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## Q. 10.6

Use the Bohr magneton $\mu_{ B }=e \hbar / 2 m=9.27 \times 10^{-24}$ J/T as a typical magnetic moment and B = 0.50 T to do the following:

(a) Find the temperature T at which $\mu B=0.1 k T$.

(b) For $\mu B=0.1 k T$ compare $\tanh (\beta \mu B) \text { with } \beta \mu B$ and thereby check the suitability of this value of T as a “classical” temperature.

(c) Make the same comparison in the expression $\tanh (\beta \mu B) \approx \beta \mu B \text { at } T=100 K$.

Strategy With the numerical values of the magnetic moment and magnetic field given, the computations are all straightforward. For the comparisons called for in parts (b) and (c), the given values suggest using two significant figures.

## Verified Solution

(a) We are given that $\mu B=0.1 k T$. We let $\mu=\mu_{ B }$ and solve for T:

$T=\frac{10 \mu B}{k}=\frac{10\left(9.27 \times 10^{-24} J / T \right)(0.5 T )}{1.38 \times 10^{-23} J / K }=3.36 K$

(b) Now

$\begin{gathered}\beta \mu B=\frac{\mu B}{k T}=0.10 \\\tanh (\beta \mu B)=\tanh (0.10)=0.10\end{gathered}$

to two significant digits. Therefore we conclude that the approximation is a good one for most purposes, even at this low temperature.

(c) Now

\begin{aligned}\beta \mu B=\frac{\mu B}{k T} &=\frac{\left(9.27 \times 10^{-24} J / T \right)(0.5 T )}{\left(1.38 \times 10^{-23} J / K \right)(100 K )} \\&=3.36 \times 10^{-3} \\\tanh (\beta \mu B) &=\tanh \left(3.36 \times 10^{-3}\right)=3.36 \times 10^{-3}\end{aligned}

so that $\beta \mu B \text { and } \tanh (\beta \mu B)$ are the same to three significant figures. At these relatively high temperatures the approximation is an excellent one.