6.4 Linear and Nonlinear Media

6.4 Linear and Nonlinear Media
6.4.1 Magnetic Susceptibility and Permeability
If the applied field is not too strong, the magnetization is proportional and parallel to
the applied field for most of the paramagnetic and diamagnetic materials.
Linear Media:
Remember that the magnetic field B means total magnetic field. The auxiliary field
H means applied field and the magnetization M is induced by H. The magnetization
induced field is then added to form the total magnetic field.
r
r
r
r
r
r
r
r
r
r
H Æ M = χ m H Æ B = µ0 H + M = µ0 H + χ m H = µ0 (1 + χ m )H = µH
(
)
(
)
χ m ~ 10−5 << 1 for para- and diamagnetic materials
χ m ~ 1 for ferromagnetic materials
µ is permeability and µ0 is the permeability of free space
Compare with The Electric Polarization:
r r r
r
r r
P = χ eε 0 E ( D = ε 0 E + P , D is electric displacement from ρ free )
χ e is called the electric susceptibility, χ e =
P
,
ε0E
χ m is called the magnetic susceptibility, χ m =
M
H
r
r r
r
r
D = ε 0 E + P = ε 0 (1 + χ e )E = εE , ε is called the permittivity.
Example: An infinite solenoid (n turns per unit length, current I)
is filled with linear material of susceptibility χ m . Find the
magnetic field inside the solenoid.
Check the symmetry firstly.
r
Hl = nlI Æ H = nIzˆ
r
r
r
r
v
B = µH = µ0 (1 + χ m )nIzˆ , M = χ m nIzˆ Æ K b = M × nˆ = χ m nIφˆ
r
r
In linear media, ∇ ⋅ H = ∇ ⋅ M = 0 , but not at the boundary.
The induced bound current is always proportional to the free
I
current for most dia- and paramagnetic material (linear
media).
r
r
r
r
J b = ∇ × M = χ m∇ × H = χ m J f
Susceptibility χ m
Material
Diamagnetic
Susceptibility χ m
Material
Paramagnetic
Bismuth
-1.6 X 10
-4
Oxygen
1.9 X 10-6
Gold
-3.4 X 10-5
Sodium
8.5 X 10-6
Silver
-2.4 X 10-6
Aluminum
2.1 X 10-5
Copper
-9.7 X 10-6
Tungsten
7.8 X 10-5
Water
-9.0 X 10-6
Platinum
2.8 X 10-4
Carbon Dioxide
-1.2 X 10-8
Liquid Oxygen(-200oC)
3.9 X 10-3
Hydrogen
-2.2 X 10-9
Gadolinium
4.8 X 10-1
6.4.2 Ferromagnetism
Domain (Bloch wall)? Permanent Magnet? Hysteresis Loop? Curie Point? Phase
transition? Transition Temperature? Demagnetizing Field (Demagnetization Factor)?
In a ferromagnet, each dipole likes to point in the same direction as its neighbors.
If all dipoles in a ferromagnet point to the same direction, the ferromagnet may
display strong magnetic field. Why isn’t it a powerful ferromagnet?
Billions of dipoles form domains and domains are randomly oriented.
5K
10 K
100 K
200 K
300 K
Hysteresis Loop:
0.00006
Magnetization
0.00004
0.00002
0.00000
-0.00002
-0.00004
-0.00006
-1000
-500
0
500
Magnetic Field (Oe)
1000
Bloch Wall:
R. Wiesendanger’s group, Science 292, 2053 (2001).
APL 86, 082503 (2005).
Curie Point?
Phase Transition?
Mi et al., JAP 97, 043903 (2005).
Demagnetization Factor?
The electric field produced by a uniformly polarized
sphere of radius R.
r
r
P
P
E=−
=−
zˆ
3ε 0
3ε 0
r, θ, φ
P
The magnetic field of a uniformly magnetized sphere.
r 2
r 2
B = µ0 M = µ0 Mzˆ
3
3
r
r
r
r
1
B
H d = Nd M =
− M = − Mzˆ
3
µ0
r
r
r
r
r
M ∝ χ m H Æ H total = H applied + H d
r, θ, φ
M
The demagnetization factor Nd need to be taking account if we are handling magnetic
properties of ferromagnetic materials.
dosing Co 1x10
MT Curves for Paramagnet:
dosing Co 2x10
0.000010
16
2
Magnetization (emu / mm )
16
0.000008
16
dosing Co 3x10
16
dosing Co 4x10
0.000006
0.000004
0.000002
0.000000
0
20
40
60
80
100
Temperature (K)
Exercise: Find answers through internet: what is superparamagnetism?
does it display as you measure the MT curves?
What
Example: A familiar toy consists of donut-shaped permanent magnets (magnetization
parallel to the axis), which slide frictionlessly on a vertical rod. Treat the magnets as
dipoles, with mass md and moment m. If you put two back-to-back magnets on the
r
rod, the upper one will float. At what height does it float?
r
r µ m
× rˆ µ0 m sin θ ˆ
0
A=
=
φ
4π r 2
4π r 2
r
1
∂
(sin θvφ )rˆ − 1 ∂ (rvφ )θˆ = rˆ µ0m 2 cos3 θ + θˆ µ0m sin3θ
B=
4π
4π r
r sin θ ∂θ
r ∂r
r
r r µ0 m 2
µm 2
dU 3µ0 m 2 1
ˆ
(
)
F
B(z ) = 0
z
,
=
−
⋅
=
,
=
−
=
U
z
m
B
dz
2π z 4
4π z 3
2πz 3
− md g +
3µ0 m 2 1
=0
2π z 4
Exercise: 6.17, 6.18, 6.25