Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

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2nd Revise Page

730 • Chapter 18 / Magnetic Properties

mutual spin alignment exists over relatively large volume regions of the crystal called

domain domains(see Section 18.7).

The maximum possible magnetization, orsaturation magnetizationMs, of a fer-

saturation

magnetization

romagnetic material represents the magnetization that results when all the magnetic

dipoles in a solid piece are mutually aligned with the external field; there is also a

corresponding saturation flux densityBs. 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 18.1

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.

Solution

(a)The saturation magnetization is just the product of the number of Bohr

magnetons per atom (0.60 as given above), the magnitude of the Bohr

magnetonμB, and the numberNof atoms per cubic meter, or

Ms= 0. 60 μBN (18.9)

Saturation

magnetization for

nickel

Now, the number of atoms per cubic meter is related to the densityρ, the

atomic weightANi, and Avogadro’s numberNA, as follows:

N= (18.10)

ρNA

ANi

=

(8. 90 × 106 g/m^3 )(6. 02 × 1023 atoms/mol)

58 .71 g/mol

= 9. 13 × 1028 atoms/m^3

For nickel,

computation of the

number of atoms per

unit volume

Finally,

Ms=

(

0 .60 Bohr magneton

atom

)(

9. 27 × 10 −^24 A-m^2

Bohr magneton

)(

9. 13 × 1028 atoms

m^3

)

= 5. 1 × 105 A/m

(b)From Equation 18.8, the saturation flux density is just

Bs=μ 0 Ms

=

(

4 π× 10 −^7 H

m

)(

5. 1 × 105 A

m

)

= 0 .64 tesla