Askeland1996_Chapter_MagneticBehaviourOfMaterials

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Chapter 19 Magnetic Behaviour of Materials 
19.1 Calculate and compare the maximum magnetisation we would expect in iron, 
nickel, cobalt, and gadolinium. There are seven electrons in the 4f level of 
gadolinium. 
Iron: The number of atoms/m3 is: 
2 atoms/cell = 0.085 X 1030 atoms/m3 
(2.866 x 10-10 m)3 
M (0.085 x 1030 ) (4 magnetons/atom) (9.27 x 10-24 A m2) 
3.15 x 106 A.m- 1 
Nickel: The number of atoms/m3 is: 
4 atoms/cell = 0.09197 x 1030 atoms/m3 
(3.5167 x 10-10 m)3 
M (0.09197 x 1030 ) (2 magnetons/atom) (9.27 x 10-24 A m2) 
1.705 x 106 A.m-1 
Cobalt: The number of atoms/m3 is: 
2 atoms/cell 0.0903 X 1030 atoms/m3 
(2.5071 x 10 10 m)2(4.0686 x 10-10)cos30 
M (0.0903 x 1030 ) (3 magnetons/atom) (9.27 x 10-24 A m2) 
2.51 x 10 6 A.m- 1 
Gadolinium: The number of atoms/m3 is: 
2 atoms/cell 0.0303 x 1030 atoms/m3 
(3.6336 x 10-10 m)2(5.781 x 10-10 m)cos30 
M (0.0303 x 1030 ) (7 magnetons/atom) (9.27 x 10-24 A m2) 
1.96 x 106 A.m- 1 
19.2 An alloy of nickel and cobalt is to be produced to give a magnetisation 
of 2 x 106 A.m-l. The crystal structure of the alloy is FCC with a lattice 
parameter of 0.3544 nm. Determine the atomic percent cobalt required, 
assuming no interaction between the nickel and cobalt. 
Let fNi be the atomic fraction of nickel,' 1 - fNi is then the atomic 
fraction of cobalt. The numbers of Bohr magnetons per cubic metre due 
to nickel and to cobalt atoms are: 
Ni: (4 atoms/cell) (2 magnetons/atom)fNj(3.544 x 10-10 m)3 
= 0.1797 x 1030 fNi 
Co: (4 atoms/cell) (3 magnetons/atom) (1-fNi )/(3.544 x 10-10 m)3 
= 0.2696 X 1030 (1 - f Ni ) 
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D. R. Askeland, The Science and Engineering of Materials 
© Springer Science+Business Media Dordrecht 1996 
The magnetisation, M, is given by: 
M [(0.1797 X 103°)fNi + (0.2696 x 1030 ) (l-fNi )) (9.27 x 10-24 ) 
M -833373 fNi + 2499192 2 x 106 
fNi = 0.60 feo = 0.40 
19.3 Estimate the magnetisation that might be produced in an alloy containing 
nickel and 70 at% copper, assuming that no interaction occurs. 
We can estimate the lattice parameter of the alloy from those of the 
pure nickel and copper and their atomic fractions: 
a o = ( 0 . 3) (0. 3 2 94 ) + ( 0 . 7) (0. 3 6 151 ) = O. 3 519 nm 
If the copper does not provide magnetic moments that influence 
magnetisation, then 
M = (4 atoms/cell) (0.3 fraction Ni) (2 maqnetons/Ni atom) (9.27 x 10-24 ) 
(3.519 x 10-10 m)3 
M 5. lOx 105 A. m- 1 
19.4 An Fe-80% Ni alloy has a 
an inductance of 0.35 tesla is 
coil that is 20 mm in length. 
coil to obtain this field? 
maximum relative permeability of 300,000 when 
obtained. The alloy is placed in a 20 turn 
What current must flow through the conductor 
Magnetic permeability ~ 
but, 
Then, 
B 
H 
I 
~r~O 
300000 X 4rr x 10- 7 = 0.377 T.m.A- 1 • 
~H 
(0.35T) / (0.377 T.m.A- 1 ) = 0.928 A.m.- 1 
HI/n 
(0.928 A.m- 1 ) (20 x 10-3 m) / (20 turns) 
0.00093A 
19.5 An Fe-49% Ni alloy has a maximum permeability of 64,000 when a magnetic 
field of 10 A.m- 1 is applied. What inductance is obtained and what current is 
needed to obtain this inductance in a 200 turn, 30 mm long coil? 
~ 
B 
I 
~rf.1.0 = 64,000 X 4rr x 10- 7 = 0.0804 T.m.A- 1 
f.1.H = (0.0804 T.m.K 1 ) (10 A.m- 1 ) = 0.804 T 
Hi/n = (10 A.m-1 ) (30 x 10-3m) / (200 turns) 1.5 mA 
19.6 The following data describe the effect of the magnetic field on the 
inductance in a silicon steel. Calculate (a) the initial permeability and (b) 
the maximum permeability for the material. 1.4 
H (A.m- 1 ) B (T) 
0 0 
20 0.08 
40 0.3 
60 0.65 
80 0.85 
100 0.95 
150 1.10 
250 1.25 
203 
1.2 
E I 
III f 0.8 
_ 0.6 
~ 
] 0.4 
0.2 
50 100 150 200 250 
M.gnetic fodd. H (Aim) 
The data is plotted; from the graph, the initial and maximum 
permeability are calculated from the slopes of the plots: 
(a) initial permeability 0.003 T.m.K 1 
(b) maximum permeability 0.011 T.m.K 1 
19.7 A magnetic material has a coercive field of 167 A.m- 1 , a saturation 
magnetixation of 0.616 tesla, and a residual inductance of '0.3 tesla. Sketch 
the hysteresis loop for the material. 
0.616 T 
0.6 B(T) 
Br 0.3 T 
-600 400 600 
H(Alm) 
-0.6 
19.8 A magnetic material has a coercive field of 10.74 A.m- 1 , a saturation 
magnetization of 2.158 tesla, and a remanence induction of 1.183 tesla. 
Sketch the hysteresis loop for the material. 
Msat = 2.158 T 
B(T) 
Br 1.183 T 
-20 -10 10 20 
H(Alm) 
-2 
19.9 Using Figure 19.16, determine the following properties of the magnetic 
material. 
a. remanence d. initial permeability 
b. saturation magnetisation e. maximum permeability 
c. coercive field f. power (maximum BH product) 
(a) remanence = 1.3 T 
(b) saturation magnetisation = 1.4 T 
(c) coercive field = 63500 A.m-1 
(d) initial permeability 0.7 T / 100000 A.m-1 = 7 x 10- 6 T.m.Kl 
(e) maximum permeability 1.4 T / 75000 A.m-1 = 1.9 X 10- 5 T.m.A- 1 
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(f) we can try several BH products in the 4th quadrant: 
1.2 T x 35800 
1.0 T x 54600 
0.8 T x 57600 
42960 T.A.m-l 
54600 T.A.m-l 
46080 T.A.m-l 
The maximum BH product, or power, is about 54600 T.A.m- 1 • 
19.10 Using Figure 19.17, determine the following properties of the magnetic 
material. 
a. remanence d. initial permeability 
b. saturation magnetisation e. maximum permeability 
c. coercive field f. power (maximum BH product) 
(a) remanence = 0.55 T 
(b) saturation magnetisation = 0.58 T 
(c) coercive field = 44,000 A.m-1 
(d) initial permeability 0.2 T / (50000 A.m-1 ) 
4 x 10-6 T.m.A- 1 
(e) maximum permeability 0.5 T / (35000 A.m-1 ) 
1.4 x 10-5 T.m.A- 1 
(f) we can try several BH products in the 4th quadrant: 
0.45 T x 23000 A.m- 1 10350 T.A.m-l 
0.40 T x 32800 A.m-1 13120 T.A.m-l 
0.35 T x 36100 A.m-1 12635 T.A.m- 1 
0.30 T x 38975 A.m-1 11692 T.A.m- 1 
The maximum BH product, or power, is about 13120 T.A.m-l. 
19.11 Estimate the power of the CosCe material shown in Figure 19.14. 
H B BH 
0 A.m-1 0.75 T 0 T.A.m- 1 
200000 A.m- 1 0.65 T 130000 T.A.m- 1 
225000 A.m-1 0.45 T 101250 T.A.m- 1 
250000 A.m-1 0 T 0 T.A.m- 1 
19.12 Why are eddy current losses important design factors in ferromagnetic 
materials but less important in ferrimagnetic materials? 
The ferrimagnetic materials are ceramics, which have a very low elec­
trical conductivity; consequently eddy currents are not introduced into 
the magnetic material. The ferromagnetic materials are metals or 
alloys; they are conductors, permitting eddy currents to be introduced 
and thus causing eddy current losses. 
19.13 What advantage does the Fe-3% Si material have compared to Supermalloy 
for use in electric motors? 
The Fe-3% Si has a larger saturation inductance than Supermalloy, 
allowing more work to be done. However Fe-3% Si does require larger 
fields, since the coercive field for Fe-3% Si is large, and the per-
205 
meability of Fe-3% Si is small compared with that of Supermalloy. 
19.14 The coercive field for pure iron is related to the grain size of the 
iron by the relationship He = 1.83 + 4.14/VA, where A is the area of the grain 
in two dimensions (mm2 ) and He is in A.m- I • If only the grain size influences 
the 99.95% iron given in Table 19.3, estimate the size of the grains in the 
material. What happens when the iron is annealed to increase the grain size? 
He = 71.60 A.m-1 
Thus, from the equation, 
71.60 1.83 + 4.14 IVA 
4.14 / 69.77 = 0.0593 or A = 0.0035 mm2 
When the iron is annealed, the grain size increases, A increases, and 
the coercive field He decreases. 
19.15 Suppose we replace 10% of the Fe 2 + ions in magnetite with Cu2 + ions. 
Determine the total magnetic moment per cubic metre. 
From Example 19.6, the lattice parameter is 8.37 x 10- 10 m. 
Vuniteell = (8.37 X 10-10 m)3 = 5.86 x 10-28 m3 
In the octahedral sites, the fraction of copper atoms is 0.1, while the 
fraction of Fe2 + ions is 0.9. The magnetic moment is then: 
moment = (8 subcells) [0.1 Cu(l maqneton) + 0.9 Fe(4 maqneton)) (9.27x10-24Am- 1 ) 
5.86 x 10-28 m3 
moment 4.68 X 105 A.m2 per m3 
19.16 Suppose thatthe total magnetic moment per cubic metre in a spinel 
structure in which Ni 2+ ions have replaced a portion of the Fe 2 + ions is 4.6 X 
105 A.m- I • Calculate the fraction of the Fe2 + ions that have been replaced and 
the wt% Ni present in the spinel. 
From Example 19.6, the volume of the unit cell is 5.86 x 10-28 m3. 
let fm be the fraction of the octahedral sites occupied by nickel, 
(l-fNi ) is the fraction of the sites occupied by iron. Then: 
moment = 4. 6x10 5 
(8 subcells) [(l-fNi ) (4 maqnetons) + (fNi ) (2 maqnetons)) (9.27 x 10-24 ) 
5.86 x 10-28 m3 
3.6348 = 4 - 4fm + 2fNi 
:. fNi = 0.183 
Thus the number of each type of atom or ion in the unit cell is: 
oxygen: (4 atoms/subcell) (8 subcells) = 
Fe3+ (2 ions/subcell) (8 subcells) = 16 
Fe2 + (0.817) (1 ion/subcell) (8 subcells) 
Ni 2 + (0.183) (1 ion/subcell) (8 subcells) 
32 
6.54 
1.46 
If we 
then 
The total number of ions in the unit cell is 56; the atomic fraction of 
each ion is: 
206 
foxygen = 32/56 = 0.5714 
f Fe2+ = 6. 54/56 = O. 1168 
f Fe3 + = 16/56 = 0.2857 
f Ni2 + = 1.48/56 = 0.0261 
The weight percent nickel is (using the molecular weights of oxygen, 
iron and nickel): 
%Ni (0.0261) (58.7l) 
(0.5714) (16) + (0.2857) (55.847) + (0.1168) (55.847) + (0.0261) (58.71) 
= 4.62 wt% 
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