Calculate the magnetic energies of Cu and Al

In summary, the question is asking for the calculation of the magnetic energies of Cu and Al in an applied field of 5mT at 0K. However, since Cu is not a superconductor and Al has a critical field higher than the applied field, it will act as a superconductor. The formula for magnetic energy in Kittel is (Square(H))/(2μ0), but it cannot be applied to a superconductor as there can be no magnetic field. The question may be referring to a magnetic method of measuring stabilization free energy, found in Kittel p.271 of the eighth edition.
  • #1
Syed Bilal
4
0
Hi,
The Question i am trying to solve is " Calculate the magnetic energies of Cu and Al in an applied field of 5mT at 0K."
As far as i can find is that Cu is not a superconductor and Al is a superconductor and has a critical field of 15mT, which is higher than the applied applied field so it will act as super conductor and the formula for magnetic energy in Kittel is (Square(H))/ (2μ0). But how to find the magnetic energy of Cu.
 
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  • #2


Syed Bilal said:
Hi,
" Calculate the magnetic energies of Cu and Al in an applied field of 5mT at 0K."

I am wondering about the meaning of that question. In a superconductor there can be no magnetic field, so you can't calculate it's magnetic energy in that field.
 
  • #3


DrDu said:
I am wondering about the meaning of that question. In a superconductor there can be no magnetic field, so you can't calculate it's magnetic energy in that field.
I am sorry i did not explain it. Consider it as a magnetic method of measuring stabilization free energy. Its in Kittel p.271 eighth edition.
 

1. What is the formula for calculating magnetic energy?

The formula for calculating magnetic energy is E = 1/2 * μ * M^2, where E is the magnetic energy, μ is the permeability of the material, and M is the magnetic moment.

2. How do I find the permeability value for copper and aluminum?

The permeability values for copper and aluminum can be found in scientific databases or by consulting a materials handbook. The value for copper is typically around 1.2566 x 10^-6 H/m and for aluminum it is around 1.000022 x 10^-6 H/m.

3. Do I need to convert the magnetic moment into a specific unit?

Yes, the magnetic moment must be converted into the SI unit of ampere-meter squared (A*m^2). If your magnetic moment is given in other units, such as electron-volt per tesla (eV/T), you will need to convert it using the conversion factor of 1 eV/T = 1.6022 x 10^-19 A*m^2.

4. Can I calculate the magnetic energy for any shape of copper and aluminum?

Yes, the formula for calculating magnetic energy is applicable to all shapes of copper and aluminum as long as the magnetic moment and permeability values are known. However, the shape and orientation of the material may affect the distribution of magnetic energy within it.

5. Why is it important to calculate the magnetic energy of materials?

Calculating the magnetic energy of materials is important for understanding their properties and behavior in magnetic fields. It can also help in designing and optimizing devices that utilize magnetic materials, such as motors and generators.

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