Magnetisation due to conduction electrons

In summary: Ok. I have some hints for this Q, but they're confusing me.My vague guess would have been to try and calculate the number of electrons in the higher energy state (which, I think, would be the ones aligned with the field) and multiply by \frac{\mu_{\beta}}{V} to obtain the overall magnetisation (the magnetic moment per unit volume).The hint I have here is thatN_{\pm} = \frac{1}{2}\int_{0}^{\epsilon_{F}+\mu_{B}H} \rho(\epsilon) d\epsilon = \frac{4\pi V}{3h^3} (2m)^{
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Homework Statement



Due to its spin, the electron possesses a magnetic moment μB.

Treating the conduction electrons in a metal as a free electron gas, obtain an expression for the magnetization due to the magnetic moments of the conduction electrons, when placed in a magnetic field. Evaluate this expression at zero temperature.

The Attempt at a Solution



Ok. I have some hints for this Q, but they're confusing me.

My vague guess would have been to try and calculate the number of electrons in the higher energy state (which, I think, would be the ones aligned with the field) and multiply by [tex]\frac{\mu_{\beta}}{V}[/tex] to obtain the overall magnetisation (the magnetic moment per unit volume).

The hint I have here is that

[tex]N_{\pm} = \frac{1}{2}\int_{0}^{\epsilon_{F}+\mu_{B}H} \rho(\epsilon) d\epsilon = \frac{4\pi V}{3h^3} (2m)^{3/2} (\epsilon_{F}\pm\mu_{B}H)^{3/2}[/tex]

All this seems to be doing (to a beginner) is summing the possible states, not obtaining the actual number of either spin parallel/anti-parallel electrons (ie. occupants of states). And presumably the no. of states is vast. If we took the N+ version of that, I take it we'd be getting the no. of states up to the energy level of the electrons with the higher spin energy (and thus including the states of electrons with lower spin energy); if we took the N- version, just the lower energy states. [tex]N_{+} - N_{-}[/tex] would then, I presume, give us the number of states of the higher energy spins (most of which would be unoccupied).

I am not sure why [tex]\rho(\epsilon)[/tex] has been written like this. By thinking about points in a positive octant (derivation not given here) one arrives at a no. of points

[tex]G(\epsilon) = \frac{4\pi V}{3h^{3}}(2m\epsilon)^{3/2}[/tex]

which can be expressed (I'll switch to his rho here)

[tex]\rho(\epsilon) d\epsilon = \frac{dG(\epsilon)}{d\epsilon} = \frac{4\pi V}{h^{3}}(2m\epsilon)^{1/2} d\epsilon[/tex]

[tex]\rho(\epsilon) d\epsilon[/tex], I take it, gives us the density of states. Surely we need some integral that multiplies that by a distribution function ([tex]e^{-\beta\epsilon}[/tex]?) in order to state the no. of states up to some energy [tex]\epsilon[/tex]

Well, I'm a bit confused (isn't it obvious) and not really sure what I'm doing. If someone could shed light on this question and show me how to put this together, I think it would open up some of the other problems on the sheet. (I haven't seen any examples of this sort of thing, unfortunately).

Cheers!
 
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  • #2
There is a derivation of this (I think) in Kittel, Pauli magnetism.
 
  • #3
Also, if you want an internet resource, try the online lecture notes 20 and 21 from http://spider.pas.rochester.edu:8080/phy418S05/lectures" [Broken] which is about Pauli paramagnetism of a free en gas.
 
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1. What is "magnetisation due to conduction electrons"?

Magnetisation due to conduction electrons refers to the phenomenon where a material becomes magnetised due to the motion of conduction electrons in the material. This occurs when the conduction electrons align their spins in the same direction, creating a net magnetic moment and resulting in magnetisation.

2. How does magnetisation due to conduction electrons differ from other types of magnetism?

Magnetisation due to conduction electrons is a type of paramagnetism, which is different from other types of magnetism like ferromagnetism and diamagnetism. In paramagnetic materials, the magnetic moments of individual atoms are randomly oriented, but when an external magnetic field is applied, the conduction electrons align their spins and the material becomes magnetised.

3. What factors affect the strength of magnetisation due to conduction electrons?

The strength of magnetisation due to conduction electrons depends on several factors, including the number of conduction electrons in the material, the strength of the external magnetic field, and the temperature. Higher temperatures can disrupt the alignment of the conduction electrons, leading to a weaker magnetisation.

4. How is magnetisation due to conduction electrons measured?

Magnetisation due to conduction electrons can be measured using a magnetometer, which is a device that measures the strength and direction of a magnetic field. By applying an external magnetic field to a material and measuring the resulting magnetisation, the magnetic susceptibility (a measure of how easily a material can be magnetised) can be determined.

5. What are some practical applications of magnetisation due to conduction electrons?

Magnetisation due to conduction electrons has many practical applications, including in the production of magnets and electromagnets, as well as in magnetic storage devices like hard drives. It is also used in magnetic resonance imaging (MRI) machines, which use strong magnetic fields to produce detailed images of the body's internal structures.

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