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JEE Main 2020 (4 Sep, Shift 2) · Magnetic Materials & Curie Law

Magnetic Materials & Curie Law question from JEE Main 2020 (4 Sep, Shift 2)

Q26JEE Main 2020 · 4 SEP · SHIFT 2Medium

Magnetic Materials & Curie Law

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

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Answer: (c)

Curie's law: $M = \chi H$ with $\chi \propto \dfrac{1}{T}$ , so $M \propto \dfrac{B}{T}$ .
$\dfrac{M_2}{M_1} = \dfrac{B_2}{B_1} \cdot \dfrac{T_1}{T_2} = \dfrac{0.3}{0.4} \times \dfrac{4}{24} = \dfrac{1}{8}$
$M_2 = \dfrac{6}{8} = 0.75\text{ A/m}$
So the answer is (c) 0.75 A/m.

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