91Ó°ÊÓ

In a magnetic field $B$, a system of $N$magnetic dipoles arrange themselves so that their magnetic moment $\mathrm{μ}$points either parallel or anti parallel to the field.

Below is a table with the information provided.

The expression for total energy is given as:

$U=\mathrm{μ}B\left(N_{↓}-N_{↑}\right)……..\left(1\right)$

Where,

$N_{↑}$= the number of dipoles pointing up

$N_{↓}$= the number of dipoles pointing down

$N=\left(N_{↓}+N_{↑}\right)$= the total number of dipoles

The net magnetization is given as:

$M=\mathrm{μ}\left(N_{↓}-N_{↑}\right)=-\frac{U}{B}……...\left(2\right)$

Where,

$B$= the magnetic field

$U$= the total energy

The multiplicity of states is given as:

$\mathrm{Î\copyright }=\frac{N!}{N_{↓}!(N−N_{↓})!}=\frac{N!}{N_{!}!N_{↓}!}……….\left(3\right)$

For a small system, entropy is given as:

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)……..\left(4\right)$

By using these formulae, let's verify the table:

For $N_{↑}=98,N_{↓}=2$and $N=100$,

By substituting these values in the equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×98) \\ \frac{U}{\mathrm{μ}B}=−96\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=96\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=\frac{96}{N}=\frac{96}{100}=0.96\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{98!2!}=4950$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(4950\right)=8.5$

For $N_{↑}=97,N_{↓}=3$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×97) \\ \frac{U}{\mathrm{μ}B}=−94\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=94\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=\frac{94}{N}=\frac{94}{100}=0.94\begin{array}{}\end{array} \end{gathered}$$

Substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{97!3!}=161700$

Now, by substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(161700\right)=12$

For $N_{↑}=52,N_{↓}=48$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×52) \\ \frac{U}{\mathrm{μ}B}=−4\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=4\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=\frac{4}{N}=\frac{4}{100}=0.04\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{52!48!}=9.32×{10}^{28}$

By substituting this value in equation (4), we get,

$\begin{array}{rl}\frac{S}{k}& =\ln \left(\mathrm{Î\copyright }\right)=\ln \left(161700\right)=12\end{array}$

For $\begin{array}{rl}{N}_{↑}& =52,N_{↓}=48\end{array}$and $N=100,$

by substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×51) \\ \frac{U}{\mathrm{μ}B}=−2\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=2\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=\frac{2}{N}=\frac{2}{100}=0.02\begin{array}{}\end{array} \end{gathered}$$

Now, substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{51!49!}=9.9×{10}^{28}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(9.9×{10}^{28}\right)=66.765$

For $N_{↑}=50,N_{↓}=50$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×50) \\ \frac{U}{\mathrm{μ}B}=0\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=0 \\ \frac{M}{N\mathrm{μ}}=\frac{0}{N}=\frac{0}{100}=0\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{50!50!}=1.009×{10}^{29}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(1.009×{10}^{29}\right)=66.784$

For $N_{↑}=49,N_{↓}=51$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×49) \\ \frac{U}{\mathrm{μ}B}=2\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=−2\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=\frac{2}{N}=\frac{2}{100}=−0.02\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{49!51!}=9.9×{10}^{28}$

By substituting this value in equation (4), we get,

$\begin{array}{rl}\frac{S}{k}& =\ln \left(\mathrm{Î\copyright }\right)=\ln (9.9×{10}^{28})=66.765\end{array}$

For $\begin{array}{rl}{N}_{↑}& =48,N_{↓}=52\end{array}$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×48) \\ \frac{U}{\mathrm{μ}B}=4\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=−4\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=−\frac{4}{N}=−\frac{4}{100}=−0.04\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{48!52!}=9.32×{10}^{28}$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(9.32×{10}^{28}\right)=66.7$

For $N_{↑}=1,N_{↓}=99$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×1) \\ \frac{U}{\mathrm{μ}B}=98\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=−98\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=−\frac{98}{N}=−\frac{98}{100}=−0.98\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{1!99!}=100$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(100\right)=4.6$

For $N_{↑}=0,N_{↓}=100$and $N=100$,

By substituting these values in equation (1), we get,

$$\begin{gathered} U=\mathrm{μ}B(100−2×0) \\ \frac{U}{\mathrm{μ}B}=100\begin{array}{}\end{array} \end{gathered}$$

By substituting this value in equation (2), we get,

$$\begin{gathered} M=−\frac{U}{B}=−100\mathrm{μ} \\ \frac{M}{N\mathrm{μ}}=−\frac{100}{N}=−\frac{100}{100}=−1\begin{array}{}\end{array} \end{gathered}$$

Now substitute the values in equation (3) to get $\mathrm{Î\copyright }$

$\mathrm{Î\copyright }=\frac{100!}{0!100!}=1$

By substituting this value in equation (4), we get,

$\frac{S}{k}=\ln \left(\mathrm{Î\copyright }\right)=\ln \left(1\right)=0$

Hence, all the table values are verified.