Question

A sample of the paramagnetic salt to which the magnetization curve of Fig. 32-14 applies is immersed in a uniform magnetic field of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\text{T}}$. At what temperature will the degree of magnetic saturation of the sample be (a)$\mathbf{50}\mathbf{\%}$and (b)$\mathbf{90}\mathbf{\%}$
?

Short answer

(a) The temperature at which the degree of magnetic saturation of the sample is $50\%$is $\text{4K}$.

(b) The temperature at which the degree of magnetic saturation of the sample is $90\%$is localid="1664195147119" $\text{2K}$.

Step-by-step solution

Listing the given quantities:

Magnetic field,$B=2.0\text{T}$

Understanding the concepts of Curie Law:

To determine the temperature at which the sample is $\mathbf{50}\mathbf{\%}$magnetically saturated, use the Curie law formula. You must use the value from the graph for the $\mathbf{90}\mathbf{\%}$.

Formula:

$\frac{M}{M_{\text{max}}}=\frac{B}{T}$

Here,$\frac{M}{M_{\text{max}}}$is the magnetic saturation,$B$is the magnetic field,$T$and is the temperature.

(a) Calculations of the temperature at which the degree of magnetic saturation of the sample is 50%:

For the Curie law,

$$\begin{gathered} \frac{M}{M_{\text{max}}}=\frac{B}{T} \\ T=\frac{B}{\frac{M}{M_{\text{max}}}} \end{gathered}$$

Substitute $50\%(=0.50)$for $\frac{M}{M_{max}}$and the given data in the above equation.

$\begin{array}{rcl}T& =& \frac{2.0\text{T}}{0.50}\\ & =& 4\text{K}\end{array}$

Hence, the temperature at which the degree of magnetic saturation of the sample at $50\%$is $4\text{K}$.

(b) Calculations of the temperature at which the degree of magnetic saturation of the sample is 90%:

For the Curie law,

$$\begin{gathered} \frac{M}{M_{\text{max}}}=\frac{B}{T} \\ T=\frac{B}{{\displaystyle \frac{M}{M_{\text{max}}}}} \end{gathered}$$

Substitute $90\%(=0.90)$for $\frac{M}{M_{\text{max}}}$and the given data in the above equation.

$\begin{array}{rcl}T& =& \frac{\text{2.0T}}{\text{0.90}}\\ & =& \text{2K}\end{array}$

Hence, the temperature at which the degree of magnetic saturation of the sample at $90\%$is $\text{2K}$.