A paramagnetic sample showing a net magnetization of 0 . 8   A   m - 1 , when placed in an…

Using Curie's law, $M=\mathrm{C}\frac{B}{T}$here $M$is the magnetization, $B$is the magnetic field, $T$is the absolute temperature and $\mathrm{C}$is Curie's constant.

Now, given that $M_1=0.8 \mathrm{A} {\mathrm{m}}^{-1}$ if $B_1=0.8 \mathrm{T}$ and $T_1=5 \mathrm{K}$therefore, $M_1=\mathrm{C}\frac{B_1}{T_1}\Rightarrow 0.8=\mathrm{C}\frac{0.8}{5}\Rightarrow \mathrm{C}=5 \mathrm{K} \mathrm{A} {\mathrm{T}}^{-1} {\mathrm{m}}^{-1}$

If the temperature becomes $T_2=20 \mathrm{K}$then magnetization will be, $M_2=\mathrm{C}\frac{B_2}{T_2}\Rightarrow {\mathrm{M}}_2=5\times \frac{0.8}{20}=0.2 \mathrm{A} {\mathrm{m}}^{-1}$