Question

For a magnetising field of intensity $2\times {10}^3{\text{Am}}^{-1}$, aluminium at $280\text{K}$ acquires intensity of magnetisation of $4.8\times {10}^{-2}{\text{Am}}^{-1}.$ If the temperature of the metal is raised to $320\text{K}$, what will be the intensity of magnetisation ?

• A. $2.1\times {10}^{-2}{\text{Am}}^{-1}$
• B. $4.2\times {10}^{-2}{\text{Am}}^{-1}$
• C. $4.2\times {10}^{-3}{\text{Am}}^{-1}$
• D. $2.1\times {10}^{-3}{\text{Am}}^{-1}$

Answer 1

The correct option is B $4.2\times {10}^{-2}{\text{Am}}^{-1}$

Given:

At, $T_1=280\text{K}$
$H_1=2\times {10}^3{\text{Am}}^{-1};I_1=4.8\times {10}^{-2}{\text{Am}}^{-1}$

Using the relation, ${\chi }_1=\frac{I_1}{H_1}$, we get

${\chi }_1=\frac{4.8\times {10}^{-2}}{2\times {10}^3}=2.4\times {10}^{-5}$

Now, according to the Curie's law,

$\chi \propto \frac{1}{T}$

$\Rightarrow \frac{{\chi }_2}{{\chi }_1}=\frac{T_1}{T_2}$

$\Rightarrow {\chi }_2=\frac{280}{320}\times 2.4\times {10}^{-5}[\because T_2=320\text{K}]$

$\therefore {\chi }_2=2.1\times {10}^{-5}$

Now, the intensity of magnetisation,

$I_2={\chi }_2{H}_2$

Since, $H$ is independent of the temperature, i.e., $H_2=H_1$

$\therefore I_2=(2.1\times {10}^{-5})\times (2\times {10}^3)=4.2\times {10}^{-2}{\text{Am}}^{-1}$

Hence, option $\left(B\right)$ is correct.