A paramagnetic sample shows a net magnetization of 6 A/m, when it is placed in an external magnetic field of 0.4T at a temperature of 4K. when the sample is placed in an external magnetic field of 0.3T at temperature of 24K, then the magnetization will be

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4 A/m

0.75 A/m

1 A/m

2.25 A/m

answer is B.

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Detailed Solution

$\begin{array}{l} χ_{m} \propto \frac{1}{T} \Rightarrow \frac{I}{H} \propto \frac{1}{T} \\ \Rightarrow \frac{I}{(B_{0} / μ_{0})} \propto \frac{1}{T} \Rightarrow \frac{μ_{0} I}{B_{0}} \propto \frac{1}{T} \\ \Rightarrow I \propto \frac{B_{0}}{μ_{0} T} \Rightarrow\propto \frac{B_{0}}{T} (∵ μ_{0} = constant) \\ \Rightarrow \frac{I_{1}}{I_{2}} = \frac{{(B_{0})}_{1}}{{(B_{0})}_{2}} \times \frac{T_{2}}{T_{1}} \end{array}$
On putting the given values in above equation we get
$\begin{array}{l} \frac{6}{I_{2}} = \frac{0 .4}{0 .3} \times \frac{24}{4} \Rightarrow \frac{6}{I_{2}} = \frac{4}{3} \times \frac{6}{1} \\ \Rightarrow \frac{6}{I_{2}} = 8 \Rightarrow I_{2} = \frac{6}{8} \\ \Rightarrow I_{2} = \frac{3}{3} A / m \Rightarrow I_{2} = 0 .75 A / m \end{array}$

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