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QJEEM20Ph06c8: JEE Main 2020 Physics - Class 12 JEEM20Ph06c8 Solution: A paramagnetic sample shows a net magnetisation of 6 A/m when…

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:

Multiple Choice Options

  • A 4 A/m
  • B 1 A/m
  • C 0.75 A/m
  • D 2.25 A/m

Step-by-Step Verified Solution

According to curies law

χ\chi =CBextT= {{C{B_{ext}}} \over T}

⇒\Rightarrow 6=C×0.446 = {{C \times 0.4} \over 4}

⇒C=60\Rightarrow C = 60

∴\therefore Case - II : χ\chi =60×0.324= {{60 \times 0.3} \over {24}} =60×3240=34=0.75 A/m= {{60 \times 3} \over {240}} = {3 \over 4} = 0.75\,A/m

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