Question

Every iron atom in a ferromagnetic domain in iron has a magnetic dipole moment equal to $9.27\times {10}^{-24}A-m^2$. A ferromagnetic domain in iron has the shape of a cube of side $1\mu m$. The maximum dipole moment occurs when all the dipoles are aligned. The molar mass of iron is $55g$ and its specific gravity is $7.9$. The magnetisation of the domain is:

• A. $8.0\times {10}^{5}A/m$
• B. $8.0\times {10}^{8}A/m$
• C. $8.0\times {10}^{11}A/m$
• D. $8.0\times {10}^{14}A/m$

Answer 1

The correct option is A $8.0\times {10}^{5}A/m$

Dipole moment, $p=9.27\times {10}^{-24}A-m^2$

$a={10}^{-6}m$

$m=55g$

$N_A=6.023\times {10}^{24}atom$

Volume of the domain, $V=a^3$

$V={10}^{-12}cm={10}^{-18}m$

Density, $\rho =7.9g/c{m}^3$

Mass of domain, $M=V\times \rho$

$M={10}^{-12}\times 7.9$

$M=7.9\times {10}^{-12}g$

The number of atom in domain,

$n=M\frac{N_A}{m}$

$n=7.9\times {10}^{-12}\times \frac{6.023\times {10}^{24}}{55g}$

$n=8.65\times {10}^{10}atom$

The total dipole moment, $P=np$

$P=8.65\times {10}^{10}\times 9.27\times {10}^{-24}$

$P=8\times {10}^{-13}A-m^2$

The magnetization of the domain, $M=\frac{P}{V}$

$M=\frac{8\times {10}^{-13}}{{10}^{-18}}$

$M=8\times {10}^{5}A/m$

The correct option is A.