Electromagnetism
Zhu Jiongming
Department of Physics
Shanghai Teachers University
Electromagnetism
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Electric Field
Conductors
Dielectrics
Direct-Current Circuits
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Magnetic Field
Electromagnetic Induction
Magnetic Materials
Alternating Current
Electromagnetic Waves
Chapter 7 Magnetic Materials
§1. Basic Laws in Magnetic Materials
§2. Paramagnetism and Diamagnetism
§3. Ferromagnetism and Ferromagnetics
§4. Magnetic Field Energy
§1. Basic Laws in Magnetic Materials
1. Magnetizing and Magnetization of Materials
● Magnetizing
● Magnetization
2. Magnetization Current
3. Magnetic Field Intensity
Ampere’s Law in Magnetic Materials
4. Examples
1. Magnetizing and Magnetization
magnetic material  magnetized  loop model for
electron orbits
Comparison：
dielectric material  polarized  electric dipole
Magnetizing
 Magnetization

Magnetizing of Materials
loop model  atomic magnetic dipole moment pm 
aligned （magnetized）
i
 cancel out inside
pm
 bound current on outline I
free current（conducting） I0  B0
bound current（magnetized）I’  B’
B = B0 + B’
Magnetization

Magnetization：vector sum of pm per unit volume
M   pmi V
i
compare： P   pi  V
i
Uniform：M is same at every point in material
ferromagnetic
 magnetic
paramagnetic
materials
non-ferromagnetic
diamagnetic

Experiments：in isotropic non-ferromagnetics
M = gB
compare： P = 0E
 paramagnetics ：g > 0 M and B same direction
 diamagnetics ： g < 0 M and B opposite

2. Magnetization Current（1）
surface S，border L
bound current I’
passing through S
S
Only the loop currents
round L contribute to I’
L
2. Magnetization Current（1）
surface S，border L
bound current I’
passing through S
S
Only the loop currents
round L contribute to I’
L
2. Magnetization Current（2）
N ：number of molecules per unit volume
M
Im ： molecule’s loop current
S

inclined cylinder：
length dl , base area S of loop
dl
 volume： Sdl cos
 number of loop： NSdl cos
（center of which in the cylinder）
 current： dI’ = ImNSdl cos  Npmdl cos  M  dl
 I '   M  dl
L
Compare：q'    P  dS
S
2. Magnetization Current（3）
dI '
j' 
dS 
dI '
i' 
dl
dl
dI’
Surface current density —— charges perpendicularly
passing through unit length per unit time ( on surface )
 It is proved（skip）
 Uniformly magnetized： j’ = 0
 Interface between two materials：
 ’ = ( M2 - M1 )  n （n：2  1）
Compare：uniform ’ = 0 ’ = ( P2 - P1 ) · n
3. Ampere’s Law in Magnetic Materials
 B  dl   I   (I
  ( I   M  dl )
0
L
0

0
B
 (
L
0
0
1
 I ')
 E  dS  
S
 M )  dl  I 0
Definition ： H 
B
L
M
Definition：D  0E + P
Electric displacement
 D  dS  q
0
Vacuum：
M =0， B = 0H
(q0   P  dS )
S
0
Magnetic field intensity
 H  dl  I
0
S
0
  ( 0 E  P )  dS  q0
L
0

1
(q0  q' )
0
S

 B  dl   I
L
0 0
Magnetic Field Intensity H
H
B
0
H (
M
B
D  0E + P
M
1
0
 g )B 
1

B
D   0 E   0 E  E
 H  (r  1) H   m H
0
B  0 ( H  M )
 0 (1   m ) H
 0  r H
 H
D   0 E   0 E
  0 (1   ) E
  0 r E
 E
Magnetic Field Intensity H
Relationship between B and H
B = 0 ( H + M )
M = m H

B = 0 ( H + m H )  0 (1   m ) H  0  r H  H
Magnetic susceptibility ：
m
Relative permeability ：
r = 1 + m
Permeability ：
 = 0 r
Permeability of vacuum ：
0
Example 1（p.288/[Ex.]）
Toroidal solenoid，r << R
r
Known：I0, n, , V Find：H, B, L
Sol.：inside：take a loop as shown
LH  dl  Hin  2R  nI 0  2R
 H in  nI 0 (tangent )
R
Bin  H in  nI 0 (tangent )
（ outside：Hout = 0 , Bout = 0 ）
S = BinS = nI0S ，S= NS= n2R·nI0S = n2V I0
 L = S/I0 = n2V （vacuum： L0 = 0n2V ）
= rL0
Example 2（p.315/ 7 - 1 - 1）（1）
Uniformly magnetized sphere
R
Known：R, M（along z axis）
Find：  ’ （on surface），pm
Sol.： '  ( M  M )  nˆ
2

z
1
 M  rˆ
 M sin  ˆ
（direction：right hand rule）
dpm  dI ' Szˆ   ' dl  Sẑ
  'Rd   ( R sin  ) 2 ẑ
 MR  sin dẑ
3
M o
3
Material 2，vacuum 1
n=r
(21)
M2 = M, M1 = 0
Example 2（p.315/ 7 - 1 - 1）（2）
dpm  MR 3 sin 3 dzˆ

pm   dpm   MR 3 sin 3 d

0
 MR    (1  cos 2  )dcos
0

1
3
3
 MR  ( cos  cos  )
3
0
4
 MR 3  MV
3
（ direction：along z ）
Consider：a vacuum sphere in an infinite magnetic
material （find  ’ on surface etc.）
3
Exercises
p.315 / 7 - 1 -
2, 3, 4
§2. Paramagnetism and Diamagnetism
1. Paramagnetism
2. Diamagnetism
1. Paramagnetism
Molecules, atoms  electrons orbital/spin magnetic
dipole moment
not cancel：permanent magnetic moment pm
cancel out： pm is 0 ( of a molecule )
pm  paramagnetism
external field B = 0 ， thermal motion  pm
orient randomly，cancel out  M = 0
 B  0，pm tend to line up with B  M  0
paramagnetic material： pm  0
include：aluminium, natrium, oxygen etc.
2. Diamagnetism（1）
Exhibited by all materials, very weak, opposite to B
 material with pm  0，exhibit paramagnetism
 material with pm = 0，exhibit diamagnetism
electron moving in a circular path，B introduced，r
no change,  increased  pm changed, opposite to B
 diamagnetism
0
e
e 0
v
I

T0
2
-e
2
er
2
pm0
pm 0  Ir  
0
2
2. Diamagnetism（2）
Assume B and  0 in the same direction
B
L EI  dl  S t  dS
B0
2
EI  2r  
 r
t
r B
er B
 EI   
F  eEI  
2 t
2 t
erB
eB
v 
 
Ft  mv
2m
2m
Lorentz force： FL = e ( v0 + v ) B
2
mv
Centripetal force：F  0
0
r
m(v0  v) 2
F
r
B
F
-e
EI
2. Diamagnetism（3）
m
F  F0  (2v0 v  v 2 )
r
2m

(v0  v)v
r
2m
erB

(v0  v)
r
2m
 (v0  v)eB  FL
eB
er 2
 
pm 0  
0
2m
2
 r unchanged，  > 0
B 
-e
pm
2
er
pm  
  0
2
  pm opposite to B  diamagnetism, very weak
§3. Ferromagnetism
1. Magnetizing Properties of Ferromagnetic Material
2. Classification and Application
3. Magnetic Domains
B
non-ferromagnetism：
M = gB = mH
B = H
ferromagnetism：？
—— experiment
0
H
Experiment
Experiment：
A
test B
1
2
R
Adjust R to control I
Electromagnetic
obtain H = nI
induction
SKIP EXPERIMENT
1. Magnetizing Properties（1）
open switch，H = 0，B = 0，O
B
S
 R maximum，switch to 1，A
R
C
 R，H，B ，AC
A
 continue，B changing slow down
o
H
 beyond S，H，B almost not change
（saturation magnetic intensity HS）
 H  to 0，B- H curve not retraced
 H = 0，B  0（residual field）magnetic hysteresis
 switch to 2，I reversed，H：0  -HS
 H = -HD ，B = 0，HD rectification force
 H：-HS  HS ，closed，hysteresis loop

1. Magnetizing Properties（2）
Saturation magnetic intensity HS
 Residual field BR
 Magnetic hysteresis
 Rectification force HD
D
 Hysteresis loop
（Symmetry about O）
 Small hysteresis loop
 H corresponding to many B S’
Magnetization curve
H and B ： 1 to 1

B
S
R
o
D’
R’
H
1. Magnetizing Properties（3）
Magnetization curve
B,
H and B ：1 to 1
 Permeability of a
ferromagnetic material：
B

H
O
 not a constant
 very large  r ~ 10 4

B

H
2. Classification and Application
Three characteristics：
 high  ：strong field by weak current，
motors, transformers
B
Soft
 Non-linearity：non-linear elements
 Magnetic hysteresis：
permanent magnets
o
 Classification：
Hard
 Soft magnetism
 Hard magnetism

H
3. Magnetic Domains
（Quantum theory）
 Magnetic domains：magnetized regions
 Effect of external magnetic field：
 growth in size of the domains oriented along B
 shift of the orientation of dipoles in a domain
Magnetizing：H，M，BT> T paramagnetism
C
B = 0 (H + M )
T < TC ferromagnetism
saturation： M stop increasing
TC ：critical temperature
 Not reversible：as H removed
Curie Point
 Temperature：thermal motion
H = 0， M = 0
Exercises
p.316 / 7 - 3 -
1
§4. Magnetic Field Energy
Energy density wm at every point in the magnetic field
Consider a solenoid
inside： H = nI，B = nI （uniform field）
outside：H = 0，B = 0
Self-inductance：L    NBS  nl  nI  S  n 2V
I
I
I
1 2 1 2 2 1
Magnetic energy：Wm  LI  n VI  HBV
2
2
2
Wm 1
1
 HB  H  B
Energy density： wm 
V
2
2
1
Non-uniform： wm( x, y, z ) Wm  VH  BdV
2
Example
A long coaxial cable made of cylinders of radius r1< r2
and material of permeability  carries a current I .
Find magnetic energy and self-inductance of a length l.
Sol.：H = I / 2r ( r1< r <r2 )，
H=0
( r < r1, r > r2 )
2
1

I
wm  H 2  2 2
2
8 r
Wm   wmdV  
V
I 2l r2

ln
4
r1
r2
r1
I 2l r dr
I 2
 l  2rdr 
2 2
4 r r
8 r
2Wm l r2
ln
L 2 
2 r1
I
2
1
Exercises
p.316 / 7 - 6 -
1, 2, 3