Question: Figuregives the magnetization curve for a paramagnetic material. The vertical axis scale is set by$\mathbf{}\mathit{a}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{15}$, and the horizontal axis scale is set by$\mathit{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{2}\mathit{T}\mathbf{/}\mathit{K}$. Let${\mathit{\mu }}_{\mathbf{s}\mathbf{a}\mathbf{m}}$be the measured net magnetic moment of a sample of the material and${\mathit{\mu }}_{max}$be the maximum possible net magnetic moment of that sample. According to Curie’s law, what would be the ratio${\mathit{\mu }}_{\mathbf{s}\mathbf{a}\mathbf{m}}\mathbf{/}{\mathit{\mu }}_{\mathbf{m}\mathbf{a}\mathbf{x}}$ were the sample placed in a uniform magnetic field of magnitude, at a temperature of 2.00 k?

1. The vertical axis scale is set by$a=0.15$
2. the horizontal axis scale is set by$b=0.2T/K$
3. Magnitude of the magnetic field at temperature T = 20 K is $B_{ext}=0.800T$

The extent of an alignment within a volume is measured as the magnetization . It is given as,

$\mathit{M}\mathbf{=}\frac{\mathbf{M}\mathbf{e}\mathbf{a}\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{e}\mathbf{d}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{g}\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{i}\mathbf{c}\mathbf{}\mathbf{m}\mathbf{o}\mathbf{m}\mathbf{e}\mathbf{n}\mathbf{t}}{\mathbf{V}}$ (i)

The magnetization for the sample is given as,

${\mathit{M}}_{\mathbf{s}\mathbf{a}\mathbf{m}}\mathbf{=}\frac{\mathbf{N}{\mathbf{\mu }}_{\mathbf{s}\mathbf{a}\mathbf{m}}}{\mathbf{V}}$

(ii)

Here, M is the magnetization of the sample, N is the number of dipoles, ${\mathit{\mu }}_{\mathbf{s}\mathbf{a}\mathbf{m}}$is the sample’s magnetic moment, V is volume.

The saturation magnetization is given as,

${\mathit{M}}_{\mathbf{s}\mathbf{a}\mathbf{t}}\mathbf{=}\frac{\mathbf{N}{\mathbf{\mu }}_{\mathbf{max}}}{\mathbf{V}}$ (iii)

Here, M is saturation magnetization,N is the number of dipoles,${\mathbf{\mu }}_{\mathbf{max}}$ is the maximum magnetic moment, V is volume.

At low values of the ratio${\overrightarrow{\mathbf{B}}}_{\mathbf{ext}}\mathbf{/}\mathbf{T}$ Curie’s law is written as,

role="math" localid="1663408637837" $\mathbf{M}\mathbf{=}\frac{{\mathbf{CB}}_{\mathbf{ext}}}{\mathbf{T}}$ (iv)

Here, T is temperature in kelvins and C is a material’s Curie constant.

We can find the ratio of the magnetic moment ofthesample to themaximum possible net magnetic moment by using the formula for saturation magnetization, Curie’s law, and the data given in the graph.

Taking the ratio of magnetization of the sample to the saturation magnetization, i.e. equation (ii) and (iii)

$$\begin{gathered} \frac{M_{sam}}{M_{max}}=\frac{\left(\frac{N{\mu }_{sam}}{V}\right)}{\left(\frac{N{\mu }_{max}}{V}\right)} \\ =\frac{{\mu }_{sam}}{{\mu }_{\max }} \end{gathered}$$

Therefore, the relation between magnetization and dipole moment can be written as,

(v)

$\frac{M_{sam}}{M_{max}}=\frac{{\mu }_{sam}}{{\mu }_{\max }}$

The slope of the graph in Fig. 32-39 is given as

$$\begin{gathered} \frac{M_{sam}/M_{max}}{B_{ext}/T}=\frac{a}{b} \\ =\frac{0.15T/K}{0.2\text{K/T}} \\ =0.75 \end{gathered}$$

Rearranging the equation for the ratio of magnetization of sample and maximum magnetization, we get

$\frac{M_{sam}}{M_{max}}=0.75\times \frac{B_{ext}K/T}{T}$

Substitute the given values of$B_{ext}$ and T.

$$\begin{gathered} \frac{M_{sam}}{M_{max}}=0.75\times \frac{0.800\text{T}}{2.00\text{K}} \\ =0.30 \end{gathered}$$

Using the relation from equation (v), we can write that

$$\begin{gathered} \frac{{\mu }_{sam}}{{\mu }_{max}}=\frac{M_{sam}}{M_{max}} \\ =0.30 \end{gathered}$$

Therefore, the ratio of the net magnetic moment of the sample material to the maximum possible net magnetic moment is 0.30 .