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(B) The volume of the cube is $V = L^3 = (1 \mu m)^3 = (10^{-6} \ m)^3 = 10^{-18} \ m^3$.The total magnetic moment $M_{total}$ is the product of the n

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$7.2 	imes 10^{5} \ A/m$

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(B) The volume of the cube is $V = L^3 = (1 \mu m)^3 = (10^{-6} \ m)^3 = 10^{-18} \ m^3$.The total magnetic moment $M_{total}$ is the product of the number of atoms and the dipole moment per atom:$M_{total} = (8 	imes 10^{10}) 	imes (9 	imes 10^{-24} \ A \cdot m^2) = 72 	imes 10^{-14} \ A \cdot m^2 = 7.2 	imes 10^{-13} \ A \cdot m^2$.The magnetisation $I$ (or $M$) is defined as the magnetic moment per unit volume:$I = \frac{M_{total}}{V} = \frac{7.2 	imes 10^{-13} \ A \cdot m^2}{10^{-18} \ m^3} = 7.2 	imes 10^5 \ A/m$.

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

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