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2. 1. Solve for the magnetization vs. field for a) a thin film of amorphous iron boron silicon
(assume µoMs = 1.6T and K = 0) with the field applied normal to the film surface. b) a
single crystal sphere of Ni with the field applied along the [111] direction.
2. Does the energy required for complete saturation in the Stoner-Wohlfarth model as
described by Eq. 9.13 diverge for 0 < θo < π/2 or take on a finite value?
3. Consider the magnetization process in a single-domain particle having cubic
anisotropy using a field applied along an easy axis orthogonal to the initial
magnetization. (Because of the uniqueness of this initial condition, i.e. the special way
the sample is prepared before application of the measuring field, the result you derive
will not be typical for cyclic magnetization.) Write and plot the energy density vs. θ then
find the shape of the m-h curve. Locate the critical parameters by combining the
equilibrium condition with the condition that the solution to ƒ'(θ) = 0 also be an
inflection point.
4. Consider the two dimensional magnetization of a thin film with four-fold in-plane
anisotropy ƒa = K1 cos22θ with an external field and stress σxx applied collinearly as
shown below:
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3. Solve for the equation of magnetization and sketch the results for B1exx > 0 and < 0.
Compare with the results in Prob. 9.4.
5. Consider a thin film with in-plane cubic anisotropy K1 and a superimposed uniaxial
anisotropy with easy axis along one of the four-fold easy axes. A field is applied
inplane, perpendicular to the uniaxial easy axis. a) Write the expression for the free
energy. b) Sketch the energy surfaces for the various terms. c) Write the equation of
motion and sketch m vs. H. d) How does m-H differ for Ku > 5 K1 or Ku < 5 K1?
6. Work out the steps to derive the field dependence of magnetization M(H) for a material
with uniaxial magnetic anisotropy and H applied perpendicular to the easy direction of
magnetization.
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4. 7. Explain why the coercivity of amorphous alloys goes through a minimum as the
magnetostriction constant λs goes through zero. Give formula(s) to support your
explanation.
8. Derive an expression for rotational permeability in cubic anisotropy, and compare with
Eq. 9.8 for uniaxial anisotropy.
9. Determine the g-factor of the YIG sphere in Fig. 9.46 using Eq. 9.15.
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5. 1 a) For each surface of the film, Hmagstat = - (1/2)M = -Ms cosθ, so Hmagstat = -M.
fmagstat = -µo Ms . Hmagstat
f = -µo MsH cosθ + (µo/2) Ms 2 cos2θ
∂f/∂θ = 0 gives: H = Ms cosθ = M⊥ (after division by sinθ, which is zero
only at and above saturation). Thus:
H/Ms = M⊥/Ms = m
The system saturates when H = Ms =1.27 MA/m or when B = Bs = 1.6T
b) [111] is the easy axis, so the only anisotropy is shape. Answer is same as a)
but Ha = Hmagstat = -NM with N = 1/3 instead of 1. m = 3H/Ms
Saturation is achieved at (1/3) µoMs ≈ 0.2T.
2. Putting m = 1 in Eq. 9.13 gives sin 2θo = 0 which can only be satisfied for θo = 0 or
π/2. So the m(H) curves in Fig. 9.3 never reach m = 1 except for the two limiting cases,
for θo = 0 or π/2.
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6. 3. The energy density, f = K1 sin2θ cos2θ - MsH cosθ , is plotted below. f is minimized
for the equation of motion: (m - 2m3) - h = 0, where h = MsH/2K1 . This cubic equation
may have up to three different solutions. The physically meaningful one(s) can be
discerned by considering the energy as a function of θ.
The equilibrium orientation θο decreases toward zero, i.e. m = cosθ increases as H
increases. At a field h ≈ 0.25, the energy minimum near θ = 1.2 vanishes and the
magnetization may jump abruptly to θ = 0, m = 1.
The calculated form of m versus h is shown. The discontinuous change in m is a
first order transition; it corresponds to what is called a switching field. It can be
determined from the derivatives of the equation of motion, either ∂h/∂m = 0 or ∂m/∂h =
∞. Thus, the critical magnetiztion at switching is given by: 1 - 6mc 2 = 0. Thus mc =
0.408... (or θc = 66o) which occurs for hc = 2/(3√ 6) = 0.272...
Fig. for Sol. 9.3. Left, normalized energy density as a function of θ (radians)for different
values of reduced field, h = H/Ha. Shift in equilibrium orientation with h is indicated.
Right, calculated m-h behavior: m increases with increasing h then at mc = 0.408, jumps to
m = 1.0. The dashed line in the m-h curve shows the continuation of the mathematical
solution. Note that the initial slope ∂m/∂h)o = 1 or ∂M/∂H = Ms/Ha gives the value for the
anisotropy field Ha. Thus, the value of K1 can be determined by measuring m(h) and using
either the initial slope or the critical field hc.
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7. Prob. 3. Left, normalized energy density as a function of θ (radians) for different
values of reduced field, h = H/H1. Shift in equilibrium orientation with h is indicated.
Right, calculated m-h behavior: m increases with increasing h then jumps to m = 1.0 at
mc = 0.408.
4. The energy density is f = -MsBo cosθ + K1 cos22θ + B1exx (cos2θ - υ sin2θ)
∂f/ ∂θ = 0 gives m[8K1 (2m2 - 1) + 2B1 exx (1 + υ)] = Ms Bo where m = cosθ and 1-
m2 = sin2θ. For exx = 0 this gives the result sketched as the solid line: mr = 1/√ 2,
saturation (m = 1) occurs at H = (8K1/µ Ms) = Ha.
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8. For B1exx > 0, mr < 1/√ 2
B1exx < 0, mr > 1/√2,
and Ha = [8K1 + 2 B1exx (1 + υ)]/(µ Ms).
Thus, saturation occurs at higher fields
for B1exx > 0 and lower fields for
B1exx < 0.
5. a) From Eq. 6.6, setting θ = 90o , we have
f = Kusin2 φ + (K1/4)sin2 2φ. −µοMs Ηsinφ.
b) Energy surfaces:
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9. c) Zero torque gives 2K sinφ cosφ + K1 sin2φ cos2φ = µο Ms Η cosφ. Divide by cos u φ
which is zero only at and above saturation. The parameter of interest is the component of
magnetization along the field direction, sinφ, which we define as m. The equation of
motion is then expressed:
2Km + 2K m (1 − 2m 2 ) = µ M H
d) Numerical solutions are shown at the right
for three values of the ratio of uniaxial to cubic
anisotropy constants, Ku/K1 = 2, 5 and 8. K1 =
104 J/m3 and the field scale is µoH (T). Note
that for Ku/K1 = 5, the infinite slope point
occurs at m = 1. For smaller Ku, the
magnetization shows a discontinuity as was
found in Prob. 9.3. For larger Ku, the m-h
curve approaches a linear form typical of pure
uniaxial, hardaxis magnetization.
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10. 7 Coercivity goes inversely as permeability. More exactly Hc is proportional to (Ku +
(3/2)λs σ)/µοMs. In amorphous materials there is no magnetocrystalline anisotropy so K
is very small. The coercivity then vanishes or goes through a minimum when
magnetostriction vanishes.
9. g = 2.11.
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11. 1. Assume a 180o domain wall exists in a demagnetized, uniaxial magnetic material.
a) Sketch what happens to the domain magnetization and domain wall in
the two cases described below for H > 0 but less than saturation, i.e.
applied field parallel to the easy axis,
ii) applied field perpendicular to the easy axis
b) Sketch the M-H loops in each case.
c) Describe how a defect might pin or impede domain wall motion.
i)
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12. c) See Text, Fig. 9.12. Domain wall energy density is σdw = 4(AK)1/2. If this is uniform
throughout the material, the wall moves under application of slightest field H. If material
is inhomogeneous, wall area or A or K may be a function of position. Non-magnetic
defects lower wall energy; magnetic defects can raise or lower all energy. This gives Hc ≠
0 and the loop in (a) above, opens up.
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