Discussion
Consider 5 independent magnetic moments. Each moment has only two directions +m and -m. Hamiltonian of such system is given by \(x = \sum_{i=1}^{5} m_{i}\). If the total magnetization of the system is equal to m, the entropy is equal to :
Answer: A
The entropy is k ln(10).
Let \(n_{+}\) moments point along +m and \(n_{-}\) along -m. A total magnetization of m requires
\[ \begin{aligned} n_{+} + n_{-} &= 5, \quad n_{+} - n_{-} = 1 \\ n_{+} &= 3, \quad n_{-} = 2 \end{aligned} \]
The number of microstates is the number of ways to choose which 2 of the 5 moments point down:
\[ \begin{aligned} \Omega &= \binom{5}{2} = 10 \\ S &= k\ln\Omega = k\ln 10 \end{aligned} \]
Zero would require a unique arrangement, as for total magnetization 5m. The value k ln 5 counts arrangements with only one moment reversed (total 3m), and k ln 20 double counts the arrangements.
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