Question

The saturation magnetization $M_{max}$ of the ferromagnetic metal nickel is $4.70\times {10}^{5}A/m$. Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is $8.90g/c{m}^3$ and its molar mass is $58.71g/mol$.)

Answer 1

The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by $M_{sat}=\mu n$, where n is the number of atoms per unit volume and μ is the magnetic dipole moment of an atom. The number of nickel atoms per unit volume is $n=\rho /m$, where ρ is the density of nickel. The mass of a single nickel atom is calculated using $m=M/N_A$, where M is the atomic mass of nickel, and $N_A$ is Avogadro’s constant. Thus,
$n=\frac{\rho N_A}{M}=\frac{\left(8.90g/c{m}^3\right)(6.02\times {10}^{23}atoms/mol)}{58.71g/mol}=9.126\times {10}^{22}atoms/c{m}^3$
$=9.126\times {10}^{28}atoms/m^3$
The dipole moment of a single atom of nickel is
$\mu =\frac{M_{sat}}{n}=\frac{4.70\times {10}^{5}A/m}{9.126\times {10}^{28}{m}^3}=5.15\times {10}^{-24}A.m^2$