Calculate the magnetic energies of Cu and Al

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SUMMARY

The discussion focuses on calculating the magnetic energies of Copper (Cu) and Aluminum (Al) in an applied magnetic field of 5mT at absolute zero (0K). It is established that Copper is not a superconductor, while Aluminum is a superconductor with a critical field of 15mT, allowing it to exhibit superconducting properties under the given conditions. The formula for magnetic energy referenced is derived from Kittel's textbook, specifically (H²)/(2μ0). However, the calculation for Copper's magnetic energy remains unresolved due to its non-superconducting nature.

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  • Understanding of superconductivity and critical magnetic fields
  • Familiarity with magnetic energy calculations
  • Knowledge of Kittel's "Introduction to Solid State Physics" eighth edition
  • Basic concepts of magnetic fields and their effects on materials
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  • Study the principles of superconductivity and its implications on magnetic fields
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Syed Bilal
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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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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.
 


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.
 

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