A paramagnetic sample shows a net magnetisati | vedclass

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(B) Given: $M_{1} = 8 	ext{ Am}^{-1}$,$B_{1} = 0.6 	ext{ T}$,$T_{1} = 4 	ext{ K}$,$B_{2} = 0.2 	ext{ T}$,and $T_{2} = 16 	ext{ K}$.According to C

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$\frac{2}{3} 	ext{ Am}^{-1}$

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(B) Given: $M_{1} = 8 	ext{ Am}^{-1}$,$B_{1} = 0.6 	ext{ T}$,$T_{1} = 4 	ext{ K}$,$B_{2} = 0.2 	ext{ T}$,and $T_{2} = 16 	ext{ K}$.According to Curie's law for paramagnetic materials,the magnetization $M$ is given by $M = \frac{C B}{T}$,where $C$ is the Curie constant.Therefore,$M_{1} = \frac{C B_{1}}{T_{1}}$ and $M_{2} = \frac{C B_{2}}{T_{2}}$.Taking the ratio: $\frac{M_{2}}{M_{1}} = \frac{B_{2}}{B_{1}} 	imes \frac{T_{1}}{T_{2}}$.Substituting the values: $\frac{M_{2}}{8} = \frac{0.2}{0.6} 	imes \frac{4}{16}$.$\frac{M_{2}}{8} = \frac{1}{3} 	imes \frac{1}{4} = \frac{1}{12}$.$M_{2} = \frac{8}{12} = \frac{2}{3} 	ext{ Am}^{-1}$.

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