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(A) Magnetization $(M)$ is defined as the net magnetic moment per unit volume.$M = \frac{M_{	ext{net}}}{V}$Given:Number of atoms $(N)$ = $4 	imes 10

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$8 	imes 10^4 	ext{ A m}^{-1}$

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(A) Magnetization $(M)$ is defined as the net magnetic moment per unit volume.$M = \frac{M_{	ext{net}}}{V}$Given:Number of atoms $(N)$ = $4 	imes 10^{14}$Magnetic moment of each atom $(m)$ = $16 	imes 10^{-24} 	ext{ A m}^2$Side length of the cube $(a)$ = $2 \mu m = 2 	imes 10^{-6} 	ext{ m}$Volume of the cube $(V)$ = $a^3 = (2 	imes 10^{-6})^3 = 8 	imes 10^{-18} 	ext{ m}^3$Total magnetic moment $(M_{	ext{net}})$ = $N 	imes m = (4 	imes 10^{14}) 	imes (16 	imes 10^{-24}) = 64 	imes 10^{-10} 	ext{ A m}^2$Now,calculating magnetization:$M = \frac{64 	imes 10^{-10}}{8 	imes 10^{-18}}$$M = 8 	imes 10^8 	imes 10^{-4} = 8 	imes 10^4 	ext{ A m}^{-1}$

$A$ domain in a ferromagnetic substance is in the form of a cube of side length $2 \mu m$. It contains $4 \times 10^{14}$ atoms and each atomic dipole has a moment of $16 \times 10^{-24} \text{ A m}^2$. The magnetization of the domain is . . . . . . .

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