Magnetic Properties of Matter

JEE Main level, single correct, medium

47 of 57

Question 47 of 57

A paramagnetic sample shows a net magnetisation of 6A/m6A/m when it is placed in an external magnetic field of 0.4T0.4T at a temperature of 4K4K. When the sample is placed in an external magnetic field of 0.3T0.3T at a temperature of 24K24K, then the magnetisation will be :

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Answer: (D) 0.75A/m0.75A/m

According to Curie's Law,
χ∝1T\chi \propto \dfrac{1}{T}
Therefore, χ1T1=χ2T2{\chi }_{1}{T}_{1}={\chi }_{2}{T}_{2}
Also, χ=MB\chi =\dfrac{M}{B}
So,
M1T1B1=M2T2B2\dfrac{{M}_{1}{T}_{1}}{{B}_{1}}=\dfrac{{M}_{2}{T}_{2}}{{B}_{2}}
On substituting the values,
6×40.4=M2×240.3\dfrac{6\times 4}{0.4}=\dfrac{{M}_{2}\times 24}{0.3}
Therefore, M2=0.75Am−1{M}_{2}=0.75A{m}^{-1}

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