magnetism-and-matter Question 14
Question: Q. 2. A paramagnetic sample shows a net magnetization 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 magnetization will be (a) $\frac{32}{3} \mathrm{Am}$ (b) $\frac{2}{3} \mathrm{Am}^{-1}$ (c) $6 \mathrm{Am}^{-1}$ (d) $2.4 \mathrm{Am}^{-1}$
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Solution:
Ans. Correct option: (b)
[NCERT Exemplar]
Explanation: According to the Curie’s law of magnetisation, the magnetic susceptibility for a substance is directly proportional to magnetic field induction $(B)$ and inversely proportional to the absolute temperature $(T)$.
$$ I \propto \frac{B}{T} $$
So that,
$$ \frac{I_{2}}{I_{1}}=\frac{B_{2} T_{1}}{B_{1} T_{2}} $$
Given,
$$ \begin{aligned} & I_{1}=8 \mathrm{Am}^{-1} \ & B_{1}=0.6 \mathrm{~T} \ & T_{1}=4 \mathrm{~K} \ & I_{2}=? \ & B_{2}=0.2 \mathrm{~T} \ & T_{2}=16 \mathrm{~K} \end{aligned} $$
Put the values,
$$ I_{2}=\frac{B_{2} T_{1}}{B_{1} T_{2}} I_{1}=\frac{0.2 \times 8 \times 4}{0.6 \times 16}=\frac{2}{3} \mathrm{Am}^{-1} $$
