Q45P - Fundamentals Of Physics

Question

Consider a solid containing N atoms per unit volume, each atom having a magnetic dipole moment $\overrightarrow{\mathbf{\mu }}$. Suppose the direction of  $\overrightarrow{\mathbf{\mu }}$ can be only parallel or anti-parallel to an externally applied magnetic field (this will be the case if is due to the spin of a single electron). According to statistical mechanics, the probability of an atom being in a state with energy U is proportional to $\stackrel{}{{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{U}}{\mathbf{KT}}}}$, where T is the temperature and k is Boltzmann’s constant. Thus, because energy U is, the fraction of atoms whose dipole moment is parallel to is proportional to ${\mathbf{e}}^{\frac{\mathbf{\mu B}}{\mathbf{KT}}}$ and the fraction of atoms whose dipole moment is anti-parallel to is proportional to ${\mathbf{e}}^{\frac{\mathbf{-}\mathbf{\mu B}}{\mathbf{KT}}}$. (a) Show that the magnitude of the magnetization of this solid is $\mathbf{M}\mathbf{=}\mathbf{\mu N}\mathbf{ }\mathbf{\text{tanh}}\left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)$. Here tanh is the hyperbolic tangent function: $\mathbf{tanh}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{)}\mathbf{/}\mathbf{(}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{)}$(b) Show that the result given in (a) reduces to $\mathbf{M}\mathbf{=}{\mathbf{N\mu }}^{\mathbf{2}}\mathbf{B}\mathbf{/}\mathbf{kT}$ for  $\mathbf{\mu B}\mathbf{\ll }\mathbf{kT}$(c) Show that the result of (a) reduces to  $\mathbf{M}\mathbf{=}\mathbf{N\mu }$ for  $\mathbf{\mu B}\mathbf{\gg }\mathbf{kT}$.(d) Show that both (b) and (c) agree qualitatively with Figure.

 

 

Step-by-Step Solution

Answer

1. Magnetization $\mathrm{M}=\mathrm{N\mu }\cdot \tanh \left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)$.
2. Magnetization $\mathrm{M}$ for $\mathrm{\mu B}\ll \mathrm{kT}$ is, $\mathrm{M}=\frac{{\mathrm{N\mu }}^2\mathrm{B}}{\mathrm{kBT}}$.
3. Magnetization $\mathrm{M}$ for $\mathrm{\mu B}\gg \mathrm{kT}$ is,$\mathrm{M}=\mathrm{N\mu }$ .
4. The plotted graph has a similar nature, so (b) and (c) agree qualitatively with figure 32-14.

 

Step 1: Listing the given quantities

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

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

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

Step 2: Understanding the concepts of magnetization

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 

$\mathrm{x}=\frac{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{x}}_{\mathrm{i}}\cdot \mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)}}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{x}}_{\mathrm{i}}\right)}}$

Average magnetic moment is given by 

$\mathrm{\mu }=\frac{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{\mu }}_{\mathrm{i}}\cdot \mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}$

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

Step 3: (a) Calculations of the magnitude of the magnetization

Average magnetic moment is given by 

$\mathrm{\mu }=\frac{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{\mu }}_{\mathrm{i}}\cdot \mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}$

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 

${\mathrm{\mu }}_1={\mathrm{\mu }}_2={\mathrm{\mu }}_3---- ={\mathrm{\mu }}_{\mathrm{N}} =\mathrm{\mu }$

Using this condition, summation becomes 

$\begin{array}{c}\mathrm{\mu }=\frac{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}{\mathrm{\mu }}_{\mathrm{i}}\cdot \mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left({\mathrm{\mu }}_{\mathrm{i}}\right)}}\\ =\frac{\mathrm{N\mu }\cdot \mathrm{p}\left(\mathrm{\mu }\right)}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left(\mathrm{\mu }\right)}}\end{array}$

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

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

Then the average magnetic moment becomes 

$\begin{array}{c}\mathrm{\mu }=\frac{\mathrm{N\mu }\cdot \mathrm{p}\left(\mathrm{\mu }\right)}{{\displaystyle {\sum }_{\mathrm{i}=1}^{\mathrm{N}}\mathrm{p}\left(\mathrm{\mu }\right)}}\\ =\frac{\mathrm{N}(+\mathrm{\mu })\mathrm{p}(+\mathrm{\mu })+\mathrm{N}(-\mathrm{\mu })\mathrm{p}(-\mathrm{\mu })}{\mathrm{p}(+\mathrm{\mu })+\mathrm{p}(-\mathrm{\mu })}\\ =\left\{\frac{\mathrm{N\mu }\cdot \mathrm{p}(+\mathrm{\mu })-\mathrm{N\mu }\cdot \mathrm{p}(-\mathrm{\mu })}{\mathrm{p}(+\mathrm{\mu })+\mathrm{p}(-\mathrm{\mu })}\right\}\end{array}$

Magnetization per unit volume is

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

$\begin{array}{c}\mathrm{M}=\frac{\mathrm{N\mu }\cdot {\mathrm{e}}^{+\frac{\mathrm{\mu B}}{\mathrm{kT}}}-\mathrm{N\mu }\cdot {\mathrm{e}}^{-\frac{\mathrm{\mu B}}{\mathrm{kT}}}}{{\mathrm{e}}^{+\frac{\mathrm{\mu B}}{\mathrm{kT}}}+{\mathrm{e}}^{-\frac{\mathrm{\mu B}}{\mathrm{kT}}}}\\ =\mathrm{N\mu }\cdot \left\{\frac{{\mathrm{e}}^{+\frac{\mathrm{\mu B}}{\mathrm{kT}}}-{\mathrm{e}}^{-\frac{\mathrm{\mu B}}{\mathrm{kT}}}}{{\mathrm{e}}^{+\frac{\mathrm{\mu B}}{\mathrm{kT}}}+{\mathrm{e}}^{-\frac{\mathrm{\mu B}}{\mathrm{kT}}}}\right\}\end{array}$

$\mathrm{M}= \mathrm{N\mu }\cdot \tanh \left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)$

Magnetization $\mathrm{M}=\mathrm{N\mu }\cdot \tanh \left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)$

Step 4: (b) Calculations of the magnetization M for role="math" localid="1662965751913" μB ≪ kT

As  $\mathrm{\mu B}\ll \mathrm{kT},$ which implies $\frac{\mathrm{\mu B}}{\mathrm{kT}}\ll 1$

When $\frac{\mathrm{\mu B}}{\mathrm{kT}}\ll 1$

${\mathrm{e}}^{\pm \frac{\mathrm{\mu B}}{\mathrm{kT}}}\approx 1\pm \frac{\mathrm{\mu B}}{\mathrm{kT}}$

$\begin{array}{c}\mathrm{M}=\mathrm{N\mu }\cdot \left\{\frac{\left(1+\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)-\left(1-\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)}{\left(1+\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)+\left(1-\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)}\right\}\\ =\frac{\left(\frac{2\mathrm{\mu B}}{\mathrm{kT}}\right)}{2}\cdot \mathrm{N\mu }\\ =\frac{{\mathrm{N\mu }}^2\mathrm{B}}{\mathrm{kBT}}\end{array}$

Hence magnetization $\mathrm{M}$ for $\mathrm{\mu B}\ll \mathrm{kT}$ is $\mathrm{M}=\frac{{\mathrm{N\mu }}^2\mathrm{B}}{\mathrm{kBT}}$.

Step 5: (c) Calculations of the magnetization M for μB ≫ kT

As $\mathrm{\mu B}\gg \mathrm{kT},$which implies $\frac{\mathrm{\mu B}}{\mathrm{kT}}\gg 1$

$\tanh \left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)\approx 1$

$\begin{array}{c}\mathrm{M}= \mathrm{N\mu }\cdot \tanh \left(\frac{\mathrm{\mu B}}{\mathrm{kT}}\right)\\ =\mathrm{N\mu }\end{array}$

Magnetization $\mathrm{M}$ for $\text{μB}\gg \text{kT}$ is $\text{M}=\text{Nμ}$.

Step 6: (d) Graphical presentation

The graph has the same nature as the plot 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