29.19: Magnetic Moment of an Electron
Electrons revolving around a nucleus are analogous to a circular current carrying loop. This current produces a magnetic dipole moment proportional to the electron's orbital angular momentum. Since the orbital angular momentum is quantized in terms of the reduced Planck's constant, the dipole moment is quantized in the Bohr Magneton. The value of the Bohr magneton is 9.27 x 10-24 Am2. Electrons also have an intrinsic spin angular momentum, and the associated spin magnetic moment is approximately one Bohr magneton. The orbital and spin angular momentum combine vectorially to give the net magnetic moment of a material. The vector sum of these moments determines the magnetic properties in materials.
Consider an example where the saturation magnetization for iron is estimated. The unknown quantity can be evaluated using the known quantities, such as the atomic weight of iron as 55.8 amu, which has a density of 7.87 g/cm3 and a net magnetic moment per atom of 2.2 Bohr magnetons.
Now, the number of atoms per unit volume of the material is given by the following expression:
number of atoms per unit volume = (density x Avogadro's number) / atomic mass
Substituting the known density values, Avogadro's number, and atomic mass into the above expression, the number of atoms per unit volume is calculated to be 0.849 x 1023 atom/cm3.
Now, the magnetic moment per atom is 2.2 Bohr magnetons. Using the Bohr magneton's value gives the magnetic moment value per atom as 20.394 x 10-24 Am2.
Finally, the saturation magnetization equals the number of atoms per unit volume multiplied by the magnetic moment per atom. As a result, the computed value of the saturation magnetization for iron is 1.7314 Am2.
