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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 $$8.90\, g/cm^3$$ and its molar mass is $$58.71\, g/mol$$.)

Solution
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The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by $$M_{sat} = μn$$, 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 $$n = ρ/m$$, 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,
$$n=\dfrac{\rho N_A}{M}=\dfrac{(8.90\,g/cm^3)(6.02 \times 10^{23} \,atoms/mol)}{58.71\,g/mol}=9.126 \times 10^{22}\,atoms/cm^3$$
$$=9.126 \times 10^{28}\,atoms/m^3$$
The dipole moment of a single atom of nickel is
$$\mu=\dfrac{M_{sat}}{n}=\dfrac{4.70 \times 10^5\,A/m}{9.126 \times 10^{28}\,m^3 }=5.15 \times 10^{-24}\,A.m^2$$

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