# Magnetic moment due to applied external magnetic field

by Kentaxel
Tags: applied, external, field, magnetic, moment
 P: 12 1. The problem statement, all variables and given/known data 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? 2. Relevant equations $B=\mu_{0}(H+M)$ $\chi=\frac{N}{V}\frac{(p\mu_{B})^{2}\mu_{0}}{3k_{B}T}$ 3. 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.
 P: 12 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