The domain in ferromagnetic material is in the form of a cube of side $2 \mu \mathrm{~m}$. Number of atoms…

The domain in ferromagnetic material is in the form of a cube of side $2 \mu \mathrm{~m}$. Number of atoms in that domain is $9 \times 10^{10}$ and each atom has a dipole movement of $9 \times 10^{-24} \mathrm{Am}^2$. The magnetisation of the domain is (approximately),
  1. $10 \times 10^4 \mathrm{Am}^{-1}$
  2. $8 \times 10^4 \mathrm{Am}^{-1}$
  3. $12 \times 10^4 \mathrm{Am}^{-1}$
  4. $9 \times 10^4 \mathrm{Am}^{-1}$

Solution

Volume of the domain, $\mathrm{V}=\left(2 \times 10^{-6}\right)^3=8 \times 10^{-18} \mathrm{~m}^3$ $\begin{aligned} & M_{\text {net }}=9 \times 10^{10} \times 9 \times 10^{-24}=81 \times 10^{-14} \mathrm{Am}^2 \\ & \therefore \text { Magnetisation, } I=\frac{M_{\text {net }}}{V}=\frac{81 \times 10^{-14}}{8 \times 10^{-18}} \\ & =10 \times 10^4 \mathrm{~A} \mathrm{m}^{-1}\end{aligned}$

Asked in: AP EAMCET 2024 (19 May Shift 2)

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