Q145 / 150 · MHT CET 2020 (19 Oct Shift 2) · Physics

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

  • A
    $7 \cdot 2 \times 10^{9} \mathrm{Am}^{-1}$
  • Correct answer
    B
    $7 \cdot 2 \times 10^{5} \mathrm{Am}^{-1}$
  • C
    $7 \cdot 2 \times 10^{12} \mathrm{Am}^{-1}$
  • D
    $7 \cdot 2 \times 10^{3} \mathrm{Am}^{-1}$

Solution

(A) Volume of cube $=\ell^{3}=\left(10^{-6}\right)^{3}=10^{-18} \mathrm{~m}^{3}$ Total magnetic moment $\mathrm{M}=8 \times 10^{10} \times 9 \times 10^{-24}$ $=72 \times 10^{-14} \mathrm{~A}-\mathrm{m}^{2}$ Magnitization $=\frac{\mathrm{M}}{\mathrm{V}}=\frac{72 \times 10^{-14}}{10^{-18}}=7.2 \times 10^{5} \mathrm{~A} / \mathrm{m}$

Hence, the correct answer is option: 'B'.

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