91Ó°ÊÓ

Probability of dipole moment to point up$\mathrm{p}(+\mathrm{μ})={\mathrm{e}}^{+\mathrm{μ\mu þ}/\mathrm{kT}}$

Probability of dipole moment to point up$\mathrm{p}(−\mathrm{μ})={\mathrm{e}}^{−\mathrm{μ\mu þ}/\mathrm{kT}}$

$\mathrm{tanhx}=\frac{{\mathrm{e}}^{\mathrm{x}}−{\mathrm{e}}^{−\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{−\mathrm{x}}}$

We will use relation for magnetization and average magnetic moment formula to derive the required expression.

Formula:

If$\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)$is the probability of occurrence of${\mathrm{x}}_{\mathrm{i}}$, then the average of x is defined as

$\mathrm{x}=\frac{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{x}}_{\mathrm{i}}â‹\dots \mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)}}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)}}$

Average magnetic moment is given by

$\mathrm{μ}=\frac{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{μ}}_{\mathrm{i}}â‹\dots \mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}$

Magnetization per unit volume is$\mathrm{M}=\left(\frac{\mathrm{μ}}{1{\mathrm{m}}^3}\right)=\mathrm{μ}$

Average magnetic moment is given by

$\mathrm{μ}=\frac{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{μ}}_{\mathrm{i}}â‹\dots \mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}$

The solid contains N atoms per unit volume, and all atoms are of the same type. Therefore, the magnetic moments due to each atom will be the same in magnitude. Hence by this argument, we set

${\mathrm{μ}}_1={\mathrm{μ}}_2={\mathrm{μ}}_3−−−−={\mathrm{μ}}_{\mathrm{N}}=\mathrm{μ}$

Using this condition, summation becomes

$\begin{array}{c}\mathrm{μ}=\frac{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{μ}}_{\mathrm{i}}â‹\dots \mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{μ}}_{\mathrm{i}}\right)}}\\ =\frac{\mathrm{\pm ·Î¼}â‹\dots \mathrm{p}\left(\mathrm{μ}\right)}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left(\mathrm{μ}\right)}}\end{array}$

Let $\mathrm{μ}=+\mathrm{μ}$if it is pointing parallel to the applied${\mathrm{B}}_{\mathrm{ext}}$field;

And$\mathrm{μ}=−\mathrm{μ}$if it is pointing anti-parallel to the applied${\mathrm{B}}_{\mathrm{ext}}$field.

Then the average magnetic moment becomes

role="math" localid="1662965537727" $\begin{array}{c}\mathrm{μ}=\frac{\mathrm{\pm ·Î¼}â‹\dots \mathrm{p}\left(\mathrm{μ}\right)}{{\displaystyle {∑}_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left(\mathrm{μ}\right)}}\\ =\frac{\mathrm{N}(+\mathrm{μ})\mathrm{p}(+\mathrm{μ})+\mathrm{N}(−\mathrm{μ})\mathrm{p}(−\mathrm{μ})}{\mathrm{p}(+\mathrm{μ})+\mathrm{p}(−\mathrm{μ})}\\ =\left\{\frac{\mathrm{\pm ·Î¼}â‹\dots \mathrm{p}(+\mathrm{μ})−\mathrm{\pm ·Î¼}â‹\dots \mathrm{p}(−\mathrm{μ})}{\mathrm{p}(+\mathrm{μ})+\mathrm{p}(−\mathrm{μ})}\right\}\end{array}$

Magnetization per unit volume is

$\begin{array}{c}\mathrm{M}=\left(\frac{\mathrm{μ}}{1{\mathrm{m}}^3}\right)\\ =\mathrm{μ}\end{array}$; it then becomes

$\begin{array}{c}\mathrm{M}=\frac{\mathrm{\pm ·Î¼}â‹\dots {\mathrm{e}}^{+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}−\mathrm{\pm ·Î¼}â‹\dots {\mathrm{e}}^{−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}}{{\mathrm{e}}^{+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}+{\mathrm{e}}^{−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}}\\ =\mathrm{\pm ·Î¼}â‹\dots \left\{\frac{{\mathrm{e}}^{+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}−{\mathrm{e}}^{−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}}{{\mathrm{e}}^{+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}+{\mathrm{e}}^{−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}}\right\}\end{array}$

$\mathrm{M}=\mathrm{\pm ·Î¼}â‹\dots \tanh \left(\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)$

Magnetization$\mathrm{M}=\mathrm{\pm ·Î¼}â‹\dots \tanh \left(\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)$

As $\mathrm{μ\mu þ}≪\mathrm{kT},$which implies$\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}≪1$

When$\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}≪1$

${\mathrm{e}}^{Â\pm \frac{\mathrm{μ\mu þ}}{\mathrm{kT}}}≈1Â\pm \frac{\mathrm{μ\mu þ}}{\mathrm{kT}}$

$\begin{array}{c}\mathrm{M}=\mathrm{\pm ·Î¼}â‹\dots \left\{\frac{\left(1+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)−\left(1−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)}{\left(1+\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)+\left(1−\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)}\right\}\\ =\frac{\left(\frac{2\mathrm{μ\mu þ}}{\mathrm{kT}}\right)}{2}â‹\dots \mathrm{\pm ·Î¼}\\ =\frac{{\mathrm{\pm ·Î¼}}^2\mathrm{B}}{\mathrm{kBT}}\end{array}$

Hence magnetization$\mathrm{M}$ for $\mathrm{μ\mu þ}≪\mathrm{kT}$is $\mathrm{M}=\frac{{\mathrm{\pm ·Î¼}}^2\mathrm{B}}{\mathrm{kBT}}$.

As $\mathrm{μ\mu þ}≫\mathrm{kT},$which implies$\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}≫1$

$\tanh \left(\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)≈1$

$\begin{array}{c}\mathrm{M}=\mathrm{\pm ·Î¼}â‹\dots \tanh \left(\frac{\mathrm{μ\mu þ}}{\mathrm{kT}}\right)\\ =\mathrm{\pm ·Î¼}\end{array}$

Magnetization$\mathrm{M}$ for $\text{μµþ}≫\text{kT}$is $\text{M}=\text{±·Î¼}$.

The graph hasthesame nature astheplot given in Fig.32.14.

We have plotted$\mathrm{M}=\mathrm{atanhx}$

Where, $\mathrm{a}$is the parameter, which can set externally to match the experimental plot