Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 10

Q. 10.6

Use the Bohr magneton \mu_{ B }=e \hbar / 2 m=9.27 \times 10^{-24} J/T as a typical magnetic moment and B = 0.50 T to do the following:

(a) Find the temperature T at which \mu B=0.1 k T.

(b) For \mu B=0.1 k T compare \tanh (\beta \mu B) \text { with } \beta \mu B and thereby check the suitability of this value of T as a “classical” temperature.

(c) Make the same comparison in the expression \tanh (\beta \mu B) \approx \beta \mu B \text { at } T=100 K.

Strategy With the numerical values of the magnetic moment and magnetic field given, the computations are all straightforward. For the comparisons called for in parts (b) and (c), the given values suggest using two significant figures.

Step-by-Step

Verified Solution

(a) We are given that \mu B=0.1 k T. We let \mu=\mu_{ B } and solve for T:

 

T=\frac{10 \mu B}{k}=\frac{10\left(9.27 \times 10^{-24} J / T \right)(0.5 T )}{1.38 \times 10^{-23} J / K }=3.36 K

 

(b) Now

 

\begin{gathered}\beta \mu B=\frac{\mu B}{k T}=0.10 \\\tanh (\beta \mu B)=\tanh (0.10)=0.10\end{gathered}

 

to two significant digits. Therefore we conclude that the approximation is a good one for most purposes, even at this low temperature.

(c) Now

 

\begin{aligned}\beta \mu B=\frac{\mu B}{k T} &=\frac{\left(9.27 \times 10^{-24} J / T \right)(0.5 T )}{\left(1.38 \times 10^{-23} J / K \right)(100 K )} \\&=3.36 \times 10^{-3} \\\tanh (\beta \mu B) &=\tanh \left(3.36 \times 10^{-3}\right)=3.36 \times 10^{-3}\end{aligned}

 

so that \beta \mu B \text { and } \tanh (\beta \mu B) are the same to three significant figures. At these relatively high temperatures the approximation is an excellent one.