1 Solutions Chapter 6 6.1 a) α1 = sin θ cos φ , α12 = sin2θ cos2 φ α2 = sin θ sin φ, α22 = sin2 θ sin2 φ α3 = cos θ, α32 = cos2 θ α12 + α22 + α32 = sin2 θ (cos2 φ + sin2 φ) + cos2 θ = 1 b) α12α22 + α22α32 + α32α12 = sin4 θ cos2 φ sin2 φ + sin2 θ sin2 φ cos2 θ + sin2 θ cos2 φ cos2 θ = sin4 θ cos2 φ sin2 φ + sin2 θ cos2 θ QED 6.2 From Eq. 6.6 ƒ100 = K0, ƒ110 = K0 + K1 /4, and ƒ111 = K0 + K1/3 + K2/27. For Fe: From Fig. 6.1 From Eq. 6.6 ƒ111 – ƒ100 ≈ 1.6 × 104 J/m3 = K1/3 + K2/27 ƒ110 – ƒ100 ≈ 1.2 × 104 J/m3 = K1/4 The second equation gives K1 = 4.8 × 104 J/m3 and using this in the first gives K2 ≈ 0, in fair agreement with the tabulated values, K1 = 4.8 × 104 J/m3 , K2 = - 1 × 104 J/m3. 2 For Ni From Fig. 6.1 From Eq. 6.6 ƒ100 – ƒ111 ≈ 2.2 × 103 J/m3 = - K1/3 – K2/27 ƒ110 – ƒ111 ≈ 1.0 × 103 J/m3 = - K1/12 – K2/27 Subtracting these two equations gives K1 = -4.8 × 103 J/m3 and, thus, K2 ≈ - 1.6 × 103 J/m3. These values compare well with the tabulated values, K1 = -4.5 × 103 and K2 = -2.3 × 103 J/m3. Clearly, there is significant opportunity for error in estimating the areas in Fig. 6.1 between the magnetization curves taken in different directions. 6.3 The energy gradient of Eq. 6.6 for small θ is given by K1θ2 + (K1 + K2) sin22φθ2/4. For Ni, both K1 and K2 are negative and K1 ≈ K2. Thus the energy gradient is given by 2|K1| θ [1 + 3/2 θ2 sin2 2 φ] which is steeper for φ = 45o. Thus M rotates toward the <111> directions, not <110>. 6.4 ƒ a 100 = K0 + K1 for small θ we get ƒa100 ≈ Ko + K1 <θ>2 and ƒ a 110 = K0 + K1 cos2 2θ ≈ K0 + K1 <1 - (2δθ)2/2 ...>2 ƒ a 110 - ƒ a 100= K1 (<1-(2δθ)2/2 ...>2 - < δθ2>) = K1 (1 - 5 δθ2> and using m(T) = <1 - δθ2/2...> for small θ as in text ∆ ƒ = (K1/4)[m]10 6.5 ƒ a easy = K0 + Ku ≈ K0 + Ku <θ2> ƒ a hard = K0 + Ku 3 ≈ K0 + Ku <(1 - θ'2/2 )2 + cos2 φ' θ'2 > Since cos2 φ' averages to 1/2 ƒ a hard = K0 + Ku <1 - θ'2/2> and ƒ a hard - ƒ a easy = Ku < 1 - 3 δθ2/2...> and m(T) = = <1 - θ2/2 . . .> for θ or θ', so ƒ a hard - ƒ a easy = Ku (T)/Ku [0] = [m(T)] 3. 6.7 In both cases the question we are asking is what is the measured magnetization in the hard direction after removal of a saturating field that was applied in the hard direction. For Fe or Ni after magnetization in the hard direction (<111> and <100>, respectively), the magnetization relaxes to the nearest easy axes, distributing itself equally among them: Ms/3 along each of the three nearest <100> directions for Fe and Ms/4 along each of the four nearest <111> directions for Ni. These axes have projections of 1/√ 3 on the original field direction in each case, so the sum over the 3 or 4 near easy axes gives a magnetization component along the hard direction of M s/√ 3 = 0.577Ms, which is observed for both Fe and Ni after magnetization in the hard direction. Cobalt on the other hand has uniaxial symmetry and after magnetization in the hard base-plane direction, the remanence is zero because the nearest easy axis is the c axis, 90 degrees from the base plane which has zero projection in the hard direction. 6.8 In the fully demagnetized state the magnetization is uniformly distributed over the six directions, ±x, ±y, ±z. Application then removal of a field along [110], assuming easy wall motion, will result in a distribution along +x and +y in H = 0. So we just use one angular 4 variable, taken as θ in the figure. To write the energy density, note that H110 = H0 (1,1, 0) and M = Ms (cosθ ,sin θ ,0) so that 2 − µ o Ms H (cosθ + sin θ ). − µo M ⋅ H = 2 The normalized component of µoM parallel to H is then given by m = (cosθ + sinθ)/√2, which gives m = 1 at saturation, θ = 45o. So the magnetic energy density is ƒ=− µ o Ms H K (cosθ + sin θ ) + sin 2 2θ 2 4 and ∂ƒ − µ o Ms H =0= (− sin θ + cosθ ) + K1 sin 2θ cos2θ . ∂θ 2 But cos2θ = (cosθ - sinθ) (cosθ + sinθ) so we can cancel the first factor here from the torque equation; it is only zero at and above saturation. Thus, µoMsH = √2K1sin2θ (cosθ + sinθ). Using m = (cosθ + sinθ)/√2 or (2m2-1) = sin(2θ), the equation of motion is µoMsH = 2K1 (2m2-1) m. This can be solved by plotting H vs. m as shown below. Here the values µoMs = 2T and K1 = 6 × 104 J/m3 have been used. 50000 0 -50000 -1 -0.5 0 0.5 1 This figure may be plotted as m vs H as shown below. From the analytic solution, 5 it is clear that saturation (m = 1) occurs for H = 2K1/µoMs = Ha = 60 kA/m. The same equation of motion applies to the y component of magnetization. The initial magnetization curve of the component, Ms/3, along ± z will involve rotation of that component into + x and + y by 90o wall motion. Thereafter, all of the magnetization proceeds by the derived equation for Mx and My. If wall motion is not easy, one would have to minimize the free energy including the full anisotropy in θ and φ. The case for the field applied along [111] is now treated. H = Ho(1, 1, 1, )/√3 and the magnetization process is the same for each Cartesian component of M. We treat the component of M that initially lies along z. At arbitrary field it is given by M(H ) = M s (sinθ /√2, sinθ /√2, cosθ ). The Zeeman energy is -µo M . H = -(µoMsH) (√2sinθ + cosθ)/√3. The cubic anisotropy for φ = 45o is given by sin 4 θ fa = K1 ( + sin 2 θ cos2 θ ) 4 which has absolute minima at θ = 0 and π as well as at θ = π/2 with φ = 0, ±π/2 and π. Saddle points can also be identified from Fig. 6.6a). The zero-torque condition is given by: ∂ƒ/∂θ = 0 = - (µoMsHo/√3) (√2cosθ - sinθ) + K1 sin2θ (1 + 3 cos2θ)/4, 6 which gives the equation of motion H= 3K1 sin(2θ )[1 + 3cos(2θ )] . 4 µ o Ms 2 cosθ − sin θ This equation can be plotted parametrically with m = (√2sinθ + cosθ)/√3 to give the result shown below. Alternatively, it can be solved analytically (with little further insight) as shown in Cullity, p. 227. The zero-torque solution shown as dashed lines below can be excluded by looking at the stability condition, d2f/dθ2 > 0, which is negative for the dashed solutions. Note that the approach to saturation accelerates as m → 1. The remanence (at H = 0 or θ = 0) is, from the definition of m, given by 1/√3 = 0.577. As H decreases from positive saturation, the magnetization reaches the extremum in the second quadrant. At this point, it is energetically favorable to jump to the third quadrant solution - if domain wall motion has not already taken the system to that branch. 7 6.9 The energy surface is described by E = +2π Ms2 cos2θ - Ku cos2θ where θ is the angle between M and the surface normal. Energy minimization gives (Ku - 2π Ms2) sinθ cosθ = 0 which has solutions at θ = 0 and π /2 or at K u = 2π Ms. Consideration of the stability condition (Ku - 2π Ms2) cos2θ > 0 indicates that θ = 0 is the stable condition for Ku > 2π Ms2 and θ = π/2 for K u < 2π Ms2. Only if K u is exactly equal to 2π Ms2 could any intermediate orientation exist. There are other forms of anisotropy for which 0 < θ < π/2 is stable for a range of values of K and M s.

3.45 Problem Set 6 (Solutions)

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