Callister - An introduction - 8th edition

808 • Chapter 20 / Magnetic Properties
H = 0 Figure 20.7 Schematic illustration of the mutual alignment of atomic
dipoles for a ferromagnetic material, which will exist even in the
absence of an external magnetic field.
is also a corresponding saturation flux density B s . The saturation magnetization is
equal to the product of the net magnetic moment for each atom and the number
of atoms present. For each of iron, cobalt, and nickel, the net magnetic moments
per atom are 2.22, 1.72, and 0.60 Bohr magnetons, respectively.
EXAMPLE PROBLEM 20.1
Saturation Magnetization and Flux Density Computations
for Nickel
Calculate (a) the saturation magnetization and (b) the saturation flux density
for nickel, which has a density of 8.90 g/cm 3 .
Saturation
magnetization for
nickel
For nickel,
computation of the
number of atoms per
unit volume
Solution
(a) The saturation magnetization is just the product of the number of Bohr
magnetons per atom (0.60 as given earlier), the magnitude of the Bohr
magneton B , and the number N of atoms per cubic meter, or
(20.9)
Now, the number of atoms per cubic meter is related to the density , the
atomic weight A Ni , and Avogadro’s number N A , as follows:
N rN A
A Ni
M s 0.60m B N
(20.10)
Finally,
M s a
5.1 10 5 A/m
18.90 106 gm 3 216.022 10 23 atomsmol2
58.71 gmol
9.13 10 28 atoms/m 3
0.60 Bohr magneton
atom
(b) From Equation 20.8, the saturation flux density is just
B s m 0 M s
a 4p 107 H
m
0.64 tesla
ba 9.27 1024 A# m
2
Bohr magneton
ba 5.1 105 A
b
m
ba9. 13 1028 atoms
b
m 3