Magnetic moment due to applied external magnetic field

AI Thread Summary
To determine the externally applied magnetic field (B_{0}) necessary for 51% of metal ions in CuSO4 to align their magnetic moments, the relationship between magnetization (M), magnetic susceptibility (χ), and the applied field must be analyzed. The calculation involves establishing that M equals 0.51H and recognizing that χ is independent of B_{0}. The discussion highlights the need to calculate the ratio of atoms in the spin-up state (p) to the total number of atoms (N), leading to the equation p = n1/N. By combining this with statistical mechanics, the final expression for B_{0} can be derived as B_{0} = (kB T/μ)(p - q), where μ is determined using the gyromagnetic ratio. This approach provides a method to achieve the desired magnetic alignment in the given scenario.
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Homework Statement



How large externally applied magnetic filed (B_{0}) is necessary in otder for 51% of the metal ions in CuSO(_{4}) to have their magnetic moments oriented in the same direction as the applied field when the salt is kep at room temperature?

Homework Equations



B=\mu_{0}(H+M)

\chi=\frac{N}{V}\frac{(p\mu_{B})^{2}\mu_{0}}{3k_{B}T}

The Attempt at a Solution



rearanging the relation

As i understand it i want

M=0,51H

then, using

\chi B_{0}=\mu_{0}M

i find that

\chi=0,51

But \chi is, as i understand it, independent on applied magnetic field B_{0} so this reasoning can't be right. However i fail to see what I'm missing.
 
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Maybe i should say this is Introductory Solid state physics. I'd modify the title but I'm not sure i can?
 
I think i figured this out, so i'll post the results here in case someone would find it useful.

To calculate the amount of atoms with a spin state corresponding to the applied magnetic field one would calculate the amounting magnetisation due to the number of atoms existing in that desired state.

So that the ratio of atoms per volume existing in spin up n_{1} to the total amount of atoms per volume N becomes:

p=\frac{n_{1}}{N}

Where in the above stated problem one would have p=0.51, and the remaining atoms existing in a spin down state is simply

q=1-p

The the resulting magnetisation would be

M=μN(p-q)

Combining this with the statistical calculation of the number of states one gets

μN(p-q)=μNtanh(\frac{μB_{0}}{k_{B}T})\approx μN(\frac{μB_{0}}{k_{B}T})

\Rightarrow B_{0}=\frac{K_{B}T}{μ}(p-q)

And after determining μ (with the gyromagnetic ratio etc.) one would obtain the desired result
 
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