Q148 / 180 · KCET 2020 · Physics
A paramagnetic sample shows a net magnetisation of $8 \mathrm{Am}^{-1}$ when placed in an external magnetic field of $0.6 \mathrm{~T}$ at a temperature of $4 \mathrm{~K}$. When the same sample is placed in an external magnetic field of $0.2 \mathrm{~T}$ at a temperature of $16 \mathrm{~K}$, the magnetisation will be
- A$\frac{32}{3} \mathrm{Am}^{-1}$
- Correct answerB$\frac{2}{3} \mathrm{Am}^{-1}$
- C$6 \mathrm{Am}^{-1}$
- D$2.4 \mathrm{Am}^{-1}$
Solution
Given, $M_{1}=8 \mathrm{Am}^{-1}, B_{1}=0.6 \mathrm{~T}, T_{1}=4 \mathrm{~K}$, $B_{2}=0.2 \mathrm{~T}$ and $T_{2}=16 \mathrm{~K}$
According to Curie's law, for paramagnetic materials,
$M=\frac{C B}{T}$
$\Rightarrow \quad M_{1}=\frac{C B_{1}}{T_{1}}$
and $\quad M_{2}=\frac{C B_{2}}{T_{2}}$
$\therefore \quad \frac{M_{2}}{M_{1}} \frac{\frac{C B_{2}}{T_{2}}}{\frac{C B_{1}}{T_{1}}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$
$=\frac{0.2}{0.6} \times \frac{4 \mathrm{~K}}{16 \mathrm{~K}}=\frac{1}{3} \times \frac{1}{4}$
$\Rightarrow \quad \frac{M_{2}}{M_{1}}=\frac{1}{12} \Rightarrow M_{2}=\frac{M_{1}}{12}=\frac{8}{12}$
$\Rightarrow \quad M_{2}=\frac{2}{3} \mathrm{Am}^{-1}$
Hence, the correct answer is option: 'B'.