$A$ domain in ferromagnetic iron is in the form of a cube of side length $1\; \mu m$. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is $55\; g/mol$ and its density is $7.9\; g/cm^3$. Assume that each iron atom has a dipole moment of $9.27 \times 10^{-24}\; A m^2$.

(N/A) The volume of the cubic domain is $V = (10^{-6}\; m)^3 = 10^{-18}\; m^3 = 10^{-12}\; cm^3$.
The mass of the domain is $\text{volume} \times \text{density} = 7.9\; g/cm^3 \times 10^{-12}\; cm^3 = 7.9 \times 10^{-12}\; g$.
Using Avogadro's number $(N_A = 6.023 \times 10^{23}\; mol^{-1})$,the number of iron atoms $N$ is given by $N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{7.9 \times 10^{-12}\; g}{55\; g/mol} \times 6.023 \times 10^{23}\; mol^{-1} \approx 8.65 \times 10^{10}$ atoms.
The maximum possible dipole moment $m_{\max}$ is achieved when all atomic moments are perfectly aligned: $m_{\max} = N \times (9.27 \times 10^{-24}\; A m^2) = (8.65 \times 10^{10}) \times (9.27 \times 10^{-24}) \approx 8.0 \times 10^{-13}\; A m^2$.
The magnetisation $M_{\max}$ is $m_{\max} / V = (8.0 \times 10^{-13}\; A m^2) / (10^{-18}\; m^3) = 8.0 \times 10^5\; A/m$.