91Ó°ÊÓ

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

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.

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

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

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

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

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

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

Thus,

$\begin{array}{c}{\mathrm{B}}_{\mathrm{ext}}=\frac{\mathrm{M}}{{\mathrm{M}}_{\mathrm{mex}}}â‹\dots \mathrm{T}\\ ≈2.0×300\mathrm{K}\\ =6.0×{10}^2\mathrm{T}\end{array}$

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

Yes, these fields attainable in the laboratory.