 m^2 (pole area with magnet radius of .0127 m)

Plugging everything into the equation for Force, I get F= 5,360 Newtons which I've converted to 1,204.97 lbs.

7. May 19, 2008

### NoTime

Kinda hard to check your work if you don't show it.
You're obviously wrong by several orders of magnitude.

8. May 20, 2008

### esander

NoTime,

I took the values listed, beginning with mu(air) and ending with L, and plugged them into the equation for B. I then used the answer I got for B (listed) and plugged the necessary values into the force (F) equation, which yielded the 1,204.97 lbs. Is there something in the equations that is missing which would bring my value down to the correct order of magnitude??

9. May 20, 2008

### marcusl

Your value for mu is way off. Typical mu for iron is 100 - 500, but you are going to run it in or near saturation so the effective value will drop towards 1. I'd be surprised if your B value exceeds about 0.1 or 0.2T.

10. May 20, 2008

### esander

marcusl,

thanks a lot. i had found another site that said mu was around 200. wikipedia has it at 5000

Is there a way to find out how big the magnet needs to be (how big the core needs to be) in order to get a higher value of B?

To give you an idea of what I am doing (i should have done this at the beginning), I need an electromagnet to launch a permanent NdFeB magnet in a horizontal direction a distance of about 3-4 inches within a fraction of a second. The final velocity of the magnet is around 2 ft./s. The only problem is I'm limited by space. The electromagnet can't have a large diameter (a couple inches) but can be longer. Do you have any idea of how strong of a magnet I would need? Any information is helpful. Thanks again guys.

Eddie

11. May 20, 2008

### esander

I guess an easier way to word my question is, how do i determine the correct value for the iron's relative permeability??

12. May 20, 2008

### esander

sorry for the multiple posts, but I've done some more work and slightly redesigned the magnet. I will need a relative permeability of either 53.6 or 6.8 depending on if I use .533 amps or 4.2 amps respectively. Could someone tell me if this is possible with the following magnet criteria?? The magnet is 3" (.0762 m) long using AWG 22 wire. There are 1062 turns (9 layers wrapped the total 3" of the magnet). The current will be either .533 or 4.2 amps as noted above.

The core is an iron core shaped like a hollow tube. The inner diameter is 5/8" while the outer diameter of the core is 1.5".

With this information i've calculated B to be .5. From that, i've calculated the Force to be about 22 lbs. Much less, and more reasonable sounding, than the original 1205 lb. force i calculated. Does this new force of 22 lbs seem somewhat accurate??

13. May 20, 2008

### NoTime

After thinking about this a while I think the equation you used is for a closed magnetic circuit.
$$F = B^{2} A / 2\mu_{0}$$
$$B = \mu NI/L$$.
Ie: The pull on a keeper bar with a U or E shape iron core.
With your arrangement the relative permeability is about 1 or the same as air.

You might be better off using a long air core (longer is better).
Insert the NdFeB magnet in the core tube.
Apply current until magnet exits core area.

Edit. You don't need a magnet. Plain iron will do, but then the initial position needs to be the far side of the coil.

Last edited: May 20, 2008
14. May 21, 2008

### esander

Thanks guys. but i'm a little confused about the relative permeability. I'm under the impression, from everything i've read, that using an iron core should increase the magnet's strength by a lot. Why in this arrangement does the iron have the same relative permeability of air??

15. May 21, 2008

### esander

This will seem like a stupid question at this point, but when i use the area of the pole to calculate the force do i use the area of the core or the area to the outside of the wire (using multiple layers)??

16. May 21, 2008

### NoTime

The key here is magnetic circuit.
The principals are much the same as for electric circuits.
For a crude approximation half your magnetic circuit is air and the other half is the iron core.
You need to add the two impedances.
It gets even more complicated when you realize that the field lines take the path of least resistance and will simply leak around the coil loops.
Yes, the iron core helps but not anywhere near what an inappropriate use of the equations might indicate.

17. Sep 28, 2008

### wd102

hi, does anyone know how many Tesla is the magnetic flux density of NdFeB and AlNiCo? is there any formula to calculate those?

thanx

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