Q64P - Fundamentals Of Physics

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) $\mathbf{90}\mathbf{\%}$
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Step-by-Step Solution

Answer

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

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

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.  

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

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