Cullity B.D. Introduction to Magnetic Materials. Second Edition (2008)
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7.5 ANISOTROPY MEASUREMENT |
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For smaller fields, and for small rotations of the wire away from the easy direction, F. E. Luborsky and C. R. Morelock [J. Appl. Phys., 35 (1964) p. 2055] give
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pM2H |
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M2H |
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L ¼ |
s |
2u (cgs) or |
L ¼ |
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2u (SI) |
(7:21) |
(H þ 2pMs) |
(4H þ 2Ms) |
where u is in radians and L is the torque per unit volume of wire.
An alternative standard torque sample might be a disk of a highly anisotropic, highcoercive field permanent magnet, such as one of the rare-earth transition metal compounds discussed in Chapter 14. These materials can have magnetic moments that are almost unchanged in fields over the range +20 kOe (2T) or more. The energy of such a sample, of magnetic moment m in a field H as function of the angle between m and H, is
E ¼ mH cos u, |
(7:22) |
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and the torque is given by |
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L ¼ |
dE |
¼ mH sin u: |
(7:23) |
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du |
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Here m is the absolute magnetic moment of the sample, in emu (cgs) or A/m (SI), which must be determined experimentally in a separate measurement.
It is important that the component parts of a torque magnetometer be properly aligned. Each of three axes should coincide with the center of the magnet air gap: the axis of the specimen, the axis of rotation of the torsion head (upper dial of Fig. 7.16), and the axis of the instrument (the line from the upper support of the torsion wire through the center of gravity of all the suspended parts). Improper alignment will introduce spurious torques which distort the experimental curve. The alignment can be checked by determining the torque curve of a specimen having a known, and simple, anisotropy, such as uniaxial, and comparing this curve with that theoretically expected.
7.5.4Torsion-Pendulum Method
In this case, the specimen is a disk, suspended by a torsion wire in the air gap of an electromagnet, just as in a torque magnetometer. Suppose the crystal is uniaxial, with the easy axis c in the plane of the disk. The initial, minimum-energy position of the specimen is one with c parallel to the field H and no twist in the torsion wire. The specimen is then rotated away from H by a small angle, released, and allowed to oscillate back and forth about the field direction at the natural frequency of the suspended system. This frequency is measured, by counting the number of oscillations in a known time interval. When the specimen is in the deflected position, two restoring torques act on it: (1) the torque in the wire, and (2) the crystal anisotropy torque which tries to rotate c back into parallelism with Ms and H. The period T of the oscillatory motion is still given by an equation of the general form of Equation 7.21, but modified to include the anisotropy torque:
f |
¼ T ¼ 2prk k |
(7:24) |
1 |
I |
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w þ s
218 MAGNETIC ANISOTROPY
where f is oscillation frequency, I is the moment of inertia of the suspended system, kw is the torsion constant of the wire, and ks is the torsional stiffness of the specimen. The quantity ks is the rate of change of torque with angle and is therefore given by dL/du. But jLj ¼ dE/du, so that ks ¼ d2E/du2, where E is the anisotropy energy. From the measured frequency, ks may be calculated, because I and kw are known, and the anisotropy constant may then be determined from ks.
Anisotropy can be measured by the torsion-pendulum method in a few seconds, but only the slope of the torque curve at the equilibrium angle is measured, not the full torque curve. The method is rarely used.
7.6ANISOTROPY MEASUREMENT (FROM MAGNETIZATION CURVES)
Anisotropy constants may determined from the magnetization curves of single crystals in two ways:
1.By fitting a calculated magnetization curve to the observed one.
2.By measuring, on a graph of M vs H, the area included between the magnetization curves for two different crystal directions.
7.6.1Fitted Magnetization Curve
This is not a very common method. However, it is instructive to see how magnetization curves are calculated, because the calculation tells us something about the magnetization process. In such calculations we ignore everything but the crystal anisotropy forces; that is, we assume that domain walls will move in negligibly small fields, but that Ms can be rotated out of the easy direction only by fields strong enough to overcome the anisotropy forces.
The simplest case is that of a crystal magnetized in one of its easy directions, e.g. an iron crystal magnetized in one of the k100l directions, as illustrated in Fig. 7.3, or a uniaxial crystal (Fig. 7.6), magnetized parallel to its easy axis. Here the whole process, from the demagnetized state to saturation, occurs by wall motion only, at an (assumed) negligibly small field. The magnetization curve, shown in Fig. 7.18, is simply a vertical line, and the hysteresis loop encloses zero area.
Fig. 7.18 Idealized hysteresis loop of a crystal measured parallel to an easy axis.
7.6 ANISOTROPY MEASUREMENT (FROM MAGNETIZATION CURVES) |
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When a field is applied to an iron crystal in the [110] direction, wall motion occurs until there are only two kinds of domains left, namely those with Ms vectors in the [010] and
[100] directions, the two easy directions closestpto the field (Fig. 7.4). The magnetization |
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of the sample is then |
M ¼ Ms cos 458 ¼ Ms= |
2 ¼ 0.707 Ms. Further increase in field |
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rotates the Ms vectors |
away from the easy |
directions by an angle d in the (001) plane |
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(Fig. 7.19a). The direction cosines of Ms relative to the crystal axes are then a1 ¼ cos d, a2 ¼ cos (908 2 d), and a3 ¼ 0, for the [100] domains. (These comprise half the volume of the crystal. It is enough to base the following calculation on them alone, because the behavior of the [010] domains is exactly similar.) The anisotropy energy is then, from Equation 7.13,
Ea ¼ K0 þ K1 sin2 2d:
4
The magnetic potential energy is, from Equation 1.5,
Ep ¼ MsH cos (458 d):
The larger the angle d, the larger is the anisotropy energy and the smaller the potential energy. The angle d will therefore be such as to minimize the total energy Et.
Et ¼ K0 þ K1 sin2 2d MsH cos(458 d): 4
To minimize Et we put
dEt |
¼ [K1 sin 2d cos 2d] [MsH sin(458 d)] ¼ 0: |
(7:25) |
dd |
This problem may be thought of in terms of torques, rather than energies. The first term in Equation 7.25 is the torque exerted on Ms by the crystal, the second term is the torque exerted on Ms by the field, and the equation states that these torques are equal and opposite. The component of Ms in the field direction is the measured magnetization:
M ¼ Ms cos(458 d): |
(7:26) |
Fig. 7.19 Magnetization of an iron crystal in a [011] direction.
220 MAGNETIC ANISOTROPY
Eliminating d from Equations 7.21 and 7.22, we find
H |
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4K1 |
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M |
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M |
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(7:27) |
¼ |
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Ms |
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Ms Ms |
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which gives the field required to reach any given level of magnetization. This field is directly proportional to K1 and independent of K2. The field required to make M ¼ Ms, which is to saturate the magnetization in the [110] direction, is
H ¼ |
2K1 |
: |
(7:28) |
Ms |
Figure 7.18b shows the magnetization curve for iron at room temperature, calculated for Ms ¼ 1714 emu/cm3 and K1 ¼ 4.5 105 erg/cm3 (1.714 106 A/m and 4.5 104 J/m3).
The magnetization curve in the [111] direction of a crystal having k100l easy directions is calculated in similar fashion. Wall motion in low fields will eliminate all but three kinds
of domains—[100], [010], and [001]—and Ms in each will be equally inclined, at 54.78, to |
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the [111] field direction. The magnetization will then be M ¼ Ms |
cos 54.78 ¼ Ms=p3 ¼ |
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shown |
0.577Ms. Further increase in field will rotate the Ms vectors in |
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planes, as |
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for one of these in Fig. 7.20. The equation for the magnetization curve is complex: |
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K |
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HMs ¼ |
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hp2 2m2 (4m2 1) þ m(7m2 3)i |
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K2 |
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hp2 2m2(10m4 9m2 þ 1) m(23m4 16m2 þ 1)i, |
(7:29) |
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where m ¼ M/Ms. The field required to saturate in the [111] direction is |
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H |
¼ |
4(3K1 þ K2) |
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(7:30) |
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9Ms |
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For a crystal like iron, this is less than the field required for saturation in the [110] direction, in agreement with the experimental results shown in Fig. 7.2a.
Fig. 7.20 Magnetization of an iron crystal in a [111] direction.
7.6 ANISOTROPY MEASUREMENT (FROM MAGNETIZATION CURVES) |
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It is also of interest to calculate the magnetization of a uniaxial crystal like cobalt, when the field is applied in a direction at right angles to the easy axis. When the field is strong enough to rotate Ms away from the easy axis by an angle u, the anisotropy energy is, from Equation 7.3,
Ea ¼ K0 þ K1 sin2u þ K2 sin4u:
The magnetic potential energy is
Ep ¼ MsH cos(908 u):
The condition for minimum total energy is
2K1 sin u cos u þ 4K2 sin3u cos u MsH cos u ¼ 0:
Also,
M ¼ Ms cos(908 u):
Elimination of u from these two equations gives
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2K |
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M |
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4K |
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M |
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H ¼ |
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þ |
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Ms |
Ms |
Ms |
Ms |
with saturation (M/Ms ¼ 1) attained in a field of
H ¼ 2K1 þ 4K2 :
Ms
If K2 is zero, the magnetization curve becomes a straight line,
H ¼ 2K1M ,
Ms2
and the saturating field becomes
(7:31)
(7:32)
(7:33)
H ¼ |
2K1 |
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(7:34) |
Ms |
Figure 7.21 shows the room-temperature magnetization curve for cobalt, calculated for Ms ¼ 1422 emu/m3, K1 ¼ 4.5 106 erg/cm3, and K2 ¼ 1.5 106 erg/cm3 (cgs), or Ms ¼ 1.422 106 A/m, K1 ¼ 450 103 J/m3, and K2 ¼ 150 103 J/m3 (SI). Fields in excess of 10,000 Oe or 0.8 MA/m are needed for saturation. The dashed line shows the magnetization behavior if K2 is zero.
For iron, nickel, and cobalt crystals the general features of the experimental magnetization curves are well reproduced by the calculated ones. Figure 7.22 shows this kind of a comparison for crystals of 3.85% silicon iron. The fact that nonzero fields are required for saturation in the easy direction, evident in Figs 7.2, 7.5, and 7.22, shows that domain walls encounter obstacles to their easy motion. This topic will be pursued in Chapter 9.
Up to this point we have examined the magnetization curve of a single crystal only when the field is applied parallel to an axis of symmetry in the crystal. If the field is in some arbitrary direction, the magnetization process becomes more complicated. The demagnetizing field Hd must then be considered and a distinction made between the applied field Ha and
222 MAGNETIC ANISOTROPY
Fig. 7.21 Calculated magnetization curves for cobalt single crystal with field perpendicular to the easy axis.
the true field H, equal to the vector sum of Ha and Hd, inside the specimen. H and M are no longer always parallel, and neither is necessarily parallel to Ha. For the case of an iron crystal, see H. Lawton and K. H. Stewart [Proc. R. Soc., 193 (1948) p. 72], and for a cobalt crystal see Y. Barnier, R. Pauthenet, and G. Rimet [Cobalt, 15 (1962) p. 1] or [J. Phys. Soc. Japan, 17 (suppl. B1) (1962) p. 309].
7.6.2Area Method
This method of determining anisotropy constants from magnetization curves is based directly on the definition of the anisotropy energy E, namely, the energy stored in a crystal when it is magnetized to saturation in a noneasy direction. If we can determine
Fig. 7.22 Calculated and measured magnetization curves for Fe–Si crystals. [R. M. Bozorth, Ferromagnetism, reprinted by IEEE Press (1993).]
7.6 ANISOTROPY MEASUREMENT (FROM MAGNETIZATION CURVES) |
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W, the work done on the crystal to bring it to saturation, we can equate E and W and so determine the anisotropy constants.
One way of finding an expression for W is to calculate the electrical work done in magnetizing a rod specimen by means of a current in a solenoidal coil wound on the rod. Assume that the rod is so long that the demagnetizing field can be neglected. Let the rod be of length l and cross-sectional area A, wound with n turns. When the current increases by an amount di, the induction increases by dB and the flux by df ¼ AdB. This change in flux causes a back emf e in the coil, and work must be done to overcome this emf. We ignore the work done in producing heat in the coil, equal to i2R, where R is the resistance, because this work does not contribute to the magnetization of the rod. The total work done in time dt is
VdW ¼ ei dt joule, |
(7:35) |
where V is the volume of the rod and W the work per unit volume. From Equation 2.6 we have (using cgs units)
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¼ |
10 8n |
df |
¼ |
10 8 n A |
dB |
volt: |
dt |
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The field produced by the current is, from Equation 1.12,
H ¼ 4p ni Oe:
10 l
Combining the last three equations and noting that V ¼ Al, we obtain
dW ¼ |
10 7 |
H dB |
joule |
or dW ¼ |
H |
dB |
erg |
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4p |
cm3 |
4p |
cm3 |
(7:36)
(7:37)
(7:38)
Then the work done per unit volume in changing the induction from 0 to B is
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B |
erg |
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W ¼ |
ð0 |
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(7:39) |
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H dB cm3 |
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4p |
where H is in oersteds and B in gauss. Because B ¼ H þ 4pM, at the same field dB ¼ 4pdM, and
M |
erg |
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W ¼ ð0 |
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(7:40) |
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H dM cm3 |
where M is in emu/cm3. The work done in magnetization is simply the area between the M, H curve and the M-axis, shown shaded in Fig. 7.23a.
The SI equivalent of Equation 7.39 is
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W ¼ ð0 |
H dB |
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m3 |
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Fig. 7.23 Work done in magnetization.
and of Equation 7.40 is
M |
J |
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W ¼ ð0 |
(7:41) |
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m0H dM m3 |
Hysteresis has no direct connection with the measurement of anisotropy from magnetization curves. In fact, complete reversibility (no hysteresis) is usually assumed in such measurements. But if hysteresis is present, then removal of the magnetizing field will return energy equal to the area shaded in Fig. 7.23b to the magnetizing circuit. The energy stored in the material at its remanence point Mr is the area shaded in Fig. 7.23c.
When a specimen is driven through one complete cycle, the total work done on the specimen is the hysteresis loss Wh, which is equal to 1/4p times the area enclosed by the B, H loop (erg/cm3, cgs), or unity times this area (J/m3, SI) as shown in Fig. 7.23d. This work appears as heat in the specimen.
For a substance of constant permeability m, such as a dia-, para-, or antiferromagnetic, B ¼ mH, dB ¼ mdH, and Equation 7.41 becomes
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mrm0H2 |
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W ¼ |
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W ¼ |
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¼ |
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(SI): |
(7:42) |
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8p |
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If the substance is air, then m or mr 1, and |
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H2 |
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W ¼ |
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or |
W ¼ |
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(SI): |
(7:43) |
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This is the energy per unit volume stored in a magnetic field in air (or vacuum).
The magnetization of a specimen can increase either by domain rotation or domain wall motion, or both. Consider a small volume of the specimen with a magnetic moment m
7.6 ANISOTROPY MEASUREMENT (FROM MAGNETIZATION CURVES) |
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oriented at an angle u to the magnetizing field H. This moment has a component m cos u parallel to the field. Summing over unit volume of the specimen, we have
X
m cos u ¼ M: |
(7:44) |
When the field increases from H to H þ dH, the moment of the small volume considered will rotate from orientation u to u 2 du. The work done by the field is (couple)(angle) ¼ (mH sin u)(2du) (see Section 1.3). Summed over unit volume, the work done is
dW ¼ XmH sin u du: |
(7:45) |
From Equation 7.43, |
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dM ¼ d X m cos u ¼ X m sin u du: |
(7:46) |
Combination of Equations 7.45 and 7.46 gives
dW ¼ H dM:
If W is the area between a particular M, H curve and the M-axis, then W equals the anisotropy energy E stored in a crystal magnetized in that particular direction. We have already worked out these energies for cubic crystals, and they appear in Table 7.1. Therefore,
W100 |
¼ E100 |
¼ K0 |
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K1 |
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W110 |
¼ |
E110 |
¼ |
K0 |
þ |
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(7:47) |
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W111 |
¼ E111 |
¼ K0 |
þ |
K1 |
þ |
K2 |
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= |
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These equations may be solved for the anisotropy constants:
K0 |
¼ |
W100, |
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K1 |
¼ |
4(W110 W100), |
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(7:48) |
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K2 |
¼ |
27(W111 W100) 36(W110 W100): |
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Here an expression like (W110 –W100) is to be understood as the area included between the M, H curves for the [110] and [100] directions. As mentioned earlier, experimental M, H curves in the easy direction usually show a nonzero area between the curve and the M-axis, indicating that the field has had to overcome hindrances to domain wall motion. These hindrances are assumed to be the same for any direction of the applied field relative to the crystal axes. Therefore, equations like 7.48, which are based on the area between certain curves, yield anisotropy constants which should be largely free of the effects of domain wall motion.
It is rare to find more than one experimental method used in a single investigation, and the literature therefore contains almost no comparisons of alternative techniques. Williams, however, measured the anisotropy constants of his silicon iron crystals by three methods, with the results shown in Table 7.3 [H. J. Williams, Phys. Rev., 52 (1937) p. 747]. Differences between the three results reflect not only experimental error, but also differences in what is actually being measured. The torque measurement is the most fundamental, because a high-field torque measurement involves only the rotation of Ms relative to the
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MAGNETIC ANISOTROPY |
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TABLE 7.3 Anisotropy Constants of Fe 1 3.85 wt% Si |
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Method |
K1 (105 erg/cm3) |
K2 (105 erg/cm3) |
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Torque curves |
2.87 |
1.0 |
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Fitting magnetization curves |
2.80 |
1.0 |
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Areas between magnetization curves |
2.72 |
1.5 |
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axes of a single-domain crystal; no wall motion is included. The other two methods include (or ignore) the effects of domain wall motion at low fields, which are not directly related to the anisotropy.
7.6.3Anisotropy Field
The crystal anisotropy forces which hold the spontaneous magnetization Ms of any domain in an easy direction can also be expressed in an indirect but often useful way that does not explicitly involve anisotropy constants. For small rotations of the magnetization away from an easy direction, the crystal anisotropy acts like a magnetic field trying to hold the magnetization parallel to the axis. This field is called the anisotropy field and is given the symbol HK. The anisotropy field is parallel to the easy direction and of a magnitude such that for small angular deviations u it exerts the same torque on Ms as the crystal anisotropy itself. The torque due to the anisotropy field is HKMs sin u, or HKMsu for small values of u. The torque due to crystal anisotropy depends on the crystal structure, the anisotropy constants, the easy axis direction, and in some cases the crystallographic plane in which Ms rotates away from the easy axis. For example, in a cubic crystal with k100l easy directions, the torque exerted on Ms by the crystal when Ms rotates away from k100l is, from Equation 7.14, þK1/2 sin 4u, or 2K1u for small u.
Equating these torques, we have
HKMsu ¼ |
2K1u |
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HK ¼ |
2K1 |
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HK ¼ |
2K1 |
(SI): |
(7:49) |
Ms |
m0Ms |
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If k111l is the easy direction, similar reasoning shows that
H |
K ¼ |
4(3K1 þ K2) |
(cgs) |
H |
K ¼ |
4(3K1 þ K2) (SI) |
(7:50) |
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9Ms |
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9m0Ms |
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The last two equations are valid whatever the plane of rotation of Ms away from the easy direction. For a uniaxial crystal we find, through Equation 7.4, that
HK ¼ |
2K1 |
(cgs) |
HK ¼ |
2K1 |
(SI): |
(7:51) |
Ms |
m0Ms |
From Equation 7.34 this is also the value of the field that is required to reach magnetic saturation in the hard direction when K2 is zero. Thus for a uniaxial crystal the anisotropy field has the added physical significance of being the magnitude of the field, applied at 908 to the easy axis, which can completely overcome the anisotropy forces by rotating Ms through 908.