Ferromagnetism — Question 6

Question 6

A supplied critical law determines how spontaneous magnetization vanishes

For temperatures below Tc, a simplified model gives $M_s(T)=M_0(1-T/T_c)^\beta$ with positive beta. Measurements give $M_s(0.75T_c)=M_0/2$ and $M_s(0.96T_c)=M_0/5$. Assume the supplied model is valid at all temperatures requested below.

1. Infer beta from each measurement and check consistency.

2. Find the temperature where Ms is M0/10 and calculate dMs/dT there in units of M0/Tc.

Question 6 - Solution

A supplied critical law determines how spontaneous magnetization vanishes

(a) Method and calculation

$$1/2=(1/4)^\beta\Rightarrow\boxed{\beta=1/2}.$$$$1/5=(0.04)^{1/2}\Rightarrow\boxed{\text{the second measurement agrees}}.$$

(b) Interpretation and further calculation

$$0.1=\sqrt{1-T/T_c}\Rightarrow\boxed{T=0.99T_c}.$$$$\frac{dM_s}{dT}=-\frac{M_0}{2T_c}(1-T/T_c)^{-1/2}\Rightarrow\boxed{\left.\frac{dM_s}{dT}\right|_{0.99T_c}=-5M_0/T_c}.$$

Physical check

The model describes spontaneous magnetization below Tc; it must not be continued to a negative square-root argument above Tc.