Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am interface designer that occasionally steps out and does some product/concept design. The last one I'm working on is something that would require me to make a matrix of little electromagnets, and i am just starting to inform my self about all that.

For the project I'm playing with right now, I used 1x1x1cm neodymium magnet for experimenting, and it did the job just perfectly. I tried bigger, I tried smaller, but this one is perfect. The problem is that i need an electromagnet to do the same thing in future. (probably lots of them)

The question is, how large does an electromagnet needs to be in volume to have the same/similar power as neodymium magnet sized 1x1x1cm. Or even better question, how small can it be, because I really need it small but powerful.

The point of the question is just for me to hear someone experienced on the subject with some theoretical answer (Because i assume it will be enough for this stage of exploration). The real work will be done in collaboration with some professional guy just like you that I will look for if i decide to dive into all this.

Now I am aware of the fact that there are probably different ways you can measure one magnets strength, and that not all neodymium magnets are the same, so it can't be treated as a constant, but I would just like to know in general, is it like 2ccm electromagnet or is it like 20ccm electromagnet.

share|improve this question

2 Answers 2

On the one hand, @sai's reasoning seems sound, on the other hand, we should not forget that an electromagnet can have a ferromagnetic core. I cannot offer any specific data right now, but I would suspect that an electromagnet with a suitable core might have comparable strength to that of a permanent magnet. The core can be magnetically soft, so it can have little average permanent magnetization when the electromagnet is switched off.

Using an electromagnet, you need extra power and/or cooling, but you also have extra control (you can vary or switch on/off the magnetic field at will).

share|improve this answer

It's probably very very late and I don't understand why this question has been unanswered for so long but I had a similar issue a while ago, so here goes.

To estimate the field at the surface of the Nd magnet, I used the K&J Magnetics calculator. For a 1 cm x 1 cm x 1 cm block, the surface field is ~0.6 T or 6000 Gauss.

Now let's consider a simple solenoid of outer diameter 1 cm and height 1 cm. If we use 50 µm copper wire, we can make about 200 turns per wind. If we make about 100 windings, we'll have about 20,000 turns, which is crazy. I'd say that you can't put more than 1 mA through a coil like that without any cooling. Let's use the infinite solenoid formula, $$B=\mu_0 \frac{NI}l,$$ to get an order of magnitude for the field. Our real field is going to be smaller.

I get ~10-3 T or 10 Gauss. So it's pretty much hopeless.. Permanent magnets are way too strong.

share|improve this answer
    
That's true, conventional solenoids a problematic, but you can either cool it and get way beyond 1 mA, but not way (if any) beyond 1A, or you can use superconductors, and that way you can get way higher than with copper wire. –  user23873 Jul 13 '13 at 13:42
    
Superconductors will start to conduct once the magnetic field gets beyond a critical point, which is why high magnetic fields are created with copper. –  lionelbrits Nov 10 '13 at 18:08
    
@lionelbrits: You probably mean they lose superconductivity (although after that they can indeed conduct, but as a normal metal, not a superconductor). –  akhmeteli Nov 10 '13 at 19:14
    
Yes, I meant that. –  lionelbrits Nov 11 '13 at 2:43
    
@lionelbrits The critical fields are way larger than the fields we're talking about here. These days superconducting solenoids reach ~20T. In any case the cryogenic cooling needed for superconductors makes them impractical for OP's use. –  sai Nov 11 '13 at 5:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.