Question

Every iron atom in a ferromagnetic domain in iron has a magnetic dipole moment equal to 9.27*10^-24A-m². A ferromagnetic domain in iron has the shape of a cube of side 1micrometer. 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--

1. 8.0*10⁵A/m
2. 8.0*10⁸A/m
3. 8.0*10^{11 A/m}
4. 8.0*10¹⁴ A/m

Answer 1

Here the volume of one domain is ${\left(1\mu m\right)}^3={10}^{-12}c{m}^3={10}^{-18}{m}^3$
So the mass of the domain is ${10}^{-12}\times 7.9\text{g}$. So the no of atoms in one domain is $\frac{6.023\times {10}^{23}}{55}\times 7.9\times {10}^{-12}=8.65\times {10}^{10}$
Now the total dipole moment is $8.65\times {10}^{10}\times 9.27\times {10}^{-24}$
Now the magnetization is given by $\frac{dm}{dV}=\frac{8.65\times {10}^{10}\times 9.27\times {10}^{-24}}{{10}^{-18}}=8\times {10}^5$