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(A) The volume of the domain $V$ is given by the cube of its side length: $V = (2 	imes 10^{-6} m)^3 = 8 	imes 10^{-18} m^3$. The net magnetic dipol

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$10 	imes 10^4 A m^{-1}$

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(A) The volume of the domain $V$ is given by the cube of its side length: $V = (2 	imes 10^{-6} m)^3 = 8 	imes 10^{-18} m^3$. The net magnetic dipole moment $M_{	ext{net}}$ is the product of the number of atoms and the dipole moment per atom: $M_{	ext{net}} = (9 	imes 10^{10}) 	imes (9 	imes 10^{-24} A m^2) = 81 	imes 10^{-14} A m^2$. Magnetization $I$ is defined as the net magnetic moment per unit volume: $I = \frac{M_{	ext{net}}}{V} = \frac{81 	imes 10^{-14} A m^2}{8 	imes 10^{-18} m^3}$. Calculating the value: $I = 10.125 	imes 10^4 A m^{-1} \approx 10 	imes 10^4 A m^{-1}$.

The domain in a ferromagnetic material is in the form of a cube of side $2 \mu m$. The number of atoms in that domain is $9 \times 10^{10}$ and each atom has a dipole moment of $9 \times 10^{-24} A m^2$. The magnetization of the domain is (approximately):

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