Question

Consider a ferrimagnet with two inequivalent sublattices such that the molecular fields B1 and B2 on sublattice 1 and 2 are given by B1 = MOH – AM B2 = MOH - 1M2, where M1 and M2 are the magnetization on each sub- lattice and H is the applied field. However, the two sublattices have different Curie constants C1 and C2 (i.e. different values of 8,1). Show that the magnetic susceptibility is given by x=77_07[1C1 +c-97 – 2002) and find a value for the transition temperature 0. Show that if C1 = C2 = C the susceptibility reduces to that for an antiferromagnet with = XC/MO.

Answer 1

Given Curie constant $C_1$ and $C_2$ for sublattice 1 and 2.

Given

$\\B_1=\mu_0H- \lambda M_1\\ B_2=\mu_0H- \lambda M_2\\ \Rightarrow\\ H_1=H- \frac{\lambda}{\mu_0} M_1\\ H_2=H- \frac{\lambda}{\mu_0} M_2$

Now we consider antiparallel interaction between sublattice 1 and 2. Then from Curie law for each sublattice we can write

$\\TM_1=C_1H_2\\ TM_2=C_2H_1\\ \Rightarrow\\ TM_1=C_1(H- \frac{\lambda}{\mu_0} M_2)\\ TM_2=C_2(H- \frac{\lambda}{\mu_0} M_1)$

This is called ferrimagnetic interaction. We can write this equations as

These equations have a nonzero solution for $M_1$ and $M_2$ in zero applied $(H=0)$ field if

$\begin{vmatrix} T &\frac{\lambda C_1}{\mu_0} \\ \frac{\lambda C_2}{\mu_0}& T \end{vmatrix}=0$

from this we get the transition temperature

$\theta=\frac{\lambda\sqrt{C_1C_2}}{\mu_0}$

For non zero applied field $H\neq0$ we solve the previous two equations and we get solutions for $M_1$ and $M_2$ . Which are given by

$\\ M_1=\frac{\left(TC_1- \frac{\lambda C_1C_2}{\mu_0}\right)}{T^2-\frac{\lambda^2 C_1C_2}{\mu_0^2}}H\\ \\ M_2=\frac{\left(TC_2- \frac{\lambda C_1C_2}{\mu_0}\right)}{T^2-\frac{\lambda^2 C_1C_2}{\mu_0^2}}H\\$

Now magnetic susceptibility

$\\ \chi=\frac{M_1+M_2}{H}=\frac{\left[(C_1+C_2)T- \frac{2\lambda C_1C_2}{\mu_0}\right]}{T^2-\frac{\lambda^2 C_1C_2}{\mu_0^2}}=\frac{1}{T^2-\theta^2}\left[(C_1+C_2)T- \frac{2\lambda C_1C_2}{\mu_0}\right]$

Where

$\theta=\frac{\lambda\sqrt{C_1C_2}}{\mu_0}$

This proves the result.

Since antiferromagnet is a special case of ferrimagnet. Then

$C_1=C_2=C$

Then we get

$\theta=\frac{\lambda C}{\mu_0}$

This proves the result.