JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Magnetic Properties of Matter

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:

(A) $1 A/m$
(B) $0.75 A/m$ ✓ Correct
(C) $2.25 A/m$
(D) $4 A/m$

Answer: (B)

Solution

By Curie's law a paramagnet has susceptibility

χ = \frac{C}{T}

, so its magnetisation is

I = χH \propto \frac{H}{T}

, with the driving field

H

proportional to the applied field. Comparing the two states,

\frac{I _{2}}{I _{1}} = \frac{B _{2} / T _{2}}{B _{1} / T _{1}} = \frac{0.3/24}{0.4/4} = \frac{1}{8}

, so

I_{2} = \frac{6}{8} = 0.75 A m^{- 1}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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