GATE 2021 PH – Question 48
The free energy of a ferromagnet is $F=F_0+a_0(T-T_C)M^2+bM^4$, where $F_0,a_0,b$ are positive constants, $M$ is magnetization and $T_C$ is the Curie temperature. Which graph best depicts $M^2$ versus $T$?
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Correct answer: (B) Figure (B)
Explanation
At equilibrium $\partial F/\partial M=2a_0(T-T_C)M+4bM^3=0$. Below $T_C$ the stable nonzero solution satisfies $M^2=a_0(T_C-T)/(2b)$, a straight line decreasing to zero at $T_C$. Above $T_C$, the stable solution is $M=0$. The linear descending branch is B.