How to Calculate Work of Isothermal Magnetization?

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SUMMARY

The discussion focuses on calculating the work of isothermal magnetization for a magnetic salt obeying the Curie Law, represented by the equation W = VμoT(M22-M12)/2C. The user initially attempts to derive this formula using W = -∫Hdm from M1 to M2, where H = MT/C. Clarifications are sought regarding the inclusion of μo and volume V in the work equation, as well as the interpretation of the negative sign in relation to the work done by or on the system.

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  • Understanding of Curie Law and its implications in magnetization
  • Familiarity with the concepts of work and energy in thermodynamics
  • Basic knowledge of calculus for integration
  • Proficiency in magnetic properties and SI units, particularly permeability (μo)
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  • Research the role of permeability (μo) in magnetic field calculations
  • Explore the differences between work done by the system versus work done on the system in thermodynamic contexts
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TristanJones
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Homework Statement



A magnetic salt obeys the Curie Law
μoM/Bo = C/T
Where M is the magnetization, Bo is the applied magnetic field in the absence of the specimen, C is a constant, and μo is the permeability of free space. The salt is magnetized isothermally from a magnetization M1 to M2. You may assume that the magnetization is uniform over the volume V of the salt. Show that the work of magnetization is:
W = VμoT(M22-M12)/2C

Homework Equations


W = --∫Hdm, from M1 to M2. H= MT/C

The Attempt at a Solution


To start, I use W = -∫Hdm, from M1 to M2. using H = MT/C and integrating, I end up with W = -T(M22-M12)/2C. How do I also get μo and V in the numerator, as well as removing the negative sign? What am I missing here?
(not necessarily looking for a blatant answer so much as a nudge in the right direction.)

Cheers.
 

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TristanJones said:

Homework Equations


W = --∫Hdm, from M1 to M2.
Are you sure that this equation shouldn't also include μ0 and V?
When written with the negative sign as you did above, does this equation represent the work done by the system or the work done on the system?
 
I'm using an equation provided in my notes. If the equation is supposed to already include V and μ0 then... well, that would be great, but I need the basis for it, which I'm having trouble figuring out.

And with the negative sign, I think it represents the work performed by the material itself, so perhaps I should treat it as external instead?
 
well, I guess that answers it! I must've been interpreting what the question was asking incorrectly and using an incorrect formula. the one described in that link fits this much better, I think.
 
Your notes might have been dealing with the work per unit volume. But, still, it should have had the μ0 if working in the SI system. The wording of the problem is not very clear about whether they are asking for work done on the system or by the system.
 

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