## Q. 20.1

Theoretical and Actual Saturation Magnetization in Fe
Calculate the maximum, or saturation, magnetization that we expect in iron. The lattice parameter of BCC iron is 2.866 Å. Compare this value with 2.1 tesla (a value of saturation flux density experimentally observed for pure Fe).

## Verified Solution

Based on the unpaired electronic spins, we expect each iron atom to have four electrons that act as magnetic dipoles. The number of atoms per m³ in BCC iron is
Number of Fe atoms/m³ = $\frac{2 \ atoms/cell}{(2.866 × 10^{-10} \ m)^3}= 8.496 × 10^{28}$

The maximum volume magnetization ($M_{sat}$) is the total magnetic moment per unit volume:

$M_{sat} = (8.496 × 10^{28} \ \frac{atoms}{m^3})(9.274 × 10^{-24} \ A \cdot m^2)(4 \ \frac{Bohr \ magnetons}{atom})$
$M_{sat} = 3.15 × 10^{6} \ \frac{A}{m}$

To convert the value of saturation magnetization M into saturation flux density B in tesla, we need the value of $\mu _{0}$M. In ferromagnetic materials $\mu _{0}$M >> $\mu _{0}$H and therefore, $B_{sat}\cong \mu _{0}M$.
Thus, the saturation induction or saturation flux density in tesla is given by
$B_{sat}= \mu _{0}M_{sat}$.

$B_{sat}=(4 \pi × 10^{-7} \ \frac{Wb}{A \cdot m})(3.15 × 10^6 \ \frac{A}{m})$
$B_{sat}=3.96 \ \frac{Wb}{m^2}=3.96 \ tesla$

This is almost two times the experimentally observed value of 2.1 tesla. Reversing our calculations, we can show that the each iron atom contributes only about 2.1 Bohr magneton and not 4. This is the difference between behavior of individual atoms and their behavior in a crystalline solid. It can be shown that in the case of iron, the difference is due to the 3d electron orbital moment being quenched in the crystal.