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Magnetic Properties of Matter question

2019 · 11 Jan · Shift 2 · Q62
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Magnetic Properties of Matter question

2019 · 11 Jan · Shift 2 · Q62

JEE MainPhysicsMagnetic Properties of MatterMCQ+4 / −1
A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 ×\times× 10–6 J/ T when a magnetic intensity of 60 ×\times× 103 A/m is applied. Its magnetic susceptibility is
  1. A
    3.3 ×\times× 10–4
  2. B
    2.3 ×\times× 10–2
  3. C
    4.3 ×\times× 10–2
  4. D
    3.3 ×\times× 10–2
View written solutionFree

Correct answer: A

  1. Given data
  • Side of cube: 1 cm=10−2 m1\,\text{cm} = 10^{-2}\,\text{m}1cm=10−2m
  • Volume of cube: V=(10−2)3=10−6 m3V = (10^{-2})^3 = 10^{-6}\,\text{m}^3V=(10−2)3=10−6m3
  • Magnetic dipole moment: m=20×10−6 J/Tm = 20\times 10^{-6}\,\text{J/T}m=20×10−6J/T Since 1 J/T=1 A⋅m21\,\text{J/T} = 1\,\text{A}\cdot\text{m}^21J/T=1A⋅m2, so m=20×10−6 A⋅m2m = 20\times 10^{-6}\,\text{A}\cdot\text{m}^2m=20×10−6A⋅m2
  • Magnetic intensity: H=60×103 A/mH = 60\times 10^3\,\text{A/m}H=60×103A/m
  1. Find magnetization

Magnetization is magnetic dipole moment per unit volume: M=mVM = \frac{m}{V}M=Vm​

Substitute values: M=20×10−610−6=20 A/mM = \frac{20\times 10^{-6}}{10^{-6}} = 20\,\text{A/m}M=10−620×10−6​=20A/m

  1. Use relation between magnetization and susceptibility

For a paramagnetic substance: M=χHM = \chi HM=χH Therefore, χ=MH\chi = \frac{M}{H}χ=HM​

Substitute values: χ=2060×103\chi = \frac{20}{60\times 10^3}χ=60×10320​

χ=13×103=3.3×10−4\chi = \frac{1}{3\times 10^3} = 3.3\times 10^{-4}χ=3×1031​=3.3×10−4

  1. Match with options

χ=3.3×10−4\chi = 3.3\times 10^{-4}χ=3.3×10−4

So the correct option is A.

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