Calculating Work for Permanent Magnet Fields

In summary, the magnet loses energy when a sample is placed in the field and the energy needed to change the magnetization is quantified. There is not much energy in small magnets, 10 J doesn't sound so wrong (you did not give the volume).This magnet is used for magnetic refrigeration where a magnetocaloric material is heated up to produce cold.
  • #1
pomekrank
13
0
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
 
Physics news on Phys.org
  • #2
pomekrank said:
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
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
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
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
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
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
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
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 doesn't it need energy to change orientation ?
 
  • #10
pomekrank said:
You wrote about dipole reorientation, but doesn't 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
and how do you quantify it ?
 
  • #12
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.
 

1. What is work done by a permanent magnet?

The work done by a permanent magnet refers to the energy required to move an object against the force of the magnet. This force is known as the magnetic force and is created by the alignment of magnetic domains within the magnet.

2. How is work done by a permanent magnet calculated?

The work done by a permanent magnet is calculated by multiplying the magnetic force by the distance the object is moved against the force. This can be represented by the equation W = Fd, where W is the work done, F is the magnetic force, and d is the distance.

3. What is the relationship between work done and the strength of the permanent magnet?

The work done by a permanent magnet is directly proportional to the strength of the magnet. This means that the stronger the magnet, the more work it can do in moving an object against its force.

4. Can work be done by a permanent magnet without moving an object?

Yes, work can be done by a permanent magnet without moving an object. This is known as magnetic potential energy, which is the energy stored in a magnetic field. This energy can be released when the magnet interacts with another object or magnetic field.

5. What are some real-life applications of work done by permanent magnets?

Permanent magnets are used in a variety of applications that require work to be done. Some examples include electric motors, generators, and MRI machines. They can also be found in everyday objects such as speakers, refrigerator magnets, and magnetic door latches.

Similar threads

  • Electromagnetism
Replies
3
Views
993
Replies
2
Views
470
  • Electromagnetism
Replies
23
Views
2K
  • Electromagnetism
Replies
13
Views
1K
Replies
5
Views
801
  • Electromagnetism
Replies
7
Views
1K
Replies
24
Views
2K
Replies
14
Views
2K
  • Electromagnetism
Replies
1
Views
649
Replies
15
Views
3K
Back
Top