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Magnetic Properties of Matter question

2020 · 4 Sep · Shift 2 · Q64
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Magnetic Properties of Matter question

2020 · 4 Sep · Shift 2 · Q64

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
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:
  1. A
    4 A/m
  2. B
    1 A/m
  3. C
    0.75 A/m
  4. D
    2.25 A/m
View written solutionFree

Correct answer: C

  1. For a paramagnetic substance, Curie’s law gives M∝BTM \propto \frac{B}{T}M∝TB​ where MMM is magnetisation, BBB is external magnetic field, and TTT is absolute temperature.

  2. Therefore, M2M1=B2/T2B1/T1=B2T1B1T2\frac{M_2}{M_1}=\frac{B_2/T_2}{B_1/T_1}=\frac{B_2 T_1}{B_1 T_2}M1​M2​​=B1​/T1​B2​/T2​​=B1​T2​B2​T1​​

  3. Given: M1=6 A/m,B1=0.4 T,T1=4 KM_1=6\ \text{A/m},\quad B_1=0.4\ \text{T},\quad T_1=4\ \text{K}M1​=6 A/m,B1​=0.4 T,T1​=4 K B2=0.3 T,T2=24 KB_2=0.3\ \text{T},\quad T_2=24\ \text{K}B2​=0.3 T,T2​=24 K

  4. Substitute: M2=6×0.3×40.4×24M_2=6\times \frac{0.3\times 4}{0.4\times 24}M2​=6×0.4×240.3×4​

  5. Simplify: M2=6×1.29.6=6×0.125=0.75 A/mM_2=6\times \frac{1.2}{9.6}=6\times 0.125=0.75\ \text{A/m}M2​=6×9.61.2​=6×0.125=0.75 A/m

  6. Hence, the correct option is: C: 0.75 A/m\boxed{\text{C: }0.75\ \text{A/m}}C: 0.75 A/m​

  7. Option check:

    • A: 444 A/m — incorrect
    • B: 111 A/m — incorrect
    • C: 0.750.750.75 A/m — correct
    • D: 2.252.252.25 A/m — incorrect

The derived answer matches the stored correct answer.

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