 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-b/b1/b/solution

Write

y = g μ_{B} B / (k_{B} T)

Z_{1} = \sum_{m = - J}^{J} e^{- y m} = \frac{s i n h [( J + 1/2 ) y ]}{s i n h ( y /2 )} .

Since

⟨ m ⟩ = - \partial_{y} lo g Z_{1}

⟨ μ_{z} ⟩ = g μ_{B} [(J + \frac{1}{2}) coth ((J + \frac{1}{2}) y) - \frac{1}{2} coth (y /2)], M = N ⟨ μ_{z} ⟩ .

Here

M

is the total magnetic moment used by the question; the magnetization per unit volume is

M / V

. For independent distinguishable atoms,

Z_{N} = Z_{1}^{N}

, with no indistinguishability factor.

Putting

J = 1/2

yields

M = N g μ_{B} [coth y - \frac{1}{2} coth (y /2)] = \frac{N g μ _{B}}{2} tanh (y /2) .

B = 0

cancel; the finite sum makes the zero-field limit explicit.^[ref]

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