91Ó°ÊÓ

The strength of the magnetic field at the location of a dipole, due to its neighboring dipoles.

GIVEN:

Magnetic field of a dipole has a strength $\left(\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\right)\left(\frac{\mathrm{μ}}{r^3}\right)$.A paramagnetic salt contains dipoles each having a magnetic moment of one Bohr magneton.

FORMULA:

Write the expression for a dipole's magnetic field.

$B=\frac{{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{π}{r}^3}$

Where $r$is the distance from dipole

$\mathrm{μ}$is the magnetic moment and

${\mathrm{μ}}_0$is the magnetic permeability of free space.

CALCULATION:

To account for the differences in directions, multiply the number of neighbours by three.

Write the expression of the magnetic field at a dipole $B_{\mathrm{d}}$due to three neighboring dipoles

$B_{\mathrm{d}}=\frac{3{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{π}{r}^3}…\text{(1)}$

$$\begin{gathered} \text{Substitute}{\mathrm{μ}}_0=4\mathrm{π}×{10}^{-7}\mathrm{N}/{\mathrm{A}}^2 \\ \mathrm{μ}=9.0×{10}^{-24}\mathrm{J}/\mathrm{T} \\ r=1.0×{10}^{-9}\mathrm{m}\text{in equation (1)} \end{gathered}$$$$\begin{gathered} B_{\mathrm{d}}=\frac{3\left(4\mathrm{π}×{10}^{-1}\mathrm{N}/{\mathrm{A}}^2\right)\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)}{4\mathrm{π}{\left(1.0×{10}^{-9}\mathrm{m}\right)}^3} \\ =2.7×{10}^{-3}\mathrm{T} \end{gathered}$$

Hence he strength of the magnetic field at a dipole's location is determined by the dipoles neighboring is$2.7×{10}^{-3}T$

When the external field is turned off, the temperature drops by this factor.

GIVEN:

Magnetic field of a dipole has a strength $\left(\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\right)\left(\frac{\mathrm{μ}}{r^3}\right)$.Each dipole in a paramagnetic salt has a magnetic moment of one Bohr magneton. In a magnetic cooling experiment, the external magnetic field of the starting value$1\mathrm{T}$is turned off.

FORMULA:

Write the expression of the magnetization $M$

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

$$\begin{gathered} \text{Here,}N\text{is the number of dipoles,} \\ T\text{is temperature and} \\ k\text{is Boltzmann constant.} \end{gathered}$$

As a result, when the external field is turned off, the magnetization remains constant.

$\frac{B_{\mathrm{i}}}{T_{\mathrm{i}}}=\frac{B_{\mathrm{i}}}{T_{\mathrm{i}}}$

Here,

the subscript $\mathrm{i}$denotes the initial values and

the subscript $\mathrm{f}$denotes the final values.

Rearrange the above expression

$\frac{T_{\mathrm{i}}}{T_{\mathrm{i}}}=\frac{B_{\mathrm{i}}}{B_{\mathrm{t}}}…\text{(2)}$

CALCULATION :

Substitute $B_{\mathrm{i}}=$$1\mathrm{T}$and $B_{\mathrm{f}}=$$2.7×{10}^{-3}\mathrm{T}$for in expression (2)

$$\begin{gathered} \frac{T_{\mathrm{i}}}{T_{\mathrm{i}}}=\frac{\left(\mathrm{IT}\right)}{\left(2.7×{10}^{-3}\mathrm{T}\right)} \\ =370 \end{gathered}$$

Hence the temperature drops by a factor of $370$

The temperature at which this material's entropy rises most steeply as a function of temperature without the application of an external field.

GIVEN:

Magnetic field of a dipole has a strength$\left(\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\right)\left(\frac{\mathrm{μ}}{r^3}\right)$.A paramagnetic salt contains dipoles each having a magnetic moment of one Bohr magneton.

FORMULA:

Write the expression of the entropy-temperature formula

$\frac{S}{NK}=\ln \left(2\cosh \frac{1}{x}\right)-\frac{1}{x}\tanh \frac{1}{x}$

Here, $S$is entropy,

$K$is a constant and

$x$stands for $\left(\frac{kT}{\mathrm{μ}B}\right)$.

From the above expression, write the temperature expression $T_{\mathrm{s}}$at which the entropy-temperature curve is the steepest.

$T_{\mathrm{s}}=\frac{\mathrm{μ}Bx}{k}…\text{(3)}$

CALCULATION:

$$\begin{gathered} \text{Substitute}\mathrm{μ}=9.0×{10}^{-24}\mathrm{J}/\mathrm{T}, \\ B=2.7×{10}^{-3}\mathrm{T}, \\ x=0.6 \\ k=1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\text{in expression (3)} \end{gathered}$$

$$\begin{gathered} T_{\mathrm{s}}=\frac{\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)\left(2.7×{10}^{-3}\mathrm{T}\right)(0.6)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)} \\ =1.0\mathrm{mK} \end{gathered}$$

Hence without an externally supplied field, the temperature at which this material's entropy grows most quickly as a function of temperature is $1.0mK$$$\begin{gathered} T_{\mathrm{s}}=\frac{\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)\left(2.7×{10}^{-3}\mathrm{T}\right)(0.6)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)} \\ =1.0\mathrm{mK} \end{gathered}$$

Reasons why trying to obtain a very low temperature with this material would be impractical

The cooling operations are focused on the portion of the entropy-temperature curve that has the maximum slope at low magnetic fields.

Write the expression of the heat capacity $c_{\mathrm{V}}$at constant volume

$c\mathrm{V}=T\left(\frac{∂S}{∂T}\right)$

Since the heat capacity at constant volume is proportional to the slope of the entropy-temperature curve, it is largest where the curve is steepest. For a high heat capacity, a great amount of heat is necessary to change the temperature noticeably.

It is not practical to reach a temperature lower than$1.0\mathrm{mK}$ because heat leaking from the outside cannot be entirely avoided.