Q.4.35 - An Introduction to Thermal Physics

Question

The magnetic field created by a dipole has a strength of approximately $\left({\mu }_0/4\pi \right)\left(\mu /r^3\right)$, where r is the distance from the dipole and ${\mu }_0$ is the "permeability of free space," equal to exactly $4\pi \times {10}^{-7}$ in SI units. (In the formula I'm neglecting the variation of field strength with angle, which is at most a factor of 2.) Consider a paramagnetic salt like iron ammonium alum, in which the magnetic moment $\mu$ of each dipole is approximately one Bohr magneton $\left(9\times {10}^{-24} \mathrm{J}/\mathrm{T}\right)$, with the dipoles separated by a distance of $1 \mathrm{nm}$. Assume that the dipoles interact only via ordinary magnetic forces.

(a) Estimate the strength of the magnetic field at the location of a dipole, due to its neighboring dipoles. This is the effective field strength even when there is no externally applied field.

(b) If a magnetic cooling experiment using this material begins with an external field strength of $1 \mathrm{T}$, by about what factor will the temperature decrease when the external field is turned off?

(c) Estimate the temperature at which the entropy of this material rises most steeply as a function of temperature, in the absence of an externally applied field.

(d) If the final temperature in a cooling experiment is significantly less than the temperature you found in part (c), the material ends up in a state where $\partial S/\partial T$ is very small and therefore its heat capacity is very small. Explain why it would be impractical to try to reach such a low temperature with this material.

Step-by-Step Solution

Answer

a) As a result, the strength of the magnetic field at a dipole's location is determined by the dipoles neighboring is $2.7\times {10}^{-3} \mathrm{T}$.

b) As a result, the temperature drops by a factor of 370.

c) Thus, without an externally supplied field, the temperature at which this material's entropy grows most quickly as a function of temperature is$1.0\mathrm{mK}$

d) As a result, attempting to achieve a very low temperature using this material would be impractical since heat leakage from the outside cannot be prevented, and a large heat capacity increases the amount of heat necessary to change the temperature.

Part (a) - Step 1: To determine

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

Part (a) - Step 2: Explanation

GIVEN:

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

FORMULA:

Write the expression for a dipole's magnetic field.

$B=\frac{{\mu }_0\mu }{4\pi r^3}$

Where $r$ is the distance from dipole

 $\mu$ is the magnetic moment and

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

Part (a) - Step 3: Calculation

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

$B_{\mathrm{d}}=\frac{3{\mu }_0\mu }{4\pi r^3}\dots \text{ (1) }$

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

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

Part (b) - Step 4: To find

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

Part (b) - Step 5: Explanation

GIVEN:

Magnetic field of a dipole has a strength $\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{\mu }{r^3}\right)$.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 $M$

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

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

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

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

CALCULATION :

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

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

Hence the temperature drops by a factor of  $370$

Part (c) - Step 6: To determine

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

Part (c) - Step 7: Explanation

GIVEN:

Magnetic field of a dipole has a strength$\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{\mu }{r^3}\right)$ .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 $\left(\frac{kT}{\mu B}\right)$.

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

 $T_{\mathrm{s}}=\frac{\mu Bx}{k}\dots \text{ (3) }$

CALCULATION: 

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

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

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

Part (d) - Step 8: To find

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

Part (d) - Step 9: Explanation

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

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

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.