 Solution

ID: past-exam-of-the-general-relativity-and-cosmology-and-condensed-matter-physics-paper-of-part-b-of-the-physics-course-of-the-university-of-oxford/2011/section-vi/8/ii/solution

The

1/ x

terms in the Brillouin function cancel. Keeping the linear terms gives

B_{J} (x) = \frac{( 2 J + 1 ) ^{2} - 1}{12 J ^{2}} x + O (x^{3}) = \frac{J + 1}{3 J} x + O (x^{3}) .

Consequently,

M ≃ \frac{n ( g μ _{B} ) ^{2} J ( J + 1 )}{3 k _{B} T} B .

Use the dimensionless SI magnetic susceptibility

χ = \partial M / \partial H_{applied}

and

B_{applied} = μ_{0} H_{applied}

, as in the source's molecular-field convention. The Curie law is

χ = \frac{C}{T}, C = \frac{μ _{0} n ( g μ _{B} ) ^{2} J ( J + 1 )}{3 k _{B}} .

If susceptibility is instead defined as

\partial M / \partial B

C / μ_{0}

and is dimensionful. The SI convention above makes the subsequent Curie–Weiss law and

λ

consistent.

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