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

Solution :
The volume of the cube is
$$V = {\left( {{{10}^6} - m} \right)^3} = {10^{ - 18}}{m^3}$$
Net dipole moment $${m_{{\text{net}}}}$$
$$\eqalign{ & = 8 \times {10^{10}} \times 9 \times {10^{ - 24}}A{m^2} \cr & = 72 \times {10^{ - 14}}A{m^2} \cr} $$
Intensity of magnetization is magnetic moment developed per unit volume.
∴ magnetization, $$M = \frac{{{m_{{\text{net}}}}}}{{{\text{volume}}}}$$
$$ = \frac{{72 \times {{10}^{ - 14}}A{m^2}}}{{{{10}^{ - 18}}{m^3}}} = 7.2 \times {10^5}A{m^{ - 1}}$$