How Does Magnetic Moment Change with Temperature and Field Strength?

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SUMMARY

The discussion focuses on calculating the magnetic moment of a crystal under varying temperature and magnetic field conditions. The derived mean magnetic moment formula is M = Nμ (2 sinh(μβ/kT) / (1 + cosh(μβ/kT))). At high temperatures and weak fields, the magnetic moment is approximated as M = (2Nμ2B) / (3kT), while at low temperatures and strong fields, it simplifies to M = Nμ. The discussion emphasizes the significance of the Curie law at high temperatures and the predominance of the ground state at low temperatures.

PREREQUISITES
  • Understanding of magnetic moment and its significance in physics
  • Familiarity with hyperbolic functions and their limits
  • Knowledge of the Curie law and its implications for paramagnetic materials
  • Basic grasp of statistical mechanics, particularly the Boltzmann factor
NEXT STEPS
  • Explore the derivation of the Curie law in detail
  • Study the behavior of hyperbolic functions in limit cases
  • Investigate the properties of paramagnetic materials under varying magnetic fields
  • Learn about the implications of temperature on magnetic susceptibility
USEFUL FOR

Students and researchers in physics, particularly those focusing on magnetism, statistical mechanics, and material science, will benefit from this discussion.

patrickmoloney
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Homework Statement


Find the magnetic moment of a crystal when placed

(i) in a weak field at high temperature

(ii) in a strong field at low temperature

Homework Equations



This is the last part of a question which I feel I have solved correctly up until this point.

The mean magnetic moment I found is

\langle M \rangle = N_{\mu} \frac{2 \sinh \Big{(}\dfrac{\mu \beta}{kT}\Big{)}}{1+ \cosh \Big{(}\dfrac{\mu \beta}{kT}\Big{)}}

The Attempt at a Solution



Well at high temperature and low magnetic field strength

\dfrac{\mu \beta}{kT} \ll 1

And at low temperature high magnetic field strength

\dfrac{\mu \beta}{kT} \gg 1

What happens to the hyperbolic functions as one goes to 0 and one goes to \infty?

EDIT: I know what happens as \lim_{x \rightarrow \infty}\sinh x = e^x and \lim_{x \rightarrow \infty}\cosh x = e^x
 
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I assume your crystal is paramagnetic. You might see what's going more easily if you cast the hyperbolic functions in terms of exponentials, then take the limits.
 
(i)Weak field at high temperature

\dfrac{\mu B}{kT} \ll 1 \implies e^{\frac{\mu B}{kT}} \approx 1 \pm \dfrac{\mu B}{kT}

From my mean magnetic momentM= N\mu \dfrac{e^{\frac{\mu B}{kT}}-e^{\frac{\mu B}{kT}}}{1 + e^{-\frac{\mu B}{kT}} + e^{\frac{\mu B}{kT}}}

substituting e^{\frac{\mu B}{kT}} = 1 \pm \dfrac{\mu B}{kT}

M = \dfrac{1+\frac{\mu B}{kT} -1 + \frac{\mu B}{kT}}{1+1-\frac{\mu B}{kT} + 1 + \frac{\mu B}{KT}} = \dfrac{2N\mu^2 B}{3kT}

(ii) Strong field at low temperature

\dfrac{\mu B}{kT} \gg 1 \implies e^{\frac{\mu B}{kT}} \gg 1 \pm e^{-\frac{\mu B}{kT}} \gg 1

Hence

M= N\mu \dfrac{e^{\frac{\mu B}{kT}}}{e^{\frac{\mu B}{kT}}}= N\mu
 
Looks right. At the high temperature end you get the 1/T Curie law dependence. At low temperature, only (mostly) the ground state is occupied and the magnetization is the number of magnetic moments times the value of one moment.
 
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kuruman said:
Looks right. At the high temperature end you get the 1/T Curie law dependence. At low temperature, only (mostly) the ground state is occupied and the magnetization is the number of magnetic moments times the value of one moment.
Much appreciated. Insight into these types of problems helps a lot. Thanks
 

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