Work done by permanent magnet

  • #1
13
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Hi,

I would like to evaluate the required work to create a 1 T field with a permanent magnet. I saw a few formulas on the internet, but I don't know how to use them correctly. The main one are
[itex]B={{\mu }_{0}}\left( H+M \right)[/itex]
[itex]{{W}_{mag}}=\frac{{{B}^{2}}}{2{{\mu }_{0}}}Vol[/itex]
Let's say I know all the properties of the magnet (remanence, volume, density, etc.) and also the magnetization as a function of B and the temperature. If I use the 2nd formula to compute, the magnitude of the work is about 10 J, which is insanely low.

How do you proceed to evaluate this work correctly ?

Thank you
 

Answers and Replies

  • #2
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If I use the 2nd formula to compute, the magnitude of the work is about 10 J, which is insanely low.
There is not much energy in small magnets, 10 J doesn't sound so wrong (you did not give the volume).
 
  • #3
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This magnet is used for magnetic refrigeration where a magnetocaloric material is heated up to produce cold. However, a 1T field produces about ΔT = 2.5 K, which means the energy for enthalpy change is about Q = m*cp*ΔT = (0.25kg)(240 J/kg-K)(2.5) = 250 J. So it makes no sense that I can have 250J of energy from a 10J field... Yes I didn't take into account the efficiencies and all losses, but the order of magnitude is just way to much.
 
  • #4
34,987
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A permanent magnet is permanent - it doesn't lose or gain energy in the process.

What did you use as volume and where does the cp value come from?
 
  • #5
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I got to say that I don't understand the physics of the process too well. If you put gadolinium in the field, the temperature increases. But where does that energy come from and how do you evaluate it ? I mean, it's not restore to the magnet when the sample is taken away of the field since the heat is loss in the environment...
Also, I find tricky that never the physical properties of the magnet be considerated in the equation (somehow the alloy should influence the results).
To answer your question, I use the size of the minimum possible size of sample as the volume (vol = m/ρ=200g/7900kg/m3=2.5 e-5 m3) and cp is the typical value at 293K and 1 T (see attach for the curve).
upload_2015-10-8_18-32-45.png
 
  • #6
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The magnetic dipoles inside get oriented along the external field, so they lose some energy that gets released as heat. Breaking this order needs energy again, which cools the material.
Doing that at 300 K doesn't sound interesting, as there are better cooling methods available at that temperature.
 
  • #7
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Let's make it sample and reformulate the problem. I put a sample of material into a 1T magnetic field. The polarization of the sample thus change, so what is the energy transfert from the permanent magnet to the material ? I mean, how much energy is absorbed ?
 
  • #8
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Why do you expect an energy transfer from/to the permanent magnet? Its field is just there and does not change (significantly) if you move your material around.
 
  • #9
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Ok but how is it possible that in case of magnetocaloric material the temperature increases if no energy is added to the system (system = sample) ? You wrote about dipole reorientation, but doesnt it need energy to change orientation ?
 
  • #10
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You wrote about dipole reorientation, but doesnt it need energy to change orientation ?
It needs energy or releases it, depending on the direction of change. That's also the energy source.
 
  • #11
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and how do you quantify it ?
 
  • #12
34,987
11,174
Like that
I guess it is possible to model it with quantum mechanics, but I wouldn't expect an easy way to predict the size of the effect for arbitrary materials.
 

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