20 CHAPTER MAGNETIC PROPERTIES PROBLEM SOLUTIONS
Basic Concepts
20.1 A coil of wire 0.20 m long and having 200 turns carries a current of 10 A. (a) What is the magnitude of the magnetic field strength H? (b) Compute the flux density B if the coil is in a vacuum. (c) Compute the flux density inside a bar of titanium that is positioned within the coil. The susceptibility for titanium is found in Table 20.2. (d) Compute the magnitude of the magnetization M. Solution (a) We may calculate the magnetic field strength generated by this coil using Equation 20.1 as
=
=
(200 turns)(10 A) = 10, 000 A - turns/m 0.20 m
(b) In a vacuum, the flux density is determined from Equation 20.3. Thus,
0 = 0
= (1.257 10 -6 H/m)(10, 000 A - turns/m) = 1.257 10 -2 tesla (c) When a bar of titanium is positioned within the coil, we must use an expression that is a combination of Equations 20.5 and 20.6 in order to compute the flux density given the magnetic susceptibility. Inasmuch as = 1.81 10-4 (Table 20.2), then
m
= 0 + 0 = 0 + 0 = 0 (1 + c m)
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= (1.257 10 -6 H/m) (10, 000 A - turns/m)(1 + 1.81 10 -4 ) = 1.257 10-2 tesla which is essentially the same result as part (b). This is to say that the influence of the titanium bar within the coil makes an imperceptible difference in the magnitude of the B field. (d) The magnetization is computed from Equation 20.6:
= c m H = (1.81 10 -4 )(10, 000 A - turns/m) = 1.81 A/m
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20.2 Demonstrate that the relative permeability and the magnetic susceptibility are related according to Equation 20.7. Solution This problem asks us to show that m and r are related according to m = r 1. We begin with Equation 20.5 and substitute for M using Equation 20.6. Thus,
= 0 + 0 = 0 + 0
But B is also defined in Equation 20.2 as
=
When the above two expressions are set equal to one another as = 0 + 0 This leads to = 0 (1 + c m) If we divide both sides of this expression by 0, and from the definition of r (Equation 20.4), then = mr = 1 + c m 0 or, upon rearrangement = 1 which is the desired result.
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20.3 It is possible to express the magnetic susceptibility m in several different units. For the discussion of this chapter, m was used to designate the volume susceptibility in SI units, that is, the quantity that gives the magnetization per unit volume (m3) of material when multiplied by H. The mass susceptibility m (kg) yields the magnetic moment (or magnetization) per kilogram of material when multiplied by H; and, similarly, the atomic susceptibility m(a) gives the magnetization per kilogram-mole. The latter two quantities are related to m through the relationships m = m(kg) mass density (in kg/m3) m(a) = m(kg) atomic weight (in kg)] When using the cgsemu system, comparable parameters exist, which may be designated by m, m(g), and m(a); the m and m are related in accordance with Table 20.1. From Table 20.2, m for silver is 2.38 105; convert this value into the other five susceptibilities. Solution For this problem, we want to convert the volume susceptibility of copper (i.e., 2.38 10-5) into other systems of units. For the mass susceptibility () = (kg / m3 )
=
- 2.38 10- 5 = - 2.27 10 -9 10.49 10 3 kg / m3
For the atomic susceptibility (a) = c m (kg) [ atomic weight (in kg) ] = (- 2.27 10 -9 ) (0.10787 kg/mol) = - 2.45 10 -10
For the cgs-emu susceptibilities, cm - 2.38 10- 5 = = - 1.89 10 -6 4p 4p ' cm - 1.89 10- 6 = = - 1.80 10 -7 3) r (g / cm 10.49 g/cm 3
' =
' (g) =
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' ' (a) = c m (g) [ atomic weight (in g) ] = (- 1.80 10 -7 )(107.87 g/mol) = - 1.94 10 -5
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20.4 (a) Explain the two sources of magnetic moments for electrons. (b) Do all electrons have a net magnetic moment? Why or why not? (c) Do all atoms have a net magnetic moment? Why or why not? Solution (a) The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin. (b) Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom. (c) All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.
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Diamagnetism and Paramagnetism Ferromagnetism
20.5 The magnetic flux density within a bar of some material is 0.435 tesla at an H field of 3.44 105 A/ m. Compute the following for this material: (a) the magnetic permeability, and (b) the magnetic susceptibility. (c) What type(s) of magnetism would you suggest is(are) being displayed by this material? Why? Solution (a) The magnetic permeability of this material may be determined according to Equation 20.2 as = B 0.435 tesla = = 1.2645 10 -6 H/m H 3.44 10 5 A / m
(b) The magnetic susceptibility is calculated using a combined form of Equations 20.4 and 20.7 as = mr - 1 = m -1 m0
=
1.2645 10- 6 H / m - 1 = 6.0 10 -3 1.257 10- 6 H / m
(c) This material would display both diamagnetic and paramagnetic behavior. All materials are diamagnetic, and since m is positive and on the order of 10-3, there would also be a paramagnetic contribution.
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20.6 The magnetization within a bar of some metal alloy is 3.2 105 A/m at an H field of 50 A/m. Compute the following: (a) the magnetic susceptibility, (b) the permeability, and (c) the magnetic flux density within this material. (d) What type(s) of magnetism would you suggest as being displayed by this material? Why? Solution (a) This portion of the problem calls for us to compute the magnetic susceptibility within a bar of some metal alloy when M = 3.2 105 A/m and H = 50 A/m. This requires that we solve for m from Equation 20.6 as = M 3.2 10 5 A / m = = 6400 H 50 A / m
(b) In order to calculate the permeability we must employ a combined form of Equations 20.4 and 20.7 as follows: = mr m0 = (c m + 1) m0 = (6400 + 1)(1.257 10 -6 H/m) = 8.05 10 -3 H/m
(c) The magnetic flux density may be determined using Equation 20.2 as
= mH = (8.05 10 -3 H/m) (50 A/m) = 0.40 tesla
(d) This metal alloy would exhibit ferromagnetic behavior on the basis of the magnitude of its m (6400), which is considerably larger than the m values for diamagnetic and paramagnetic materials listed in Table 20.2.
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20.7 Compute (a) the saturation magnetization and (b) the saturation flux density for cobalt, which has a net magnetic moment per atom of 1.72 Bohr magnetons and a density of 8.90 g/cm3. Solution (a) The saturation magnetization for Co may be determined in the same manner as was done for Ni in Example Problem 20.1. Thus, using a modified form of Equation 20.9
= 1.72 mB N
in which B is the Bohr magneton and N is the number of Co atoms per cubic meter. Also, there are 1.72 Bohr magnetons per Co atom. Now, N (the number of cobalt atoms per cubic meter) is related to the density and atomic weight of Co, and Avogadro's number according to Equation 20.10 as
=
r Co N A ACo
=
(8.90
10 6 g/m 3)(6.022 10 23 atoms/mol) 58.93 g/mol
= 9.10 10 28 atoms/m 3 Therefore,
= 1.72 mB N = (1.72 BM/atom)(9.27 10 -24 A - m2 /BM )(9.10 10 28 atoms/m 3)
= 1.45 10 6 A/m (b) The saturation flux density is determined according to Equation 20.8. Thus
= 0
= (1.257 10 -6 H/m)(1.45 10 6 A/m) = 1.82 tesla
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20.8
Confirm that there are 2.2 Bohr magnetons associated with each iron atom, given that the
saturation magnetization is 1.70 106 A/m, that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 nm. Solution We want to confirm that there are 2.2 Bohr magnetons associated with each iron atom. Therefore, let be the number of Bohr magnetons per atom, which we will calculate. This is possible using a modified and rearranged form of Equation 20.9that is =
Now, N is just the number of atoms per cubic meter, which is the number of atoms per unit cell (two for BCC, Section 3.4) divided by the unit cell volume-- that is,
=
2 2 =3 VC a
a being the BCC unit cell edge length. Thus ' B = Ms M a3 =s NmB 2mB
=
(1.70 10 6 A / m)[ (0.2866 10- 9 m) 3 / unit cell] (2 atoms/unit cell)(9.27 10 -24 A - m2 / BM )
= 2.16 Bohr magnetons/atom
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20.9 Assume there exists some hypothetical metal that exhibits ferromagnetic behavior and that has (1) a simple cubic crystal structure (Figure 3.24), (2) an atomic radius of 0.153 nm, and (3) a saturation flux density of 0.76 tesla. Determine the number of Bohr magnetons per atom for this material. Solution We are to determine the number of Bohr magnetons per atom for a hypothetical metal that has a simple cubic crystal structure, an atomic radius of 0.153 nm, and a saturation flux density of 0.76 tesla. It becomes necessary to employ Equation 20.8 and a modified form of Equation 20.9 as follows:
0 = = = 0
Here nB is the number of Bohr magnetons per atom, and N is just the number of atoms per cubic meter, which is the number of atoms per unit cell [one for simple cubic (Figure 3.23)] divided by the unit cell volumethat is,
=
1
which, when substituted into the above equation gives
=
0
For the simple cubic crystal structure (Figure 3.23), a = 2r, where r is the atomic radius, and VC = a3 = (2r)3. Substituting this relationship into the above equation yields
=
(2r ) 3
m0 mB
=
(0.76 tesla)(8)( 0.153 10- 9 m) 3 = 1.87 Bohr magnetons/atom (1.257 10- 6 H / m)(9.27 10- 24 A - m2 / BM)
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20.10 There is associated with each atom in paramagnetic and ferromagnetic materials a net magnetic moment. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot. Solution Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented. For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.
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Antiferromagnetism and Ferrimagnetism
20.11 Consult another reference in which Hunds rule is outlined, and on its basis explain the net magnetic moments for each of the cations listed in Table 20.4. Solution Hund's rule states that the spins of the electrons of a shell will add together in such a way as to yield the maximum magnetic moment. This means that as electrons fill a shell the spins of the electrons that fill the first half of the shell are all oriented in the same direction; furthermore, the spins of the electrons that fill the last half of this same shell will all be aligned and oriented in the opposite direction. For example, consider the iron ions in Table 20.4; from Table 2.2, the electron configuration for the outermost shell for the Fe atom is 3 d64s2. For the Fe3+ ion the outermost shell configuration is 3d5, which means that five of the ten possible 3d states are filled with electrons. According to Hund's rule the spins of all of these electrons are aligned, there will be no cancellation, and therefore, there are five Bohr magnetons associated with each Fe3+ ion, as noted in the table. For Fe2+ the configuration of the outermost shell is 3d6, which means that the spins of five electrons are aligned in one direction, and the spin of a single electron is aligned in the opposite direction, which cancels the magnetic moment of one of the other five; thus, this yields a net moment of four Bohr magnetons. For Mn2+ the electron configuration is 3d5, the same as Fe3+, and, therefore it will have the same number of Bohr magnetons (i.e., five). For Co2+ the electron configuration is 3d7, which means that the spins of five electrons are in one direction, and two are in the opposite direction, which gives rise to a net moment of three Bohr magnetons. For Ni2+ the electron configuration is 3d8 which means that the spins of five electrons are in one direction, and three are in the opposite direction, which gives rise to a net moment of two Bohr magnetons. For Cu2+ the electron configuration is 3d9 which means that the spins of five electrons are in one direction, and four are in the opposite direction, which gives rise to a net moment of one Bohr magneton.
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20.12 Estimate (a) the saturation magnetization, and (b) the saturation flux density of nickel ferrite [(NiFe2O4)8], which has a unit cell edge length of 0.8337 nm. Solution (a) The saturation magnetization of nickel ferrite is computed in the same manner as Example Problem 20.2; from Equation 20.13
=
3
Now, nB is just the number of Bohr magnetons per unit cell. The net magnetic moment arises from the Ni 2+ ions, of which there are eight per unit cell, each of which has a net magnetic moment of two Bohr magnetons (Table 20.4). Thus, nB is sixteen. Therefore, from the above equation
=
(16 BM / unit cell)(9.27 10- 24 A - m2 / BM ) ( 0.8337 10- 9 m) 3 / unit cell
= 2.56 10 5 A/m (b) This portion of the problem calls for us to compute the saturation flux density. From Equation 20.8
= 0
= (1.257 10 -6 H/m)(2.56 10 5 A/m) = 0.32 tesla
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20.13 The chemical formula for manganese ferrite may be written as (MnFe 2O4)8 because there are eight formula units per unit cell. If this material has a saturation magnetization of 5.6 105 A/m and a density of 5.00 g/cm3, estimate the number of Bohr magnetons associated with each Mn2+ ion. Solution We want to compute the number of Bohr magnetons per Mn2+ ion in (MnFe2O4)8. Let nB represent the number of Bohr magnetons per Mn2+ ion; then, using Equation 20.9, we have
=
in which N is the number of Mn2+ ions per cubic meter of material. But, from Equation 20.10
=
Here A is the molecular weight of MnFe2O4 (230.64 g/mol). Thus, combining the previous two equations
=
or, upon rearrangement (and expressing the density in units of grams per meter cubed),
=
=
(5.6 10 5 A/m) (230.64 g/mol)
(9.27
1 0- 24
A - m2 /BM )( 5.00 10 6 g/m 3)(6.022 10 23 ions / mol) = 4.6 Bohr magnetons/Mn 2 + ion
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a d 20.14 The formula for yttrium iron garnet (Y3Fe5O12) may be written in the form Y3c Fe2 Fe3 O12 , where
the superscripts a, c, and d represent different sites on which the Y 3+ and Fe3+ ions are located. The spin magnetic moments for the Y3+ and Fe3+ ions positioned in the a and c sites are oriented parallel to one another and antiparallel to the Fe3+ ions in d sites. Compute the number of Bohr magnetons associated with each Y3+ ion, given the following information: (1) each unit cell consists of eight formula (Y 3Fe5O12) units; (2) the unit cell is cubic with an edge length of 1.2376 nm; (3) the saturation magnetization for this material is 1.0 104 A/m; and (4) assume that there are 5 Bohr magnetons associated with each Fe3+ ion. Solution
c For this problem we are given that yttrium iron garnet may be written in the form Y3 Fe a Fe d O12 where 23
the superscripts a, c, and d represent different sites on which the Y3+ and Fe3+ ions are located, and that the spin magnetic moments for the ions on a and c sites are oriented parallel to one another and antiparallel to the Fe 3+ ions on the d sites. We are to determine the number of Bohr magnetons associated with each Y 3+ ion given that each unit cell consists of eight formula units, the unit cell is cubic with an edge length of 1.2376 nm, the saturation magnetization for the material is 1.0 10 A/m, and that there are 5 Bohr magnetons for each Fe3+ ion. The first thing to do is to calculate the number of Bohr magnetons per unit cell, which we will denote nB. Solving for nB using Equation 20.13, we get
4
=
3
=
(1.0 10 4 A / m)(1.2376 10- 9 m) 3
9.27 10- 24 A - m2 / BM
= 2.04 Bohr magnetons/unit cell
Now, there are 8 formula units per unit cell or
2.04 = 0.255 Bohr magnetons per formula unit. Furthermore, for 8
each formula unit there are two Fe3+ ions on a sites and three Fe3+ on d sites which magnetic moments are aligned antiparallel. Since there are 5 Bohr magnetons associated with each Fe3+ ion, the net magnetic moment contribution per formula unit from the Fe3+ ions is 5 Bohr magnetons. This contribution is antiparallel to the contribution from the Y3+ ions, and since there are three Y3+ ions per formula unit, then No. of Bohr magnetons/Y 3+ = 0.255 BM + 5 BM = 1.75 BM 3
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The Influence of Temperature on Magnetic Behavior
20.15 Briefly explain why the magnitude of the saturation magnetization decreases with increasing temperature for ferromagnetic materials, and why ferromagnetic behavior ceases above the Curie temperature. Solution For ferromagnetic materials, the saturation magnetization decreases with increasing temperature because the atomic thermal vibrational motions counteract the coupling forces between the adjacent atomic dipole moments, causing some magnetic dipole misalignment. spin coupling forces. Ferromagnetic behavior ceases above the Curie temperature because the atomic thermal vibrations are sufficiently violent so as to completely destroy the mutual
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Domains and Hysteresis
20.16 Briefly describe the phenomenon of magnetic hysteresis, and why it occurs for ferromagnetic and ferrimagnetic materials. The phenomenon of magnetic hysteresis and an explanation as to why it occurs for ferromagnetic and ferrimagnetic materials is given in Section 20.7.
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20.17 A coil of wire 0.1 m long and having 15 turns carries a current of 1.0 A. (a) Compute the flux density if the coil is within a vacuum. (b) A bar of an ironsilicon alloy, the B-H behavior for which is shown in Figure 20.29, is positioned within the coil. What is the flux density within this bar? (c) Suppose that a bar of molybdenum is now situated within the coil. What current must be used to produce the same B field in the Mo as was produced in the ironsilicon alloy [part (b)] using 1.0 A? Solution (a) This portion of the problem asks that we compute the flux density in a coil of wire 0.1 m long, having 15 turns, and carrying a current of 1.0 A, and that is situated in a vacuum. Combining Equations 20.1 and 20.3, and solving for B yields
0 = 0 =
0
=
(1.257 10- 6 H / m) (15 turns) (1.0 A)
0.1 m
= 1.89 10 -4 tesla
(b) Now we are to compute the flux density with a bar of the iron-silicon alloy, the B-H behavior for which is shown in Figure 20.29. It is necessary to determine the value of H using Equation 20.1 as
=
NI (15 turns)(1.0 A) = = 150 A - turns/m l 0.1 m
Using the curve in Figure 20.29, B = 1.65 tesla at H = 150 A-turns/m, as demonstrated below.
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(c) Finally, we are to assume that a bar of Mo is situated within the coil, and to calculate the current that is necessary to produce the same B field as when the iron-silicon alloy in part (b) was used. Molybdenum is a paramagnetic material having a solve for H
m
of 1.19 10
-4
(Table 20.2). Combining Equations 20.2, 20.4, and 20.7 we
=
=
= 0 0 (1 + c m)
And when Mo is positioned within the coil, then, from the above equation
=
(1.257 6
1.65
10- tesla = 1.312 10 6 A - turns/m H / m)(1 + 1.19 x 10- 4 )
Now, the current may be determined using Equation 20.1:
=
Hl = N
(1.312 10 6 A - turns / m) (0.1 m)
15 turns
= 8750 A
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20.18
A ferromagnetic material has a remanence of 1.25 teslas and a coercivity of 50,000 A/m.
Saturation is achieved at a magnetic field intensity of 100,000 A/m, at which the flux density is 1.50 teslas. Using these data, sketch the entire hysteresis curve in the range H = 100,000 to +100,000 A/m. Be sure to scale and label both coordinate axes. Solution The B versus H curve for this material is shown below.
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20.19 The following data are for a transformer steel:
H (A/m)
0 10
B (teslas)
0 0.03
H (A/m)
200 400
20 50 100 150
(a) Construct a graph of B versus H.
0.07 0.23 0.70 0.92
600 800 1000
B (teslas) 1.04 1.28 1.36 1.39 1.41
(b) What are the values of the initial permeability and initial relative permeability? (c) What is the value of the maximum permeability? (d) At about what H field does this maximum permeability occur? (e) To what magnetic susceptibility does this maximum permeability correspond? Solution (a) The B-H data for the transformer steel provided in the problem statement are plotted below.
(b) The first four data points are plotted below.
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The slope of the initial portion of the curve is i (as shown), is = DB (0.15 - 0) tesla = = 3.0 10 -3 H/m DH (50 - 0) A / m
Also, the initial relative permeability, ri, (Equation 20.4) is just = mi 3.0 10- 3 H / m = = 2387 m0 1.257 10- 6 H / m
(c) The maximum permeability is the tangent to the B-H curve having the greatest slope; it is drawn on the plot below, and designated as (max).
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The value of (max) is (modifying Equation 20.2) (max) = DB (1.3 - 0.3) tesla = = 8.70 10 DH (160 - 45) A - m
-3
H/m
(d) The H field at which (max) occurs is approximately 80 A/m [as taken from the plot shown in part (c)]. (e) We are asked for the maximum susceptibility, (max). Combining modified forms of Equations 20.7 and 20.4 yields () = () 1 = () 1 0
=
8.70 10- 3 H / m - 1 = 6920 1.257 10- 6 H / m
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20.20 An iron bar magnet having a coercivity of 4000 A/m is to be demagnetized. If the bar is inserted within a cylindrical wire coil 0.15 m long and having 100 turns, what electric current is required to generate the necessary magnetic field? Solution In order to demagnetize a magnet having a coercivity of 4000 A/m, an H field of 4000 A/m must be applied in a direction opposite to that of magnetization. According to Equation 20.1
=
=
(4000 A/m) (0.15 m) = 6A 100 turns
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20.21 A bar of an ironsilicon alloy having the BH behavior shown in Figure 20.29 is inserted within a coil of wire 0.20 m long and having 60 turns, through which passes a current of 0.1 A. (a) What is the B field within this bar? (b) At this magnetic field, (i) What is the permeability? (ii) What is the relative permeability? (iii) What is the susceptibility? (iv) What is the magnetization? Solution (a) We want to determine the magnitude of the B field within an iron-silicon alloy, the B-H behavior for which is shown in Figure 20.29, when l = 0.20 m, N = 60 turns, and I = 0.1 A. Applying Equation 20.1
=
NI (60 turns) (0.1 A) = = 30 A/m l 0.20 m
Below is shown the B-versus-H plot for this material. The B value from the curve corresponding to H = 30 A/m is about 1.37 tesla.
(b) (i) The permeability at this field is just B/H of the tangent of the B-H curve at H = 30 A/m. The slope of this line as drawn in the above figure is
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=
DB (1.70 - 1.04) tesla = = 1.10 10 -2 H/m DH (60 - 0) A / m
(ii) From Equation 20.4, the relative permeability is = m 1.10 10- 2 H / m = = 8751 m0 1.257 10- 6 H / m
(iii) Using Equation 20.7, the susceptibility is = mr - 1 = 8751 - 1 = 8750 (iv) The magnetization is determined from Equation 20.6 as
= c m H = (8750)(30 A/m) = 2.63 10 5 A/m
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Magnetic Anisotropy
20.22 Estimate saturation values of H for single-crystal iron in [100], [110], and [111] directions. Solution This problem asks for us to estimate saturation values of H for single crystal iron in the [100], [110], and [111] directions. All we need do is read values of H at the points at which saturation is achieved on the [100], [110], and [111] curves for iron shown in Figure 20.17. Saturation in the [100] direction is approximately 5400 A/ m. Corresponding values in [110] and [111] directions are approximately 40,000 and 47,000 A/m, respectively.
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20.23 The energy (per unit volume) required to magnetize a ferromagnetic material to saturation (E s) is defined by the following equation: Es =
0
Ms
m0 H dM
That is, Es is equal to the product of 0 and the area under an M versus H curve, to the point of saturation referenced to the ordinate (or M) axisfor example, in Figure 20.17 the area between the vertical axis and the magnetization curve to Ms. Estimate Es values (in J/m3) for single-crystal nickel in [100], [110], and [111] directions. Solution In this problem we are asked to estimate the energy required to magnetize single crystals of nickel in [100], [110], and [111] directions. These energies correspond to the products of 0 and the areas between the vertical axis of Figure 20.17 and the three curves for single crystal nickel taken to the saturation magnetization. For the [100] direction this area is about 15.8 108 A2/m2. When this value is multiplied by the value of 0 (1.257 10-6 H/m), we get a value of about 1990 J/m 3. The corresponding approximate areas for [110] and [111] directions are 9.6 108 A2/m2 and 3.75 108 A2/m2, respectively; when multiplied by 0 the respective energies for [110] and [111] directions are 1210 and 470 J/m3.
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Soft Magnetic Materials Hard Magnetic Materials
20.24 Cite the differences between hard and soft magnetic materials in terms of both hysteresis behavior and typical applications. Solution Relative to hysteresis behavior, a hard magnetic material has a high remanence, a high coercivity, a high saturation flux density, high hysteresis energy losses, and a low initial permeability; a soft magnetic material, on the other hand, has a high initial permeability, a low coercivity, and low hysteresis energy losses. With regard to applications, hard magnetic materials are utilized for permanent magnets; soft magnetic materials are used in devices that are subjected to alternating magnetic fields such as transformer cores, generators, motors, and magnetic amplifier devices.
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20.25 Assume that the commercial iron (99.95 wt% Fe) in Table 20.5 just reaches the point of saturation when inserted within the coil in Problem 20.1. Compute the saturation magnetization. Solution We want to determine the saturation magnetization of the 99.95 wt% Fe in Table 20.5, if it just reaches saturation when inserted within the coil described in Problem 20.1i.e., l = 0.20 m, N = 200 turns, and A = 10 A. It is first necessary to compute the H field within this coil using Equation 20.1 as
=
NI (200 turns)(10 A) = = 10, 000 A - turns/m l 0.20 m
Now, the saturation magnetization may be determined from a rearranged form of Equation 20.5 as
=
0
0
The value of Bs in Table 20.5 is 2.14 tesla; thus,
=
(2.14 tesla) - (1.257 10- 6 H / m)(10, 000 A / m) 1.257 10- 6 H / m
= 1.69 10 6 A/m
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20.26 Figure 20.30 shows the B-versus-H curve for a steel alloy. (a) What is the saturation flux density? (b) What is the saturation magnetization? (c) What is the remanence? (d) What is the coercivity? (e) On the basis of data in Tables 20.5 and 20.6, would you classify this material as a soft or hard magnetic material? Why? Solution The B-versus-H curve of Figure 20.30 is shown below.
(a) The saturation flux density for the steel, the B-H behavior for which is shown in Figure 20.30, is 1.3 tesla, the maximum B value shown on the plot. (b) The saturation magnetization is computed from Equation 20.8 as
=
0
=
1.3 tesla = 1.03 10 6 A/m 1.257 10- 6 H / m
(c) The remanence, Br, is read from this plot as from the hysteresis loop shown in Figure 20.14; its value is about 0.80 tesla. (d) The coercivity, Hc, is read from this plot as from Figure 20.14; the value is about 80 A/m.
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(e) On the basis of Tables 20.5 and 20.6, this is most likely a soft magnetic material. The saturation flux density (1.3 tesla) lies within the range of values cited for soft materials, and the remanence (0.80 tesla) is close to the values given in Table 20.6 for hard magnetic materials. However, the Hc (80 A/m) is significantly lower than for hard magnetic materials. Also, if we estimate the area within the hysteresis curve, we get a value of approximately 250 J/m3, which is in line with the hysteresis loss per cycle for soft magnetic materials.
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Magnetic Storage
20.27 Briefly explain the manner in which information is stored magnetically. The manner in which information is stored magnetically is discussed in Section 20.11.
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Superconductivity
20.28 For a superconducting material at a temperature T below the critical temperature TC, the critical field HC (T), depends on temperature according to the relationship T2 (T ) = HC (0) 1 - 2 TC where HC(0) is the critical field at 0 K. (a) Using the data in Table 20.7, calculate the critical magnetic fields for tin at 1.5 and 2.5 K.. (b) To what temperature must tin be cooled in a magnetic field of 20,000 A/m for it to be superconductive? Solution (a) Given Equation 20.14 and the data in Table 20.7, we are asked to calculate the critical magnetic fields for tin at 1.5 and 2.5 K. From the table, for Sn, TC = 3.72 K and BC(0) = 0.0305 tesla. Thus, from Equation 20.2
(20.14)
(0) (0) = m0
= 0.0305 tesla = 2.43 10 4 A/m 1.257 10- 6 H / m
Now, solving for HC(1.5) and HC(2.5) using Equation 20.14 yields 2 () = (0) 1 2 (1.5 K) 2 (1.5) = (2.43 10 4 A / m)1 = 2.03 10 4 A/m (3.72 K) 2 (2.5 K) 2 (2.5) = (2.43 10 4 A / m)1 = 1.33 10 4 A/m (3.72 K) 2
(b) Now we are to determine the temperature to which tin must be cooled in a magnetic field of 20,000 A/m in order for it to be superconductive. All we need do is to solve for T from Equation 20.14i.e.,
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(T ) = 1 HC (0)
And, since the value of HC(0) was computed in part (a) (i.e., 24,300 A/m), then
= (3.72 K) 1 -
20, 000 A/m = 1.56 K 24 , 300 A/m
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20.29
Using Equation 20.14, determine which of the superconducting elements in Table 20.7 are
superconducting at 3 K and in a magnetic field of 15,000 A/m. Solution We are asked to determine which of the superconducting elements in Table 20.7 are superconducting at 3 K and in a magnetic field of 15,000 A/m. First of all, in order to be superconductive at 3 K within any magnetic field, the critical temperature must be greater than 3 K. Thus, aluminum, titanium, and tungsten may be eliminated upon inspection. Now, for each of lead, mercury, and tin it is necessary, using Equation 20.14, to compute the value of HC(3)also substituting for HC(0) from Equation 20.3; if HC(3) is greater than 15,000 A/m then the element will be superconductive. Hence, for Pb T 2 BC (0) T2 (2) = (0) 1 = 12 2 m0 TC TC 0.0803 tesla (3.0 K) 2 1 = 5.28 10 4 A/m 1.257 10- 6 H / m (7.19 K) 2
=
Since this value is greater than 15,000 A/m, Pb will be superconductive. Similarly for Hg 0.0411 tesla (3.0 K) 2 1 = 1.56 10 4 A/m 1.257 10- 6 H / m (4.15 K) 2
(3) =
Inasmuch as this value is greater than 15,000 A/m, Hg will be superconductive. As for Sn 0.0305 tesla (3.0 K) 2 1= 8.48 10 3 A/m - 6 H /m 1.257 10 (3.72 K) 2
(3) =
Therefore, Sn is not superconductive.
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20.30 Cite the differences between type I and type II superconductors. Solution For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below HC, while at HC conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and uppercritical fields; so also is magnetic flux penetration gradual. Furthermore, type II generally have higher critical temperatures and critical magnetic fields.
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20.31 Briefly describe the Meissner effect. Solution The Meissner effect is a phenomenon found in superconductors wherein, in the superconducting state, the material is diamagnetic and completely excludes any external magnetic field from its interior. In the normal conducting state complete magnetic flux penetration of the material occurs.
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20.32 Cite the primary limitation of the new superconducting materials that have relatively high critical temperatures. Solution The primary limitation of the new superconducting materials that have relatively high critical temperatures is that, being ceramics, they are inherently brittle.
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DESIGN PROBLEMS
Ferromagnetism
20.D1 A cobaltnickel alloy is desired that has a saturation magnetization of 1.3 10 6 A/m. Specify its composition in weight percent nickel. Cobalt has an HCP crystal structure with c/a ratio of 1.623, whereas the maximum solubility of Ni in Co at room temperature is approximately 35 wt%. Assume that the unit cell volume for this alloy is the same as for pure Co. Solution For this problem we are asked to determine the composition of a Co-Ni alloy that will yield a saturation magnetization of 1.3 106 A/m. To begin, let us compute the number of Bohr magnetons per unit cell nB for this alloy from an expression that results from combining Equations 20.11 and 20.12. That is
B = N C = V
M s VC mB
in which Ms is the saturation magnetization, VC is the unit cell volume, and B is the magnitude of the Bohr magneton. According to Equation 3.S1 (the solution to Problem 3.7), for HCP
= 6 R2c 3
And, as stipulated in the problem statement, c = 1.623a; in addition, for HCP, the unit cell edge length, a, and the atomic radius, R are related as a = 2R. Making these substitutions into the above equation leads to the following:
= 6 R 2 c 3 = 6 R 2 (1.623a) 3 = 6 R 2 (1.623)(2R) 3
= 12 R 3 (1.623) 3 From the inside of the front cover of the book, the value of R for Co is given as 0.125 nm (1.25 10-10 m). Therefore,
= (12)(1.25 10 -10 m) 3 (1.623) 3
= 6.59 10 -29 m3
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And, now solving for nB from the first equation above, yields
B = =
M s VC (1.3 10 6 A / m)(6.59 10- 29 m3 / unit cell) = mB 9.27 10- 24 A - m2 Bohr magneton
= 9.24
Bohr magneton unit cell
Inasmuch as there are 1.72 and 0.60 Bohr magnetons for each of Co and Ni (Section 20.4), and, for HCP, there are 6 equivalent atoms per unit cell (Section 3.4), if we represent the fraction of Ni atoms by x, then
B = 9.24 Bohr magnetons/unit cell
0.60 Bohr magnetons 6 x Ni atoms .72 Bohr magnetons (6) (1 - x) Co atoms 1 = + Ni atom unit cell Co atom unit cell
And solving for x, the fraction of Ni atoms , x = 0.161, or 16.1 at% Ni. In order to convert this composition to weight percent, we employ Equation 4.7 as ' CNi ANi ' ' CNi ANi + CCo ACo
Ni =
100
=
(16.1 at %)(58.69 g/mol) 100 (16.1 at %)(58.69 g/mol) + (83.9 at %)(58.93 g/mol)
= 16.0 wt%
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Ferrimagnetism
20.D2 Design a cubic mixed-ferrite magnetic material that has a saturation magnetization of 4.6 105 A/m. Solution This problem asks that we design a cubic mixed-ferrite magnetic material that has a saturation magnetization of 4.6 105 A/m. From Example Problem 20.2 the saturation magnetization for Fe3O4 is 5.0 105 A/m. In order to decrease the magnitude of Ms it is necessary to replace some fraction of the Fe 2+ with another divalent metal ion that has a smaller magnetic moment. From Table 20.4 it may be noted that Co 2+, Ni2+, and Cu2+, with 3, 2, and 1 Bohr magnetons per ion, respectively, have fewer than the 4 Bohr magnetons/Fe2+ ion. Let us first consider Co2+ (with 3 Bohr magnetons per ion) and employ Equation 20.13 to compute the number of Bohr magnetons per unit cell (nB), assuming that the Co2+ addition does not change the unit cell edge length (0.839 nm, Example Problem 20.2). Thus,
=
3
=
(4.6 10 5 A/m)(0.839 10- 9 m) 3 /unit cell
9.27 10- 24 A - m2 /Bohr magneton
= 29.31 Bohr magnetons/unit cell If we let xCo represent the fraction of Co2+ that have substituted for Fe2+, then the remaining unsubstituted Fe2+ fraction is equal to 1 xCo. Furthermore, inasmuch as there are 8 divalent ions per unit cell, we may write the following expression:
B (Co) = 8 [ 3xCo + 4(1 - xCo )] = 29.31
which leads to xCo = 0.336. Thus, if 33.6 at% of the Fe2+ in Fe3O4 are replaced with Co2+, the saturation magnetization will be decreased to 4.6 105 A/m. For the cases of Ni 2+ and Cu2+ substituting for Fe2+, the equivalents of the preceding equation for the number of Bohr magnetons per unit cell will read as follows:
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B (Ni) = 8 [ 2 xNi + 4(1 - xNi )] = 29.31 B (Cu) = 8 [ xCu + 4(1 - xCu )] = 29.31
with the results that xNi = 0.168 (or 16.8 at%) xCu = 0.112 (11.2 at%) will yield the 4.6 105 A/m saturation magnetization.
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