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MHT CET 2020 (19 Oct, Shift 2) · Magnetisation: B, H, M, χ & μ

Magnetisation: B, H, M, χ & μ question from MHT CET 2020 (19 Oct, Shift 2)

Q42MHT CET 2020 · 19 OCT · SHIFT 2Medium

Magnetisation: B, H, M, χ & μ

A domain in a ferromagnetic substance is in the form of a cube of side 1 μm1\,\mu\text{m}. If it contains 8×10108 \times 10^{10} atoms and each atomic dipole has a dipole moment of 9×10−24 Am29 \times 10^{-24}\text{ Am}^2, then the magnetisation of the domain is

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Answer: (a)

  1. Total moment =8×1010×9×10−24=7.2×10−13 A m2= 8 \times 10^{10} \times 9 \times 10^{-24} = 7.2 \times 10^{-13}\ \text{A m}^2
  2. Volume =(10−6)3=10−18 m3= (10^{-6})^3 = 10^{-18}\ \text{m}^3
  3. Magnetisation =7.2×10−1310−18=7.2×105 A m−1= \dfrac{7.2 \times 10^{-13}}{10^{-18}} = 7.2 \times 10^{5}\ \text{A m}^{-1}
  4. So the answer is (a).
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