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Transcript of Magnetism and Magnetic Materials DTU (10313) – 10 ECTS KU – 7.5 ECTS Module 6 18/02/2011...
![Page 1: Magnetism and Magnetic Materials DTU (10313) – 10 ECTS KU – 7.5 ECTS Module 6 18/02/2011 Micromagnetism I Mesoscale – nm- m Reference material: Blundell,](https://reader036.fdocuments.us/reader036/viewer/2022062321/56649ec15503460f94bcc967/html5/thumbnails/1.jpg)
Magnetism and Magnetic MaterialsMagnetism and Magnetic Materials
DTU (10313) – 10 ECTSDTU (10313) – 10 ECTSKU – 7.5 ECTSKU – 7.5 ECTS
Module 6
18/02/2011
Micromagnetism I
Mesoscale – nm-mReference material:Blundell, section 6.7Coey, chapter 7These lecture notes
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Intended Learning Outcomes (ILO)Intended Learning Outcomes (ILO)
(for today’s module)(for today’s module)
1. Explain why and how magnetic domains form2. Estimate the domain wall width3. Calculate demagnetizing fields in simple geometries4. Describe superparamagnetism in simple terms5. List Brown’s equation in micromagnetics6. Explain how hysteresis arises in a simple Stoner-Wolfharth model
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FlashbackFlashback
€
M = M S BJ (y)
y =gJμ BJ(B + λM )
kBT
€
TC =gJμ B(J +1)λM S
3kB
=nλμ eff
2
3kB
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Edge effects and consequencesEdge effects and consequences
This is a bit misleading
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DipolesDipoles
Two interacting dipoles
€
E =μ 0
4πμ1 ⋅μ2
r 3 − 3μ1 ⋅r( ) μ2 ⋅r( )
r5
⎡
⎣ ⎢ ⎤
⎦ ⎥
€
Hdip(r) =1
4π3
μ ⋅r( )r
r5 −μr 3
⎡
⎣ ⎢ ⎤
⎦ ⎥ Dipole field
Dipolar energy
€
EZ = −μ ⋅B
Torque
€
A(r) =μ 0
4πμ × r
r 3
⎛ ⎝ ⎜
⎞ ⎠ ⎟ Dipole vector potential
1
2
H12
H21
€
τ =×B
Zeeman energy
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Energy of magnetized bodiesEnergy of magnetized bodies
This is to avoiddouble-counting
€
dE =12
μ 0
4πΜ(r) ⋅Μ(r')
r − r' 3 − 3Μ(r) ⋅ r − r'( )[ ] Μ(r') ⋅ r − r'( )[ ]
r − r' 5
⎧ ⎨ ⎩
⎫ ⎬ ⎭
drdr'
€
dμ = Μ(r)drd2
d1
€
E =12
μ 0
4πdrΜ(r) ⋅ dr'∫∫ Μ(r')
r − r' 3 − 3Μ(r') ⋅ r − r'( )[ ] r − r'( )
r − r' 5
⎧ ⎨ ⎩
⎫ ⎬ ⎭
Each dipole (magnetic moment) within a magnetized body interacts with each and every other. The sum of all that is the “self energy” of a magnetized body.
Recognize this?It’s the dipole field “density”.
€
Ed = −μ 0
2Μ(r) ⋅Hd (r)∫ dr
€
Hd (r) = hdip (r − r' )∫ dr' The demagnetization field
![Page 7: Magnetism and Magnetic Materials DTU (10313) – 10 ECTS KU – 7.5 ECTS Module 6 18/02/2011 Micromagnetism I Mesoscale – nm- m Reference material: Blundell,](https://reader036.fdocuments.us/reader036/viewer/2022062321/56649ec15503460f94bcc967/html5/thumbnails/7.jpg)
The demagnetization fieldThe demagnetization field
For spheres, ellipsoids, and a few other shapes, the demag field is uniform throughout the shape. In general, the demag field is highly non-uniform.
€
A(r) =μ 0
4πΜ(r' )× (r − r' )
r − r' 3∫ dr'
€
B(r) =∇× A(r) = μ 0 Μ(r)+Ηd (r)[ ]
= +
B M H
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Demag field for uniformly magnetized objects Demag field for uniformly magnetized objects
Introducing the characteristic function D(r), with value 1 inside the object, and 0 outside, we disentangle shape effects and get a convenient expression for the demag field.
€
A(r) =μ 0M 0
4πm ×(r − r' )
r − r' 3∫ dr'
=μ 0M 0
4πm × D(r' )
r − r'
r − r' 3∫ dr'
=μ 0M 0
4πm × D(r)⊗
rr 3
⎡ ⎣ ⎢
⎤ ⎦ ⎥
€
A(k) = −iμ 0M 0D(k)m × k
k2
B(k) = μ 0M 0D(k)k × m × k
k2
= μ 0M(k) −μ 0M 0D(k)m ⋅k
k2 k
Hd (r) = −M 0
8π 3 D(k)m ⋅k
k2 keik⋅r d∫ k Representation of the demag field for a uniformly magnetized tetrahedron
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Demag energy and Demag energy and demag factorsdemag factors
€
Hd (r) = −M 0
8π 3 D(k)m ⋅k
k2 keik⋅r d∫ k = − ˆ N (r)M
€
Ed = −μ 0
2Μ⋅Hd (r)∫ dr =
μ 0
2M i
ˆ N ij (r)∫ dr[ ]M j
=12
μ 0M 02V N imi
2
i=x,y,z
∑ = KdV N xmx2 + N ymy
2 + N zmz2
( )
€
N i = ˆ N ii (r) =1V
ˆ N ii (r)∫ dr
€
H di (r) = − ˆ N ij (r)M j
Demag field as a result of a tensor operation on the magnetization
Demag factors
€
ˆ N ij (r) = −1
8π 3 D(k)kik j
k2 eik⋅r d∫ k The demag tensor (a function of position)
The demag energy: a 2-form involving the three demag factors along main axes and the magnetization direction cosines
This is valid for any shape, provided its magnetization is uniform.
Nx
NyNz
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Domain wallsDomain walls
Large dipolar energy, no exchange energy
Snaller dipolar energy, some exchange energy
Idem
Bloch walls: bulk, thick objects
Neel walls: thin films, thin objects
Cross-over between dipolar and domain wall energies for a sphere (idealized model)
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Wall widthWall width
€
dE θ →θ + dθ( ) = JS2 dθ( )2
€
E = −2JS1 ⋅S2 = −2JS2 cosθ
€
E θ = 0 →θ =θDW( ) = NJS2 θDW
N
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
= JS2 θDW2
N
€
εK θ = 0 →θ = π( ) =NKπ
sin2 θdθ =0
π
∫ δK2a
€
εK = K sin2 θ
€
σ DWπ = JS2 π 2
δa
€
σ Kπ =
δK2
€
min JS2 π 2
δa+
δK2
⎡
⎣ ⎢
⎤
⎦ ⎥→δ = πS
2JaK
= πAK
with A =2JS2
a
The strong commercial magnet NdFeB has K=4.3e6 J/m3, and A=7.3e-12 J/m. Estimate the domain wall width in this material.
The domain wall energy is proportional to the area
€
σ DWπ = π AK
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Magnetocrystalline anisotropyMagnetocrystalline anisotropy
The crystal structure creates anisotropy: some directions are more responsive (“easier to magnetize”) to applied fields than others.
Consider a sphere of radius R magnetized along some easy axis u with anisotropy constant Ku=4.53e5 J/m3 (value for Co). If the magnetization flips to –u, the energy remains the same (up and down states are degenerate). But, in order to rotate from +u to –u, the magnetization has to go through a high energy state, i.e. when M points perpendicular to u. Suppose that the temperature is such that kBT is of the same order of the energy barrier separating the degenerate states. What happens?
€
εK = Ku sin2 θ + Ku2 sin4 θ
€
εK = K1 mx2my
2 + mx2mz
2 + my2mz
2( ) + K2mx
2my2mz
2
Uniaxial
Cubic
M
u
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Stoner-WolfharthStoner-Wolfharth
€
E = KdV (N x cos2 θ M + N y sin2 θ M ) −μ 0M 0H cos(θ M −θ H )
€
ε =sin2 θ M − 2h cos(θ M −θ H )
€
h =1
(N y − N x )HM 0
x
y€
∂ε∂θM
= 0 →θ M (h,θ H )
MH
The direction of M at any given applied field
Single-domain hysteresis is a consequence of anisotropy (shape or magnetocrystalline).
![Page 14: Magnetism and Magnetic Materials DTU (10313) – 10 ECTS KU – 7.5 ECTS Module 6 18/02/2011 Micromagnetism I Mesoscale – nm- m Reference material: Blundell,](https://reader036.fdocuments.us/reader036/viewer/2022062321/56649ec15503460f94bcc967/html5/thumbnails/14.jpg)
Brown’s equationsBrown’s equations
€
Ed = −μ 0M 0
2m(r) ⋅Hd (r)∫ dr
=Kd
8π 3
m(k) ⋅k 2
k2∫ dk
€
EK = −Ku m(r) ⋅u[ ]∫ 2dr
€
Ex = A ∇mx( )2 + ∇my( )
2+ ∇mz( )
2
[ ]∫ d 3r
= −A m(r) ⋅∇2m(r)∫ d 3r
€
EZ = −μ 0M 0 m(r) ⋅Happ(r)∫ dr
The whole set of equations provides a full description of the energy landscape of a micromagnetic system (such as the one shown above) and drives its evolution towards the ground state of minimum energy
€
Μ(r) = M 0m(r)
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Sneak peekSneak peek
Micromagnetic simulations
€
∂m∂t
= −γ m × Heff +α m ×∂m∂t
LLG equation
Magnetodynamics and evolution
Searching for ground states
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Wrapping upWrapping up
Next lecture: Tuesday February 22, 13:15, KU (A9)
Micromagnetism II (MB)
•Magnetic domains•Bloch and Neel walls, and wall widths•Dipolar/magnetostatic/demag energy•Demagnetization fields and factors•Stoner-Wolfharth hysteresis•Magnetocrystalline anisotropy•Brown’s equations
Please remember to:•Install OOMMF on your laptop•Familiarize a little bit with it•Bring your laptop to class on Tuesday, February 22