Magnetic moment due to applied external magnetic field

[Kentaxel]

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.

 

[Kentaxel]

Maybe i should say this is Introductory Solid state physics. I'd modify the title but I'm not sure i can?

 

[Kentaxel]

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