Download Chapter 6 Magnetic Fields in Matter

Chapter 6 Magnetic Fields in Matter
6.1.2. Torques and Forces on Magnetic Dipoles
6.1 Magnetization
6.1.1 Diamagnets, Paramagnets, Ferromagnets
A magnetic dipole experiences a torque in a magnetic field,
just as an electric dipole does in an electric field.
All the magnetic phenomena are due to electric charges in
motion:
Electrons orbiting around nuclei
magnetic dipoles
Electrons spinning about their axes
Any current loop could be built up from infinitesimal
rectangles, with all the “internal” side canceling. There is
no actual loss of generality in using the shape.
When a magnetic field is applied, a net alignment of these
magnetic dipoles occurs, and the medium becomes
magnetically polarized, or magnetized.
The magnetic polarization M, unlike electrical polarization
P, might be parallel to B (paramagnets) or opposite to B
(diamagnets).
A few substances (ferromagnets) retain their magnetization
1
even after the external field has been removed.
Let’s calculate the torque on a rectangular current loop in
a uniform magnetic field.
Torques and Forces on Magnetic Dipoles
Torques and Dipole Moment
Center the loop at the origin, and tilt it an angle θ from the
z axis towards the y axis. Let B point in the z direction.
N = IabB sin θxˆ = mB sin θxˆ = m × B
where m = Iab is the magnetic dipole moment of the loop.
2
This equation is identical in form to the electrical analogy.
N = p×E
The torque is again in such a direction as to line the dipole
up parallel to the field (paramagnetism).
Sloping sides: the forces cancel.
Horizontal sides: the forces cancel but they generate a
torque.
N = L × F = aF sin θxˆ
The magnitude of the force on each of these segments is:
F = IbB
3
QM: The Pauli exclusion principle dictates that the
electrons within a given atom lock together in pairs with
opposite spins, and this effectively neutralizes the torque
on the combination.
4
Forces in Nonuniform Magnetic Field
Forces on an Infinitesimal Current Loop and Model
F = ∇(m ⋅ B)
In a uniform field, the net force on a current loop is zero:
=0
F = I ∫ (dl × B) = I ∫ (dl ) × B = 0
(= −∇U , where U = −(m ⋅ B))
Identical to the electrical formula
In a nonuniform field this is no longer the case, because
the magnetic field B could not come outside the integral.
F = ∇(p ⋅ E)
Does the magnetic dipole consist of a pair of opposite
magnetic monopoles just like an electric dipole?
Fringing field effect : F = 2πIRB cos θ
5
6.1.3. Effect of a Magnetic Field on Atomic Orbits
6
Electron Speeds Up or Slows Down
v2
e2
= me
4πε 0 R 2
R
1
Electrons not only spin; they also revolve around the nucleus.
Let’s assume the orbit is a circle of radius R. The current
looks like steady (really?)
Current I =
e
e
ev
=
=
T 2πR 2πR
v
Orbital dipole moment m = −
without the magnetic field.
The centripetal force comes from two sources:
the electric force and the magnetic force.
The negative charge
of the electron
e2
v′ 2
′
+
=
e
v
B
m
e
4πε 0 R′2
R′
1
evR
zˆ (m = IπR 2 )
2
7
Assume R′ ≅ R
m (v′2 − v 2 ) me (v′ + v)(v′ − v)
=
ev′B = e
R
R
eRB
∴ ∆v = (v′ − v) ≈
When B is turn on, the electron speeds up.
8
2me
The Dipole Moment and The Diamagnetism
6.1.4 Magnetization
A change in the orbital speed means a change in the
dipole moment.
In the present of a magnetic field, matter becomes magnetized.
Upon microscopic examination, it contains many tiny dipoles,
with a net alignment along some direction.
e2 R 2
1
∆m = − e∆vR = −
B
2
2me
Two mechanisms account for this magnetic polarization:
The change in m is opposite to the direction of B.
1.Paramagnetism: the dipoles associated with the spins of
unpaired electrons experience a torque tending to line them up
parallel to the field.
In the presence of a magnetic field, each atom picks up a
little “extra” dipole moment, and the increments are all
antiparallel to the field. This is the mechanism responsible
for diamagnetism.
This is a universal phenomenon, affecting all atoms, but it is
typically much weaker than paramagnetism.
9
6.2 The Field of a Magnetized Object
6.2.1 Bound Currents
2. Diamagnetism: the orbital speed of the electrons is altered
in such a way as to change the orbital dipole moment in the
direction opposite to the field.
We describe the state of magnetic polarization by the vector
quantity:
M ≡ magnetic dipole moment per unit volume.
Vector potential and Bound Currents
Can the equation be expressed in a more illuminating form,
as in the electrical case? Yes!
Suppose we have a piece of magnetized
material (i.e. M is given). What field does
this object produce?
By exploiting the identity,
∇′
The vector potential of a single dipole m is
A(r ) =
10
µ0 m × rˆ
4π r 2
1 rˆ
=
r r2
1
∂
∂
∂
+ yˆ ′
+ zˆ ′ )
∂x′
∂y′
∂z′ ( x − x′) 2 + ( y − y′) 2 + ( z − z′) 2
xˆ ′( x − x′) + yˆ ′( y − y′) + zˆ ′( z − z ′)
rˆ
=
=
(( x − x′) 2 + ( y − y′) 2 + ( z − z ′) 2 )3/ 2 r 2
(xˆ ′
The vector potential is A(r ) =
µ0
1
M (r′) × (∇′ )dτ ′
∫
r
4π
Using the product rule ∇ × ( fA) = ∇f × A + f (∇ × A)
In the magnetized object, each volume element carries a
dipole moment Mdτ’, so the total vector potential is
and integrating by part, we have
µ0  1
M (r′)

]dτ ′
 [∇′ × M (r′)]dτ ′ − ∫ ∇′ × [
4π ∫ r
r

how? Prob. 1.60
µ
µ0  1
1

0
=
 [∇′ × M (r′)]dτ ′ +
∫ [M(r′) × nˆ ′]da′
4π ∫ r
 4π r
A(r ) =
µ M (r′) × rˆ
A(r ) = 0 ∫
dτ ′
4π
r2
11
12
Vector potential and Bound Currents
A(r ) =
µ0 1
µ 1
[∇′ × M (r′)]dτ ′ + 0 ∫ [M (r′) × nˆ ′]da′
∫
4π r
4π r
J b = ∇′ × M (r′)
K b = M (r′) × nˆ ′
volume current
surface current
bound currents
With these definitions,
Gauss's law
∫ (∇ ⋅ E)dτ = v∫ E ⋅ da
v
A(r ) =
S
 (∇ ⋅ ( v × c))dτ = c ⋅ (∇ × v )dτ
∫v
∫
Let E = v × c,  v
 v∫ ( v × c) ⋅ da = −c ⋅ v∫ v × da
S
S
Since c is a constant vector, so ∫ (∇ × v )dτ = − v∫ v × da
v
S
µ0
4π
∫
Jb
µ
dτ ′ + 0
r
4π
v
∫
S
Kb
da′
r
The electrical analogy
volume charge density ρ b = −∇ ⋅ P
13
Example 5.11 A spherical shell, of radius R, carrying a
uniform surface charge σ, is set spinning at angular velocity
ω. Find the vector potential it produce at point r.
surface charge density σ b = P ⋅ nˆ
14
µ 0 Rω ( − cosψ sin θ ′ sin φ ′) xˆ 2
R sin θ ′dθ ′d φ ′
4π ∫
r 2 + R 2 − 2 rR cos θ ′
µ Rω (cosψ sin θ ′ cos φ ′ − sinψ cos θ ′) yˆ 2
+ 0 ∫
R sin θ ′dθ ′d φ ′
4π
r 2 + R 2 − 2 rR cos θ ′
µ Rω (sin ψ sin θ ′ sin φ ′) zˆ 2
+ 0 ∫
R sin θ ′dθ ′d φ ′
4π
r 2 + R 2 − 2 rR cos θ ′
A (r ) =
− R 3σω sinψµ0 yˆ
cos θ ′
sin θ ′dθ ′dφ ′
∫
2
4π
r + R 2 − 2rR cos θ ′
π
− R 3σω sinψµ0 yˆ
− cos θ ′
=
(2π ) ∫
d cos θ ′
2
4π
r + R 2 − 2rR cos θ ′
0
A(r ) =
Sol :First, let the observer is in the z axis and ω is tilted at an angle ψ
µ K (r′)
Vector potential is A(r ) = 0 ∫
da′
4π
r
The surface current density K (r′) = σ v′
− µ0 R 3σω sinψ yˆ
u
du
∫
2
2
2
−1 r + R − 2rRu
1
=
15
16
A(r ) =
− µ0 R 3σω sinψyˆ ( R 2 + r 2 + Rr ) | R − r | −( R 2 + r 2 − Rr )( R + r )
)
(−
3R 2 r 2
2
 µ 0 Rσ
(ω × r ) inside

A(r ) =  24
 µ 0 R 3σ (ω × r ) outside
 2r
Example 6.1 Find the magnetic field of a
uniformly magnetized sphere of radius R.
Sol : Choosing the z axis along the direction of M,
J ′b = ∇ × M = 0

we have 
K ′b = M × nˆ ′ = M sin θφˆ
The surface current density is analogous to that of a spinning
spherical shell with uniform surface current density.
Reverting to the “natural” coordinate, we have
K ′b = M × nˆ ′ = M sin θφˆ ⇔ K ′ = σv′ = σRω sin θφˆ
σRω → M
Surprisingly, the field inside the spherical shell is uniform.
17
6.2.2 Physical Interpretation of Bound Current
2
µ 0 M (inside)
3
Can you find a more direct method?
B=
18
Physical Interpretation of Bound Current
Bound surface current Kb:
Bound current density Jb:
What is the current in terms of M?
What if the magnetization is not uniform?
In terms of the magnetization M, its dipole
moment is m=Mat=Ia. So, M=I/t=Kb
The adjacent current loops do not completely cancel out.
Consider a thin slab of uniformly magnetized material, with
the dipoles represented by tiny current loops.
All the “internal” currents cancel. However, at the edge there
is no adjacent loop to do the canceling.
19
∂M z
dydz
∂y
∂M y
dydz
Case (b) I x = [ M y ( z + dz ) − M y ( z )]dy =
∂z
∂M z ∂M y
∴(Jb )x =
−
⇒ Jb = ∇ × M
∂y
∂z
Case (a) I x = [ M z ( y + dy ) − M z ( y )]dz =
20
6.3 The Auxiliary Field H
6.3.1 Ampere’s Law in Magnetized Materials
The Role of H in Magnetostatics
H plays a role in magnetostatics analogous to D in the
electrostotics.
What is the difference between bound current and free
current?
J =J +J
b
f
D allows us to write Gauss’s law in terms of free change alone.
Ampere’s law can be written:
1
µ0
D = ε 0 E + P,
(∇ × B) = J = J f + J b = J f + ∇ × M
⇒ ∇×(
1
µ0
H permits us to express Ampere’s law in terms of free current
1
alone.
B − M) = J f
H=
H
In terms of H, then the Ampere’s law reads
f
µ0
B − M, ∇ × H = J f
What we can control directly.
∇ × H = J f (differential form)
∫ H ⋅ dl = I
∇ ⋅ D=ρ f
Why can’t we turn the bound currents on or off
independently?
(integral form)
21
Example 6.2 A long copper rod of radius R carries a
uniformly distributed (free) current I. Find H inside and
outside the rod.
22
Sol :
Use the Ampere's law in the integral form and
properly choose a suitable Amperian loop.
s ≤ R : H (2πs ) = I f enc = I
so H =
πs 2
πR 2
sI
φˆ
2πR 2
s > R : H (2πs ) = I , so H =
I
φˆ
2πs
How to determine the magnetic field B?
How to choose a suitable Amperian loop? Symmetry.
23
24
H and B, D and E
∇×H = J f
∇⋅D = ρf
6.3.2 A Deceptive Parallel
In free space
Which equation is more useful?
We can easily control the free current I, but not the
free charge. So H can be determined accordingly.
∇ × B = µ0 J
∇⋅B = 0
In matter ∇ × H = J f
∇⋅H = ∇⋅(
On the other hand, the potential difference V can be
read from the voltmeter, which can be used to
determine E.
B
µ0
− M ) = −∇ ⋅ M ≠0
At what condition the divergence of H is equal to zero?
The name of H: Some author call H, not B, the “magnetic
field”, but it is not a good choice. Let’s just call it “H”.
M // B i.e. M // B // H for uniform material only.
25
26
6.3.3 Boundary Conditions
The magnetostatic boundary conditions can be rewritten in
terms of H and the free surface current Kf.
∇×H = J f
⇒
//
//
− H below
= K f × nˆ
H above
⊥
⊥
⊥
⊥
∇ ⋅ H = −∇ ⋅ M ⇒ H above
− H below
= −( M above
− M below
)
Homework #11
Problems: 6.4, 6.10, 6.13, 6.15
The corresponding boundary condition in terms of B and total
surface current K.
//
//
∇ × B = µ0 J ⇒ B above
− Bbelow
= µ0 (K × nˆ )
⊥
⊥
∇⋅B = 0 ⇒
− Bbelow
=0
Babove
How to express the boundary conditions at metal or
dielectric interface?
27
28
6.4 Linear and Nonlinear Media
6.4.1 Magnetic susceptibility and Permeability
Material Susceptibility
The magnetization of paramagnetic and diamagnetic
materials is sustained by the field, i.e. when B is removed,
M disappears.
M = χ m H,
where the proportionality constant χ m is called
the magnetic susceptibility.
χm
Because M = χ m H ∝ I f
B?
µ0
Materials that obey M = χ m H are called linear media.
Β = µ 0 (H + M ) = µ 0 (1 + χ m )H = µH,
where µ = µ 0 (1 + χ m ) is call the permeabili ty of the material.
Why not use M =
29
Divergence of the Magnetization
Example 6.3 An infinite solenoid (n turns per unit
length, current I) is filled with linear material of
susceptibility χm: Find the magnetic field inside the
solenoid.
Does the linear media avoid the defect that the divergence
of M is zero? No!
Even though M, H, and B are parallel, the divergence of M
is not zero at the boundary. Consider the following example.
Sol: The problem exhibits solenoidal symmetry. Thus, we
can employ the Ampere’s law.
∫ H ⋅ dl = I
f
∫ M ⋅ da ≠ 0
(integral form)
HA = nAI ∴ H = nIzˆ
Gaussian pillbox
⇒ ∇⋅M ≠ 0
B = µ 0 (1 + χ m )nIzˆ
The enhancement of the magnetic field strength depends on
the susceptibility of the material.
Is there a material that the field is significantly enhanced?
30
31
and J b = ∇ × M = ∇ × χ m H = χ m J f
32
6.4.2 Ferromagnetism
Ferromagnetic Domains
Ferromagnets---which are not linear---require no external
fields to sustain the magnetization unlike paramagnets and
diamagnets.
In a ferromagnet, each dipole “like” to point in the same
direction as its neighbors. All the spins point the same way.
Domain boundaries: Domains parallel to the field grow, and
the others shrink.
Why isn’t every wrench and nail a powerful magnet?
If the field is strong enough, one domain takes over entirely,
and the iron is said to be “saturated”.
Domains.
33
Hysteresis Loop
34
Curie Temperature and Phase Transitions
Temperature effect: The dipoles within a given domain line
up parallel to one another. However, the random thermal
motions complete with this ordering.
Curie temperature: As the temperature increases, the
alignment is gradually destroyed. At certain temperature the
iron completely turns into paramagnet. This temperature is
called the curie temperature.
phase transitions
Hysteresis: The path we have traced out.
In the experiment, we adjust the current I, i.e. control H.
In practice M is huge compared to H.
35
36
Homework #12
Problems: 6.17, 6.21, 6.23, 6.26
37
Supplementary Material (optional):
Properties of ferrite materials (I)
Introduction to ferrite materials
鐵磁性
38
The ferrites are crystals
having small electric
conductivity compared to
ferromagnetic materials,
反鐵磁性
Thus they are useful in
high-frequency situations
because of the absence
of significant eddy
鐵氧體磁性
current losses.
39
Nonreciprocal electrical property: the transmission
coefficient through the device is not the same for different
direction of propagation.
Unequal propagation constant: The left and right
circularly polarized waves have different propagation
constant along the direction of external magnetic field B0.
Anisotropic magnetic properties: The permeability of the
ferrite is not a single scalar quantity, but instead is a tensor,
which can be represented as a matrix.
40
Classical picture of the magnetization process
Properties of ferrite materials (II)
--- By treating the spinning electron as a gyroscopic top.
Ferrites are ceramiclike materials with specific resistivities
that may be as much as 1014 greater than that of metals and
with dielectric constants around 10 to 15 or greater.
Ferrites are made by sintering a mixture of metal oxides
and have the general chemical composition MO·Fe2O3,
where M is a divalent metal such as Mn, Mg, Fe, Zn, Ni, Cd,
etc.
Relative permeabilities of several thousand are common.
The magnetic properties of ferrites arise mainly from the
magnetic dipole moment associated with the electron
spin.
41
Quantum mechanics’ viewpoint sz=±1/2
If an electron is located in a uniform static magnetic field
B0, a torque is given by
m
e
T = m × B0 = − p × B0 =
B0 × P
p
mo
dP
e
B 0 × P = ω0 × P
=T=
dt
mo
where ω0 =
eB0
is called the Larmor fequency;
mo
=
is angular momentum; and
2
e=
is magnetic dipole moment.
m=
2mo
42
P=
Saturation magnetization
In the absence of any damping forces, the actual
precession angle will be determined by the initial position
of the magnetic dipole, and the dipole will precess about
B0 at this angle indefinitely (free precession).
In reality, however, the existence of damping forces will
cause the magnetic dipole to spiral in from its initial angle
until m is aligned with B0.
This explains why sz equals ±1/2 in the Quantum
Mechanics.
As the strength of the bias field
H0 is increased, more magnetic
dipole moments will align with H0
until all are aligned, and M
reaches an upper limit.
The material is then said to be magnetically saturated, and
Ms is denoted as the saturation magnetization. Ms typically
ranges from 4πMs=300 to 5000 Gauss.
Below saturation, ferrite materials can be very lossy at
microwave frequencies, and rf interaction is reduced.
But where does the damping force come from?
43
The ferrites are usually operated in the saturated state.
44
Curie temperature
Properties of some ferrite materials
The saturation magnetization
of a material is a strong
function of temperature,
decreasing as temperature
increases.
This effect can be understood by noting that the vibrational
energy of an atom increases with temperature, making it
more difficult to align all the magnetic dipoles.
Why use 4πMs? B = 4πM + H = µH
At a high enough temperature a zero net magnetization
results. This temperature is called the Curie temperature, Tc.
45
(Gaussian unit)
The unit of B is gauss; the unit of H is Oersted. They have same
dimension.
What does ∆H and Mr mean?
Ferrite linewidth and remanent magnetization
46
Anisotropic magnetic properties (I)
Anisotropic magnetic properties (II)
If H is the applied ac field, the total magnetic field is Ht = H0 zˆ + H ,
Omitting higher order terms, the equations can be reduced to
where H << H 0 . The field produced a total magnetization is
the ferrite given by M t = M s zˆ + M .
Ms is the dc saturation magnetization and M is the
additional ac magnetization (in the xy plane) caused by
applied field.
The component equations of motion:
dM x
= −µ0γM y (H0 + H z ) + µ0γ (M s + M z )H y
dt
dM y
= −µ0γM x (H0 + H z ) − µ0γ (M s + M z )H x
dt
dM z
= −µ0γM x H y + µ0γM y H x
dt
dM x
dH y
d 2M x
= −ω0 M y + ωm H y ,
+ ω02 M x = ωm
+ ω0ωm H x ,
2
dt
dt
dt
dM y
d 2M y
dHx
= −(ω0 M x + ωm H x ),
+ ω02 M y = −ωm
+ ω0ωm H y.
dt
dt 2
dt
dM z
= 0, whereω0 = µ0γH0 and ωm = µ0γM s
dt
If M and H ∝ e jωt , the above equtions can be reduced to
the phasor equations:
 χ xx χ xy 0
M = [χ ]H = χ yx χ yy 0 H ,
2
2
(ω0 − ω )M y = − jωωm H x + ω0ωm H y .
 0
0 0
ωω
jωω
where χ xx = χ yy = 2 0 m 2 and χ xy = −χ yx = 2 m 2
48
ω0 − ω
ω0 − ω
(ω02 − ω 2 )M x = ω0ωm H x + jωωm H y ,
47
Anisotropic magnetic properties (III)
To relate B and H, we have
 µ
B = µ0 (M + H ) = [µ ]H ⇒ [µ ] = µ0 ([U ] + [χ ]) = − jκ
 0
jκ
µ
0
Forced precession of spinning electron (I)
If a small ac magnetic field is superimposed on the static
field H0, the magnetic dipole moment will undergo a forced
precession.
0
0 
µ0 
Of particular interest is the case where the ac magnetic
field is circularly polarized in the plane perpendicular to H0.
ω0ωm

µ = µ0 (1 + χ xx ) = µ0 (1 + χ yy ) = µ0 (1 + ω 2 − ω 2 )

0

ωω
m
κ = − jµ χ ) = jµ χ = µ
)
0 xy
0 yx
0

ω02 − ω 2
A right-hand circularly polarized wave can be expressed in
phasor form as
H + = H + ( xˆ − jyˆ )
A material having a permeability tensor of this form is called
gyrotropic.
and in time-domain form as
{
}
H + = Re H + e jωt = H + ( xˆ cos ωt + yˆ sin ωt )
How to apply this concept to a circularly polarized wave?
49
Forced precession of spinning electron (II)
50
Real and imaginary permeability &
propagation and attenuation constant.
propagation constant
β ± = ω εµ ±
RHCP
LHCP
ωm

H + ,
ω0 − ω
ωm

+
H+
⇒M =
−
j
ω
ω
−
ω
m
0
M y+ =
H + .

ω0 − ω
M x+ =
ωm
) RHCP
ω0 − ω
ω
µ − = µ0 (1 + m ) LHCP
51
ω0 + ω
⇒ µ + = µ0 (1 +
52
Faraday rotation --- a nonreciprocal effect
Microwave gyrator
Consider linearly polarized
electric field at z=0, represented
as the sum of a RHCP and a
LHCP wave:
E ( z =0) = xˆE0 =
Gyrator with a
twist section.
E0
E
( xˆ − jyˆ ) + 0 ( xˆ + jyˆ )
2
2
These two polarized waves propagate with different
propagation constants.
E0
E
( xˆ − jyˆ )e− jβ+A + 0 ( xˆ + jyˆ )e− jβ−A
2
2
β −β
β −β 

= E0  xˆ cos( + − )A − yˆ sin( + − )Ae− j ( β+ +β− )A / 2
2
2


E
β −β
θ = tan−1 y = −( + − )A. This effect is called Faraday rotation.
2
Ex
53
E ( z =A) =
Faraday-rotation isolator
Gyrator without
a twist section.
54
Resonance isolator
55
56
Four-port circulator
Three-port circulator
57
58