Chapter 18: Magnetic Properties
ISSUES TO ADDRESS...
• What are the important magnetic properties?
• How do we explain magnetic phenomena?
• How are magnetic materials classified?
• How does magnetic memory storage work?
• What is superconductivity and how do magnetic
fields effect the behavior of superconductors?
Chapter 18 - 1
Generation of a Magnetic Field Vacuum
• Created by current through a coil:
B0
H
I
N = total number of turns
 = length of each turn (m)
I = current (ampere)
H = applied magnetic field (ampere-turns/m)
B0 = magnetic flux density in a vacuum
(tesla)
• Computation of the applied magnetic field, H:
H=
NI
• Computation of the magnetic flux density in a vacuum, B0:
B0 = 0H
permeability of a vacuum
(1.257 x 10-6 Henry/m)
Chapter 18 - 2
Generation of a Magnetic Field -within a Solid Material
• A magnetic field is induced in the material
B
applied
magnetic
field H
current I
B = Magnetic Induction (tesla)
inside the material
B = H
permeability of a solid
m
• Relative permeability (dimensionless) mr =
m0
상대투자율
Chapter 18 - 3
Generation of a Magnetic Field within a Solid Material (cont.)
M = cmH
• Magnetization
Magnetic susceptibility
(dimensionless)
B = 0H + 0M
• B in terms of H and M
• Combining the above two equations:
B
B = 0H + 0 cmH
= (1 + cm)0H
cm > 0
vacuum cm = 0
cm < 0
H
cm = r - 1
permeability of a vacuum:
(1.26 x 10-6 Henry/m)
cm is a measure of a material’s
magnetic response relative to a
vacuum
Chapter 18 - 4
Chapter 18 - 5
Basic Concepts
18.1 A coil of wire 0.25 m long and having 400 turns carries a current of 15 A.
(a) What is the magnitude of the magnetic field strength H?
(b) Compute the flux density B if the coil is in a vacuum.
(c) Compute the flux density inside a bar of chromium that is positioned within the coil. The
susceptibility for chromium is found in Table 18.2.
(d) Compute the magnitude of the magnetization M.
Solution
(a)
NI
(400 turns)(15 A)
H =
(b)
l
=
0.25 m
= 24, 000 A - turns/m
B0 = 0H = (1.257  10 -6 H/m)(24, 000 A - turns/m) = 3.0168  10 -2 tesla

as cm = 3.13  10-4 (Table 18.2), then
 (c) Inasmuch
B = 0H +  0 M =  0H +  0c mH =  0H (1 + c m) = (1.257  10


-6
H/m) (24,000 A - turns/m) (1 + 3.13 10-4 )
=3.0177  10-2 tesla
(d)
M = c m H = (3.13  10-4 )(24,000 A - turns/m) = 7.51 A - turns/m
Chapter 18 - 6
Origins of Magnetic Moments
• Magnetic moments arise from electron motions and the
spins on electrons.
magnetic moments
electron
electron
spin
nucleus
electron orbital
motion
electron
spin
• Net atomic magnetic moment:
-- sum of moments from all electrons.
Fundamental magnetic moment
Bohr magnetron (μB)
= 9.27x10-24 A-m2
• Four types of response...
No magnetism ( ex. Inert gas )
Chapter 18 - 7
B (tesla)
Types of Magnetism
(3) ferromagnetic e.g. ferrite(), Co, Ni, Gd
(4) ferrimagnetic e.g. Fe3O4, NiFe2O4
( cm as large as 106 !)
(2) paramagnetic ( cm ~ 10-4)
e.g., Al, Cr, Mo, Na, Ti, Zr
vacuum (cm = 0)
(1) diamagnetic (cm ~ -10-5)
e.g., Al2O3, Cu, Au, Si, Ag, Zn
H (ampere-turns/m)
Chapter 18 - 8
Basic Concepts
18.2 (a) Explain the two sources of magnetic moments for electrons.
(b) Do all electrons have a net magnetic moment? Why or why not?
(c) Do all atoms have a net magnetic moment? Why or why not?
Solution
(a) The two sources of magnetic moments for electrons are the electron's orbital motion
around the nucleus, and also its spin.
(b) Each electron will have a net magnetic moment from spin, and possibly, orbital
contributions, which do not cancel for an isolated electron.
(c) All atoms do not have a net magnetic moment. If an atom has completely filled electron
shells or subshells, there will be a cancellation of both orbital and spin magnetic
moments.
Chapter 18 - 9
 Diamagnetism – The magnetite of the induced magnetic moment is extremely small and in
a direction opposite to that of the applied field.
 Paramagnetism – The orientations of atomic magnetic moments are random, such that a
piece of material possesses no net macroscopic magnetization. As the dipoles align with
the external field, they give relatively small positive magnetic susceptibility.
 Diamagnetic and paramagnetic materials  nonmagnetic materials
Chapter 18 - 10
 Ferromagnetism – It possess a permanent magnetic moment in the
absence of an external field and manifest very large and permanent
magnetization.
ex) α-ferrite, cobalt, nickel ; If H << M, B = μ0M
 Domain – The mutual spin alignment exists over relatively large-volume
regions of crystal
 Saturation magnetization - It represents the maximum possible
magnetization that results when all the magnetic dipoles in a solid piece are
mutually aligned with the external field
Ms = (net magnetic moment for each atom)x(the number of atoms present)
Net magnetic moment per atom: 2.22 μB for Fe, 1.72 μB for Co, 0.60 μB for Ni
Example 18.1. Saturation Magnetization and Flux density Computations for Nickel
Calculate (a) saturation magnetization and (b) saturation flux density for nickel (ρ=8.90 g/cm3)
Solution:
 N A (8 .90 x 10 6 g/m 3 )(6 .02 x 10 23 atoms/mol )
28
3
N=
ANi
=
= 9.13 x 10
58 .71 g/mol
0.60 B 9.27 x 10 -24 A  m
M s = 0.60 B N = (
)(
atom
B
2
atoms/m
9.13 x 10 23 atoms
)(
) = 5.1x 10 5 A/m
3
m
4  x 10 -7 H 5.1x 10 5 A
B s = o M s = (
)(
) = 0.64 tesla
m
m
Chapter 18 - 11
 Antiferromagnetism – It shows the alignment of the spin moments of
neighboring atoms or ions in exactly opposite directions.
ex) MnO
 Ferrimagnetism – Macroscopic magnetic characteristics of
ferromagnets and ferrimagnets are similar, but the distinction lies in the
source of net magnetic moment.
ex) MFe2O4 : M = Fe, Ni, Mn, Co, Cu
 Fe3O4  FeO + Fe2O3
Example Prob. 18.2. Saturation Magnetization for Fe3O4
Calculate saturation magnetization for Fe3O4 given that
each unit cell contains 8 Fe2+ and 16 Fe3+ (a = 0.839 nm)
Solution: M = N'  = n B 
s
B
B
VC
( 8 x 4 B /unitcell)(9.27 x 10 - 24 A  m 2 /B )
=
(0.839 x 10 - 9 m) 3 / unitcell
= 5.0 x 10 5 A/m
Chapter 18 - 12
Magnetic Responses for 4 Types
none
opposing
(2) paramagnetic
random
aligned
(3) ferromagnetic
(4) ferrimagnetic
aligned
Applied
Magnetic Field (H)
aligned
No Applied
Magnetic Field (H = 0)
(1) diamagnetic
Chapter 18 - 13
Concept Check 18.1 Cite major similarities and differences between ferromagnetic and ferrimagnetic
materials. Answer
The similarities between ferromagnetic and ferrimagnetic materials are as follows:
(1) There is a coupling interaction between magnetic moments of adjacent atoms/cations for both material types.
(2) Both ferromagnets and ferrimagnets form domains.
(3) Hysteresis B-H behavior is displayed for both, and, thus, permanent magnetizations are possible.
The differences between ferromagnetic and ferrimagnetic materials are as follows:
(1) Magnetic moment coupling is parallel for ferromagnetic materials, and antiparallel for ferrimagnetic.
(2) Ferromagnetics, being metallic materials, are relatively good electrical conductors; inasmuch as ferromagnetic
materials are ceramics, they are electrically insulative.
(3) Saturation magnetizations are higher for ferromagnetic materials.
Ferromagnetism
18.6 The magnetization within a bar of some metal alloy is 1.2 × 106A/m at an H field of 200 A/m.
Compute the following: (a) magnetic susceptibility, (b) permeability, and (c) magnetic flux density
within this material. (d) What type of magnetism would you suggest as being displayed by this material?
Solution
(a) This portion of the problem calls for us to compute the magnetic susceptibility within a bar of some metal alloy
when M = 1.2  106 A/m and H =200 A/m. This requires that we solve for cm from Equation 18.6 as
cm =
M
1.2  10 6 A / m
=
= 6000
H
200 A / m
(b) In order to calculate the permeability we must employ a combined form of Equations 18.4 and 18.7 as follows:

 = r  0 = (c m + 1)  0 = (6000 + 1) (1.257  10 -6 H/m) = 7.54  10 -3 H/m
(c) The magnetic flux density may be determined using Equation 18.2 as


B = H = (7.54  10 -3 H/m) (200 A/m) = 1.51 tesla

(d) This metal alloy would exhibit ferromagnetic behavior on the basis of the magnitude of its cm (6000), which is
considerably larger than the cm values for diamagnetic and paramagnetic materials listed in Table 18.2.
Chapter 18 - 14
Antiferromagnetism and Ferrimagnetism
18.12 Estimate (a) the saturation magnetization, and (b) the saturation flux density of cobalt ferrite
[(CoFe2O4)8], which has a unit cell edge length of 0.838 nm.
Solution
(a) The saturation magnetization of cobalt ferrite is computed in the same manner as Example Problem 18.2; from
nB  B
Equation 18.1
Ms =
a3
Now, nB is just the number of Bohr magnetons per unit cell. The net magnetic moment arises from the Co 2+ ions, of
which there are eight per unit cell, each of which has a net magnetic moment of three Bohr magnetons (Table 18.4).
 n is twenty-four. Therefore, from the above equation
Thus,
B
(24 BM / unit cell)(9.27  1024 A - m2 / BM)
= 3.78  10 5 A/m
(0.838  109 m)3 / unit cell
(b) This portion of the problem calls for us to compute the saturation flux density. From Equation 18.8
Ms =
Bs =  0 M s = (1.257  10 -6 H/m)(3.78  10 5 A/m) = 0.475 tesla

Concept Check 18.3 Explain why repeatedly dropping a permanent magnet on the floor will cause
it to become
 demagnetized.
Answer:
Repeatedly dropping a permanent magnet on the floor will cause it to become demagnetized because
the jarring causes large numbers of magnetic dipoles to become misaligned by dipole rotation.
Chapter 18 - 15
 The Influence of Temperature on Magnetic Behavior
– With increasing temperature, the increased thermal motion of the atoms tends to
randomize the direction of any moments that may be aligned.
i.e. Saturation magnetization is decreased for both ferro- and ferrimagnets.
 Curie temperature (Tc) – at which the saturation magnetization decreases gradually and
then abruptly drops to zero with increasing temperature.
 Ferro- and ferrimagetic materials are paramagnetic above Tc
 Neel Temperature – at which antiferromagnetism vanishes
; antiferromagnetic material a become paramagnetic above this temperature
Chapter 18 - 16
Domains in Ferromagnetic &
Ferrimagnetic Materials
• As the applied field (H) increases the magnetic domains
change shape and size by movement of domain boundaries.
B sat
H
Magnetic
induction (B)
H
H
H
H
0
• “Domains” with
aligned magnetic
moment grow at
expense of poorly
aligned ones!
Applied Magnetic Field (H)
H=0
Chapter 18 - 17
Hysteresis and Permanent
Magnetization
• The magnetic hysteresis phenomenon
Remanence (잔류자기) or
Remanent flux density, Br
Stage 3. Remove H, alignment
remains! => permanent magnet!
보자력
Stage 4. Coercivity, HC
Negative H needed to
demagnitize!
Stage 5. Apply -H,
align domains
B
Stage 2. Apply H,
align domains
H
Stage 1. Initial (unmagnetized state)
Stage 6. Close the
hysteresis loop
Chapter 18 - 18
Domains and Hysteresis
18.18 A ferromagnetic material has a remanence of 1.0 tesla and a
coercivity of 15,000 A/m. Saturation is achieved at a magnetic field
strength of 25,000 A/m, at which flux density is 1.25 teslas. Using
these data, sketch entire hysteresis curve in the range H = –25,000 to
+25,000 A/m. Be sure to scale and label both coordinate axes.
Solution
The B versus H curve for this material is shown below.
Concept Check 18.3 Schematically sketch on a single plot the
B-versus-H behavior for a ferromagnetic material
(a) at 0 K,
(b) at a temperature just below its Curie temperature, and
(c) at a temperature just above its Curie temperature. Briefly
explain why these curves have different shapes.
Chapter 18 - 19
 B-versus-H behaviors of paramagnetic, diamagnetic,
and ferromagnetic/ferromagnetic materials
 Magnetic Anisotropy
Chapter 18 - 20
Hard and Soft Magnetic Materials
Hard magnetic materials:
B
Soft
-- large coercivities
-- used for permanent magnets
-- add particles/voids to
inhibit domain wall motion
-- example: tungsten steel -Hc = 5900 amp-turn/m)
-- motor application (drill, window wiper, fan, video
recorder), speaker, earphone
H
Soft magnetic materials:
-- small coercivities
-- used for electric motors (transformer core)
-- example: commercial iron 99.95 Fe
Chapter 18 - 21
Chapter 18 - 22
An Iron- Silicon Alloy that is used in Transformer Cores
 Energy losses of transformers could be minimized if their cores were fabricated from
single crystals such that a [100]-type direction is oriented parallel to the direction of an
applied magnetic field.
 Single crystals are expensive
 Attractive alternative is to fabricate cores from polycrystalline sheets of this alloy that
are anisotropic.
 The magnetic characteristics of this alloy may be improved through a series of
deformation and heat-treating procedures that produce a (100) [001] texture
Chapter 18 - 23
Figure 18.30 shows the B-versus-H curve for a nickel–iron alloy.
(a) What is the saturation flux density?
(b) What is the saturation magnetization?
(c) What is the remanence?
(d) What is the coercivity?
(e) On the basis of data in Tables 18.5 & 18.6, would you classify this
material as a soft or hard magnetic material? Why?
Solution
(a) The saturation flux density for this nickel-iron is 1.5 tesla, the
maximum B value shown on the plot.
(b)
Bs
1.5 tesla
Ms =
0
=
1.257 
10 6
H /m
= 1.19  10 6 A/m
(c) The remanence, Br, is read from this plot as from the hysteresis loop shown in Figure 18.14; its
value isabout 1.47 tesla.

(d) The coercivity, Hc, is read from this plot ; the value is about 17 A/m.
(e) On the basis of Tables 18.5 and 18.6, this is most likely a soft magnetic material. The saturation flux
density (1.5 tesla) lies within the range of values cited for soft materials, and the remanence (1.47 tesla)
is close to the values given in Table 18.6 for hard magnetic materials. However, the Hc (17 A/m) is
significantly lower than for hard magnetic materials.
Chapter 18 - 24
Magnetic Storage
• Digitized data in the form of electrical signals are transferred to
and recorded digitally on a magnetic medium (tape or disk)
• This transference is accomplished by a recording system that
consists of a read/write head
-- “write” or record data by applying a
magnetic field that aligns domains
in small regions of the recording
medium
-- “read” or retrieve data from medium by
sensing changes in magnetization
recording medium
recording head
Chapter 18 - 25
Magnetic Storage Media Types
-- CoCr alloy grains (darker regions)
separated by oxide grain boundary
segregant layer (lighter regions)
-- Magnetization direction of each
grain is perpendicular to plane of
disk
80 nm
• Hard disk drives (granular/perpendicular media):
~ 500 nm
~ 500 nm
• Recording tape (particulate media):
-- Acicular (needle-shaped)
ferromagnetic metal alloy
particles
-- Tabular (plate-shaped)
ferrimagnetic barium-ferrite
Chapter 18 - 26
particles
Superconductivity
Found in 26 metals and hundreds of alloys & compounds
Mercury
Copper
(normal)
4.2 K
• TC = critical temperature
= temperature below which material is superconductive
Chapter 18 - 27
Critical Properties of
Superconductive Materials
TC = critical temperature - if T > TC not superconducting
JC = critical current density - if J > JC not superconducting
HC = critical magnetic field - if H > HC not superconducting
æ
T2 ö
HC (T ) = HC (0)ç1 - 2 ÷
TC ø
è
Chapter 18 - 28
Meissner Effect
• Superconductors expel magnetic fields
normal
superconductor
• This is why a superconductor will float above a
magnet
Chapter 18 - 29
Advances in Superconductivity
• Research in superconductive materials was stagnant
for many years.
– Everyone assumed TC,max was about 23 K
– Many theories said it was impossible to increase
TC beyond this value
• 1987- new materials were discovered with TC > 30 K
– ceramics of form Ba1-xKxBiO3-y
– Started enormous race
• Y Ba2Cu3O7-x
TC = 90 K
• Tl2Ba2Ca2Cu3Ox
TC = 122 K
• difficult to make since oxidation state is very important
• The major problem is that these ceramic materials
are inherently brittle.
Chapter 18 - 30
Chapter 18 - 31
Summary
• A magnetic field is produced when a current flows
through a wire coil.
• Magnetic induction (B):
-- an internal magnetic field is induced in a material that is
situated within an external magnetic field (H).
-- magnetic moments result from electron interactions with
the applied magnetic field
• Types of material responses to magnetic fields are:
-- ferrimagnetic and ferromagnetic (large magnetic susceptibilities)
-- paramagnetic (small and positive magnetic susceptibilities)
-- diamagnetic (small and negative magnetic susceptibilities)
• Types of ferrimagnetic and ferromagnetic materials:
-- Hard: large coercivities
-- Soft: small coercivities
• Magnetic storage media:
-- particulate g-Fe2O3 in polymeric film (tape)
-- thin film CoPtCr or CoCrTa (hard drive)
Chapter 18 - 32
ANNOUNCEMENTS
Reading:
Core Problems:
Self-help Problems:
Chapter 18 - 33