Force between two identical magnets vs. IRON 4

[dbostic]

Hello all.

I have a basic question about magnetism. I know very little about magnetic fields.

Let's say I have two identical (shape, size, field-strength) magnets. They are positioned to be attracted to each other (with some force N).

If I replace one of the magnets with an identically-shaped piece iron (non-magnetized), what will be the attractive force N ?

Thanks in Advance!

 

[MagMike]

If the two shapes were in contact with each other, then the attractive force will be less than 0.5N
If the two shapes were separated by an air gap, than the attractive force will be even lower. It's not possible to give a quantitative number, but it could be as low as 0.1N

 

[electricpete]

If the two shapes were in contact with each other, then the attractive force will be less than 0.5N
If the two shapes were separated by an air gap, than the attractive force will be even lower. It's not possible to give a quantitative number, but it could be as low as 0.1N

I'm lost. What is it that tells us the force is that low?

=====================================
(2B)+(2B)' ?

 

[MagMike]

electricpete: I was assuming the N was an arbitrary value. I see know that it could be interpreted at Newtons. It would be better if the OP and I called it X instead.

dbostic: Can you confirm this?

 

[IRstuff]

The force is proportional to B[sup]2[/sup]*area, see:

http://en.wikipedia.org/wiki/Force_...ween_two_nearby_magnetized_surfaces_of_area_A

There are online calculators:

http://www.intemag.com/pull.html

TTFN
faq731-376

 

[RyreInc]

Well I don't know the answer, but I just ran a quick FEA sim with two cylindrical objects having the same diameter as their lengths. With two N45 magnets, the resulting force is 17.6N, with one magnet and (non-saturating) steel it's 11.0N. So that goes against MagMike's assertion...

 

[MagMike]

RyreInc: Interesting! You motivated me to check my numbers. I modeled N42 discs, 1" diameter x .25" length, oriented through the length. 3D FEA modeling software.

Two N42 discs are in contact: 28 lbs. attraction
One N42 disc in contact with similarly sized iron (steel) disc: 12 lbs

Two N42 discs separated by .250": 7.3 lbs
One N42 disc and one similarly sized iron (steel) disc separated by .250": 1.3 lbs

Please let me know the dimensions of your model and I run them through my modeling software.

 

[RyreInc]

MagMike:

My magnets were cylindrical, 0.2 dia x 0.2 length, also magnetized through length. The units were meters since that was the default, but that doesn't really matter.

I suspect the difference is due to the aspect ratio of the magnets--the ones I simulated will have an operating point closer to Br, and will utilize more of the steel.

I ran another simulation with the same aspect ratio as yours, and with steel I indeed get a force less than half that as between two magnets: 2.05 vs. 0.91.

 

[IRstuff]

Note that the area of the steel ought to have some effect, since that would affect how the lines are terminated.

TTFN
faq731-376

 

[dbostic]

All,

You guys are great. Thanks for digging into this so quickly. In my original post (OP you guys call it?), my force N was indeed an unknown variable; we'll call it X from now on . I had no idea there was software to calculate these forces. Engineering is a big field, I guess. ;-) Neither did I suspect that shape was a significant factor.

For the purposes of this problem, let's assume each magnet (or steel blank) is a disc 8mm dia, 3mm height, with the field magnetized through the disc face. Anyone wanna run the numbers again? If so, thanks. I think it's an interesting problem for laypeople. I had previously never stopped to wonder how the forces were different between magnetized and unmagnetized objects. It's nice to know.

Actually, I'm trying to figure out if I need two magnets for this project, or if one is fine. I'm trying not to have overkill; neither am I trying to optimize prematurely. Ah, the eternal struggle of the engineer. Thanks again.

 

[MagBen]

Software doensot work well at some conditions, especially 2D is not good for "small" magnets. The important thing is to understand the principle.
for an infinite long magnet, and also the steel is infinite long, the force between two magnets, equals the force between one magnet and the steel, and force can be calculated as .577 x (Br)^2 x S (lbs). Br the remanence (kG), S contact area (inch^2). The result is the max force.
In practical, and for a limited length, the force between two identical magnets is always bigger than the force between one magnet and the steel, simply because the flux density, B, at the contact area in the former situation is bigger than in the latter situation. A 2:1 relationship doesnot exist.

 

[MagBen]

for a disc 8mm dia, 3mm height, assuming Br=13,000Gs,
magnet-steel, the force will be 1.3 lbs
magnet-magent, force=2.8 lbs

 

[MagBen]

sorry, for magnet-steel in contact, the force will be 2.8 lbs

 

[IRstuff]

Note that there are backing iron options that make huge differences in holding force of magnets:

http://www.magneticsolutions.com.au/magnet-force-calc.html

TTFN
faq731-376

 

[MagBen]

-IRstuff,
You are correct, it is about the designing a close magnetic ciucuit so the useful B can be close to Br, and therefore, a max holding force. A magnet with two iron pole pices serves this purpose. An AlNiCo hodling magnet is normally designed with multi-poles for this purpose. Also, this design is to increase the effective L/D ratio to increase Bd (operating B point), i.e. increase permanence coefficient.

 

[dbostic]

MagBen,

Thanks for your comment. I had no idea about magnet design for different purposes. I suppose it's like any engineering field - until one digs into it, one has no idea about the complexity of the field.

For my particular application, I'm convinced I won't be able to use two magnets; it will be one magnet, one piece of metal (steel, iron, soft iron). I will post a new question shortly. Thanks again.