Is the limiting case for 1/2-spin paramagnetic example circular in reasoning? In chapter 3 of F. Mandl's book, Statistical Physics, he first utilised the example of a paramagnetic solid, which particle is 1/2-spin. He considers a single dipole to be the system of interest, while all the surrounding dipoles at as the heat bath, subjected to a B-field, $\mathcal{B} = \mu_{0}\mathcal{H}+\mu_{0}\mathcal{J}$. From here he works towards obtaining the magnetic susceptibility, $\chi = \frac{\mathcal{J}}{\mathcal{H}}$.
It is obtained that the magnetization, $\mathcal{J}$ is given by the following equation:
$$\mathcal{J} = \frac{\mathcal{M}}{V} = \frac{N\bar\mu}{V} = \frac{N[(\mu{}p_{+})+(-\mu{}p_{-})] }{V} = \frac{N}{V}\mu{}\tanh{\frac{\mu\mathcal{B}}{kT}}$$
Where $\mathcal{M}$ = total magnetic moment of the field, $N$ = total no. of dipoles, $V$ = volume of solid, $\mu$ = magnetic moment of the dipole.
Mandl goes on to assume that $\mathcal{B}$ is low and $T$ is high, hence:
$$\mathcal{J} \approx \frac{N}{V}\frac{\mu^{2}\mathcal{B}}{kT}$$
Substituting $\mathcal{B} = \mu_{0}\mathcal{H}+\mu_{0}\mathcal{J}$, into the equation, and approximating that magnetisation $J$ is very small, (i.e. $\mathcal{B} \approx \mu_{0}\mathcal{H}$), he gets the magnetic susceptibility as:
$$\chi = \frac{\mathcal{J}}{\mathcal{H}} = \frac{N\mu^2\mu_0}{VkT}$$
This is what I don't understand. magnetic susceptibility is a ratio of $\mathcal{J}$ and $\mathcal{H}$, who are we able to elimate $\mathcal{J}$ at the start if so? If we don't assume that $\mathcal{J}$ is small, we will get the following expression:
$$\frac{\mathcal{J}}{\mathcal{H}} = \frac{\frac{N\mu^2\mu_0}{VkT}}{1+\frac{N\mu^2\mu_0}{VkT}}$$
The only approximation that gives us Mandl's expression is if $\frac{N\mu^2\mu_0}{VkT}$ is small. However, this is not due to the assumption that $\mathcal{J}$ is small. Please Advice, Thank You!
 A: Whenever you make statements like $B$ is small and $T$ is large, you always have to ask the question "compared to what?" Unfortunately, for the sake of brevity many times physics textbooks do not write this explicitly.
For example, when he is simplifying $J$, really the approximation he is making is $\frac{\mu B}{kT}\ll 1$. 
Now, let us come to $J$. Notice it has a $\frac{1}{kT}$ dependence. Now we assume that $kT$ is large relative to any other energy scale in the problem. What this means is that any term with $T^0$ is always bigger than any term with $T^{-1}$, no matter the coefficients. Then we can say that $\mu_0 J \ll B$. In other words, as long as temperature is sufficiently large, this approximation is good. Also, because $T$ is large, the expression $\frac{N\mu^2 \mu_0}{VkT} \ll 1$ because $T$ is sufficiently large. In summary, ultimately he is not assuming that $J$ is small, but this is a consequence of assuming $T$ sufficently large.
Another way of thinking properly about what Mandl did is the following:
$$\begin{eqnarray*}
\chi &=& \frac{J}{B - \mu_0 J} \\
&=& \frac{1}{\frac{B}{\mu_0 J} - 1} \\
&=& \frac{1}{\frac{B}{\mu_0 J}} \,\,\,\,\, \text{(Because $\frac{B}{\mu_0 J} \gg 1$)} \\
&=& \frac{N\mu^2 \mu_0}{VkT}
\end{eqnarray*}
$$
You shouldn't throw away terms until they're small compared to some other term in your problem.
