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1.Show that the magnetostriction constants of Fe and Ni are not of the sign you would
expect for dipole forces between atomic "bar" dipole magnets.
2.Consider a bar of iron with [110] along its length and the top surface is (001). It is under
tensile stress along its length.
100 = 20.5 10-6
, 111 = -21.5 10-6
, c = 2.4 1011
,
11
c 11, 11 2
12 = 1.4 10 c44 =1.2 10 N/m
a)Write the strain tensor in the bar coordinates and in the crystallographic coordinates.
b)Which terms in Eq. 7.9 are relevant to the effects of stress on the orientation of M?
c)If the stress is such as to produce a strain e of 0.1%, compare the magnitude of the
magneto-elastic and magneto-crystalline anisotropy.
d)Evaluate the appropriate energy terms to determine the direction of M at e = 0.01.
e)Derive the equation of motion of the magnetization as a function of field and strain. Plot
m-H for e = 0 and e = 1%. Qualitatively describe your results.
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3. Derive Eqs. 7.7, the equilibrium strains, from Eqs. 7.6.
4. Show that setting 111 = 100 = s in Eq. 7.16 gives Eq. 7.17.
7.5 Consider a thin epitaxial Ni film on Cu(100) (aNi = 3.524Å, aCu =
3.615Å).
Assume for Ni 100 = -46 10-6
, c = 2.5 1011
, c= 1.6
1011
and c = 1.18 1011
11 12 44
N/m2
a)Compare the relative strength of crystal anisotropy, magnetoelastic
anisotropy, and
magnetostatic energy as far as determining the direction of M.
b)Assume e 0, find the equilibrium direction of M in zero field and for
small Ni
12
thickness.
c)As the film grows, describe what happens to e (i j).
ij
d)If e = 2e , find the direction of M.
12 11

Discuss.
7.6 Show that the two alternate ways of defining the magnetoelastic coefficients (Eqs. 7.8
and 7.34) are equivalent. How are the Bs and bs related?
7.7. It has been written in a technical article. "Typically, magnetic transition metals produce
magnetostrictive strains e [at saturation] of order 10-5 to 10-4 which vary about the direction
of magnetization roughly as e = (3/2) s(cos2 - 1/3) where s is the saturation
magnetostriction constant. Conversely, stresses imposed on the material which result
in strains of order 10-5 to 10-4 (i.e. e = s) contribute significantly to the total magnetic
anisotropy: the uniaxial stress- induced anisotropy energy density is (3/2) s ." Critique
these statements by considering the energy densities in each case).
7.8 Use Eq. 7.10 to describe the field and strain directions you would use to measure the
two magnetostriction constants 100 and 111 on a Ni sample cut as shown in Fig. 7.3.
Do not
assume that the sample has randomly distributed magnetization directions in zero field.
7.9. Assume a multidomain polycrystalline sample with the domain magnetizations
randomly oriented as represented below.

Describe how you would expect the magnetization distribution to change if the
sample has positive magnetostriction s and a tensile stress xx is applied for H = 0.
Contrast this with the change caused by application of H = (Ho, 0,0) for ij = 0. Plot Mx
vs. Hx for xx = 0
and xx 0. Plot Mx vs. xx for applied fields in the range Hx < HK.
7.10. Consider the anisotropy constants, K1, and the
polycrystalline magnetostriction constants, s, shown below
for fcc NiFe alloys. This problem is about the behavior of
alloys having compositions A, B and C.
a.Sketch and label the lowest order crystal anisotropy
surfaces in the x-y plane for thin films of compositions A and C
(i.e. assume z = 0).
b.For unstrained bulk samples of compositions A, B and C, in which crystallographic
directions are the samples easily magnetized?
c.For unstrained single crystal samples A, B and C, a field is applied in the [100]
direction. Sketch the shape of the M-H curve in each case assuming zero coercivity.
Use approximately the same field scale in each case so the M-H curves can be
compared.

e. For single crystal samples A, B and C apply a field in the [100] direction as in
part c above
but now also put the sample under tensile stress parallel to the field direction.
Sketch with a dashed line over your result in c, the shape of the strained M-H
curve in each case assuming zero coercivity.
f. What is the approximate value of the magnetoelastic coupling coefficient, B1, for
sample B? If a strain is applied along [100] to cancel the crystal anisotropy in this
sample and have the magnetization lie along the x axis in zero field, what would
be the nature of that strain (tensile or compressive along [100]) and what is its
approximate magnitude?
a.If the samples are subject to strains, which composition, A, B or C, would make
the best soft magnetic material? Why?
7.1 The potential energy of two dipoles, m at the origin and 'm a distance r away, is
given by Eqs. 1.9 and 1.8. When ferromagnets are of interest, the two dipoles have
the same orientation and the second dipole moment can be expressed in polar
coordinates of its position as 'm = 'm (cos er - sin eo). The radial force between
the two dipoles (now m = 'm) is given by

F is attractive for collinear moments ( = 0) and repulsive for side-by-side moments (
= /2).
Thus one would expect strain e1 < 0 for any direction of magnetization and e2 > 0
perpendicular to any direction of magnetization. Applying Equation 9.1 to these dipole
strains implies s < 0. But Fe and Ni have very different 's in different directions and
100 for iron
is positive not negative.
3.The six Eqs. 7.6 contain six unknowns, the eij. The coefficients cij and constants
B1 i2 and B2 i j are assumed known. Solve the first three for eij (i = j) and the last
three for eij (i j) to get Eq. 7.7.
4. Eq. 7.16 then gives

Since the principal axes are no longer tied to the crystal (which is isotropic), the
coordinates
can be rotated so that the z axis coincides with the direction in which the strain is measured.
Then only one direction cosine survives, 32 = cos2 where is the angle between M and
the strain direction.
7.5. Misfit, (aCu - aNi)/aCu is = 2.5% from lattice constants given.
a) K1 = -4.5 103 J/m3 and from Eq. 7.20, B1 = -(3/2) 100(c11 - c12) = 6.2 106
N/m2. For a biaxial misfit strain, the tensor components are:

and fME = B1(e11 12 + e22 22 +e33 32) + B2(e12 1 2 + e23 2 3 +e31 3 1)
becomes fME = 1.55 105 ([cos2 + sin2 sin2 - cos2 ) ignoring any shear strain in the
film.
Now the first question is what is the relative magnitude of the energies involved,
i.e. can the magnetization point out of plane? The magnetostatic energy of Ni,
oMs2/2, (favoring in plane magnetization) is of order 1.5 105 J/m3 which is
much greater than the magnetocrystalline anisotropy (which favors <111> easy axes), but is
comparable to the ME energy B1e 1.55 105 J/m3 (whose orientation preference has yet
to determine).
fME = 1.55 105 . ([cos2 + sin2 sin2 - cos2 )
= 3.1 105 . sin2 + const.
fa = -4.5 103(cos2 sin2 sin4 + sin2 cos2 ) fMS = 1.5 105
cos2
Clearly the ME energy density dominates (as long as the strain in the film has the
full misfit value). The magnetostatic energy is a close second and the crystal anisotropy is
but 1% of the other two.
b)First, fa is the only term that contains the angle , so even though it is the weakest term,
it should be minimized with respect to to find, as expected, = ±45o, ±135o
...., i.e. the azimuths containing the <111> directions are favored. Now because fa is
so small, it is necessasry to consider only the ME and MS terms in minimization with

respect to . ftot = (3.1 105 - 1.5 105 J/m3 )sin2 const. obviously minimizes for
= 0o. This reflects the fact that the ME energy dominates the MS energy and the nature
of the strain and ME coefficient is such that perpendicular magnetization is favored. So it
no longer matters what is.
c) As the film grows, the in-plane biaxial strain decreases due to misfit dislocation formation.
The shear strain exy probably remains negligible, but the biaxial, x-y plane strain may
take on a z-dependence from the Cu/Ni interface to the top of the film, i.e. there may be
shear components eyz, ezx 0. In this case it is necessary to consider the terms
B2(e23 2 3+e31 3 1). Because B2 4.3 106 N/m2 < B1 and the shear strains are
probably small compared to the biaxial strains, there should be no effect from this term.
Also, unless this term is large, 1 and 2 will remain zero.
e12
d) If = 2e11 = 0.05, thenthe terms B2(e1212) = 2.1 
105sin cos sin2 B2 4.3 106 N/m) must be retained. This term is stronger
than the Ni magnetocrystalline anisotropy term which favors = ±45o, ±135o ...., so it
will dictate the equilibrium azimuth at = -45o, +135o ( - 2.1 105 sin2 ).
This new shear magnetoelastic anisotropy is negative at its equilibrium values. So it
combines with the magnetostatic term, -1.5 105 sin2 to compete with the ME term,
and now causes the magnetization to fall back in plane assuming an

orientation consistent with a uniaxial easy axis = -45o, +135o.
7.7. For a sample being magnetized with H = Ha, it follows that Ktot = Kxtl + Ks + ... =
(1/2)HaMs which is the effective anisotropy energy density, i.e. it includes K1 effects as
in Eq. 7.18.
For H Ha, a uniaxial material strains by anywhere from e = s at = 0 to e = -
s/2 at = /2 so the magnetoelastic energy density f1 = F/V = Be is of order:
f1 = (3/2) Be = (3/2) B s (3/2) E s2 f1 = (3/2) E s2
Now if you impose a stress s which is great enough to give e = s, then,
(see Eq. 9.19)
which is the same energy as when the material is magnetized to saturation: f1 = f2.
But the question remains, is this energy comparable to the total anisotropy energy?
The answer is yes, only if Kxtl is small:
Ktot = Ks + Kxtl + .. = (3/2)λs σ + Kxtl = f1 or f2 only if Kxtl <<
Ks
7.8 Using a 90o biaxial strain gauge, it is possible to measure the strains in [100] and
[110] simultaneously for two field directions [100] and [110]. From the strain
measurements with field parallel to the strain direction e100||, e100|| , it is possible to
calculate e100|| - 4e110||, which gives -4h2. The numerical value for h2 now gives λ111
= 1/3 h2.

The numerical value for h2 now gives 111 = 1/3 h2.
Using the h2 value in 1/6 h1 + h2 = e110|| which is measured, gives us a value for h1 and
100
= 2/3 h1. Two more independent measurements would be needed to get h3 and h4.
7.9. For s > 0 (B < 0), xx 0 and all other stresses zero, it is expected that eyy = ezz =
-
exx so the magnetoelastic free energy is
ƒME = -|B|exx [ 12 - ( 22 + 32)] -|B|exx [4 12 - 1]/3
2 2 2
The energy is lowered if a > ( 2 + 3 ) so magnetization along ± x is favored and
the
random distribution of moments would become:
For application of a field, only M in the field direction (not opposite it) lowers the
energy. So the distribution is unidirectional:
The M-H curves are linear and for xx > 0, Mx vs. Hx saturates at lower fields (see above,
right). To plot M vs. xx , the energy density must be considered:

Ku h
Mx
 cos  m

Ms Ku  (4 / 3)Bexx
with exx = xx/E, Ku = (1/2)MsH and h = H/Ha.
Thus Mx vs. H is linear with a slope of Ms/Ha for xx = 0 and an increasing slope of
Ms/[Ha - (8/3) |B| /(Ms E)] as xx increases (B < O)
In all cases, Mx saturates at Ms when = 0. We can write the equation for m as
m = h/(1-x)
with x = (4/3)|B|e/Ku. The result is plotted below after the Mathematica® program that
gives the plot.
m=h/(1-x)
Plot3D[m, {x, 0, 1}, {h, 0, 1},
AxesLabel->{"x", "h", "m"}, PlotRange->{0, 1},
ƒ = ƒME + ƒK + ƒZ
= -|B| exx [4 12 - 1]/3 + (1/2) Ms Ha 12 - Ms H 1
Choosing between M and the x axis, solve ƒ/ƒ = 0 for the reduced magnetization
along x:

PlotPoints-> 25]
7.10
a)Alloy A has an anisotropy energy surface that is a cut through Fig. 6.6a and that of B is a
cut through Fig. 6.6b.
b)For zero strain, A is easily magnetized along <100> directions, B and C along <111>.
c) The M-H[100] curve of A is like that of Fe in Fig. 6.1a; the remanence is close to unity.
Those of alloys B and C are like that of Ni in Fig. 6.1c; the remanence is given from Eq.
6.6 and m(0) = (1,1,1)/ 3 as m.
H/|H| = 1/ 3
= 0.577. Taking the field along z for convenience, the free energy is
f M H K ( 2 2
cycl.) M H cos K (1/ 4 cos2
)
0 s 1 1 2 0 s 1
giving f 0 0 Ms H 2K1 cos and cos m
0 Ms H / 2K1.
Thus saturation occurs at Ha = 2K1/ oMs.
d)Sample B would make the best soft material because both K and are close to zero.

e) With a field and tensile stress along [100], the magnetoelastic energy will favor
[100] magnetization when > 0 (cases A and B) and favor [010] and [001]
magnetization when < 0. That is, the energy surfaces determined in part a)
will be supplemented with magnetoelastic energy terms shaped like oblate spheroids
with axis along [100] for samples A and B and prolate spheroids with axis along [100]
for sample C.
f)
In the energy surfaces above, the solid lines represent the cubic magnetocrystalline
anisotropy energy surfaces. The magnetoelastic contribution adds an uniaxial term with
axis of symmetry along the strain direction. The dotted surface is the resultant energy
surface. The changes in MH are shown by the solid lines relative to the unstrained loops
(dashed).
A B C

strain =
0
e100 >
0
M
e100 > 0
strain = 0
M M
strain = 0,
e100 > 0
H H H
The strain needed to cancel the
cubic anisotropy is about
0.51%. The factor of 1.3 in the magnetoelastic energy is the 1+ term for
uniaxial deformation in Fig. 7B.3.

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