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Q.4.35

Expert-verified
Found in: Page 148

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# 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 $4\pi ×{10}^{-7}$ 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 $\left(9×{10}^{-24}\mathrm{J}/\mathrm{T}\right)$, 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.

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

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.

See the step by step solution

## 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

${B}_{\mathrm{d}}=\frac{3{\mu }_{0}\mu }{4\pi {r}^{3}}\dots \text{(1)}$

$\text{Substitute}{\mu }_{0}=4\pi ×{10}^{-7}\mathrm{N}/{\mathrm{A}}^{2}\phantom{\rule{0ex}{0ex}}\mu =9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\phantom{\rule{0ex}{0ex}}r=1.0×{10}^{-9}\mathrm{m}\text{in equation (1)}$${B}_{\mathrm{d}}=\frac{3\left(4\pi ×{10}^{-1}\mathrm{N}/{\mathrm{A}}^{2}\right)\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)}{4\pi {\left(1.0×{10}^{-9}\mathrm{m}\right)}^{3}}\phantom{\rule{0ex}{0ex}}=2.7×{10}^{-3}\mathrm{T}$

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

## 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 $M$

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

$\text{Here,}N\text{is the number of dipoles,}\phantom{\rule{0ex}{0ex}}T\text{is temperature and}\phantom{\rule{0ex}{0ex}}k\text{is Boltzmann constant.}$

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×{10}^{-3}\mathrm{T}$ for in expression (2)

$\frac{{T}_{\mathrm{i}}}{{T}_{\mathrm{i}}}=\frac{\left(\mathrm{IT}\right)}{\left(2.7×{10}^{-3}\mathrm{T}\right)}\phantom{\rule{0ex}{0ex}}=370$

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

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

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:

$\text{Substitute}\mu =9.0×{10}^{-24}\mathrm{J}/\mathrm{T},\phantom{\rule{0ex}{0ex}}B=2.7×{10}^{-3}\mathrm{T},\phantom{\rule{0ex}{0ex}}x=0.6\phantom{\rule{0ex}{0ex}}k=1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\text{in expression (3)}$

${T}_{\mathrm{s}}=\frac{\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)\left(2.7×{10}^{-3}\mathrm{T}\right)\left(0.6\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)}\phantom{\rule{0ex}{0ex}}=1.0\mathrm{mK}$

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$${T}_{\mathrm{s}}=\frac{\left(9.0×{10}^{-24}\mathrm{J}/\mathrm{T}\right)\left(2.7×{10}^{-3}\mathrm{T}\right)\left(0.6\right)}{\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)}\phantom{\rule{0ex}{0ex}}=1.0\mathrm{mK}$

## 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.