Question

A sample of the paramagnetic salt to which the magnetization curve of Figure applies is held at room temperature ($\mathbf{\text{300K}}$). (a) At what applied magnetic field will the degree of magnetic saturation of the sample be$\mathbf{\text{50\%}}$? (b) At what applied magnetic field will the degree of magnetic saturation of the sample be$\mathbf{90}\mathit{\%}$? (c) Are these fields attainable in the laboratory?

Short answer

a) The applied magnetic field will the degree of magnetic saturation of the sample be$50\%$ is $1.5\times {10}^2\mathrm{T}$.

b) The applied magnetic field will the degree of magnetic saturation of the sample be$90\%$ is$6.0\times {10}^2\mathrm{T}$ .

c) Yes, these fields attainable in the laboratory.

Step-by-step solution

Listing the given quantities 

Room temperature,$\mathrm{T}=300\mathrm{K}$

Understanding the concepts of magnetization

The degree of the magnetic saturation i.e., the ratio of magnetization to the maximum possible magnetization is given in the problem. We can find the magnetic field for different degrees of magnetic saturation by studying the graph given in.

The expression for the curie formulas is given as follows.

$\frac{\mathbf{M}}{{\mathbf{M}}_{\mathbf{max}}}\mathbf{=}\frac{\mathbf{B}}{\mathbf{T}}$

(a) Calculations of the applied magnetic field at which the degree of magnetic saturation of the sample be 50%

The magnetic saturation

$\begin{array}{c}\left(\frac{\mathrm{M}}{{\mathrm{M}}_{\max }}\right)=50\mathrm{\%}\\ =0.50\end{array}$

At this point in the graph in , we can observe that

$\frac{{\mathrm{B}}_{\mathrm{ext}}}{\mathrm{T}}=0.50\mathrm{T}/\mathrm{K}$

Thus,

$\frac{{\mathrm{B}}_{\mathrm{ext}}}{\mathrm{T}}=\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{mex}}}$

The above equation implies

$\begin{array}{c}{\mathrm{B}}_{\mathrm{ext}}=\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{mex}}}\cdot \mathrm{T}\\ =0.50\times 300\mathrm{K}\\ =1.5\times {10}^2\mathrm{T}\end{array}$

The applied magnetic field will the degree of magnetic saturation of the sample be$50\mathrm{\%}$ is $1.5\times {10}^2\mathrm{T}$.

(b) Calculations of the applied magnetic field at which the degree of magnetic saturation of the sample be 90% 

The magnetic saturation of the sample is

$\left(\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{mex}}}\right)=90\mathrm{\%}=0.90$

At this point in the graph in , we can observe that

$\frac{{\mathrm{B}}_{\mathrm{ext}}}{\mathrm{T}}\approx 2.0\mathrm{T}/\mathrm{K}$

Thus,

$\begin{array}{c}{\mathrm{B}}_{\mathrm{ext}}=\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{mex}}}\cdot \mathrm{T}\\ \approx 2.0\times 300\mathrm{K}\\ =6.0\times {10}^2\mathrm{T}\end{array}$

The applied magnetic field will the degree of magnetic saturation of the sample be$90\mathrm{\%}$ is $6.0\times {10}^2\mathrm{T}$.

(c) Explanation

Yes, these fields attainable in the laboratory.