ENE 325
Electromagnetic Fields
and Waves
Lecture 8 Scalar and Vector Magnetic
Potentials, Magnetic Force, Torque, Magnetic
Material, and Permeability
1
Review (1)

Ampere’s circuital law - the integration of around any closed path is
equal to the net current enclosed by that path.
 H d L  Ienc


‘Curl’ is employed to find the point form of Ampère’s circuital law.
Curl of H or  H is the maximum circulation of H per unit area as the
area shrinks to zero
H dL

  H  lim
J
S 0
S
2
Review (2)
Magnetic flux density B is related to the magnetic field
intensity H in the free space by

B  0 H
Weber/m2 or Tesla (T)
where 0 is the free space permeability, given in units of
henrys per meter, or
0 = 410-7 H/m.
 Magnetic flux  (units of Webers) passing through a
surface is found by    B d S
3
Outline
Scalar and vector magnetic potentials
 Magnetic force and torque
 Magnetic material and permeability

4
The scalar and vector magnetic
potentials (1)

Scalar magnetic potential (Vm) E  V is the simple
practical concept to determine the electric field.
Similarly, the scalar magnetic potential, Vm, is defined
to relate to the magnetic field H but there is no
physical interpretation.
Assume
H  Vm
 H  J   (Vm )  0
To make the above statement true, J = 0.
5
The scalar and vector magnetic
potentials (2)
From
 B  0 H  0
0 (Vm )  0
2Vm  0
Laplace’s equation
This equation’s solution to determine
the potential field requires that the potential
on the boundaries is known.
6
The scalar and vector magnetic
potentials (3)
 The difference between V (electric potential) and Vm
(scalar magnetic potential) is that the electric potential is a
function of the positions while there can be many Vm values
for the same position.
 H d L  Ienc
7
The scalar and vector magnetic
potentials (4)
While for the electrostatic case
 E  0
 E dL  0
a
Vab    E d L
does not depend on path.
b
8
The scalar and vector magnetic
potentials (5)
 Vector magnetic potential (A) is useful to find a
magnetic filed for antenna and waveguide.
From
Let assume
 B 0
B  (  A)
so
 (  A)  0
and
H
 H 
1
0
1
0
 A
 A  J  0
9
The scalar and vector magnetic
potentials (6)
 It is simpler to use the vector magnetic potential to determine
the magnetic field. By transforming from Bio-savart law,
we can write
0 Id L
A 
.
4 R
The differential form
0 Id L
dA
.
4 R
10
Ex1 Determine the magnetic field from the
infinite length line of current using the vector
magnetic potential
d L  dza z
Find A at point P(r, , z)
z
P
dA
2
2
r
z
z
y

x
r
0 Idza z
4 r 2  z 2
1
1  dAz
 d A 
then d H 

0
0  r
I r dz
dH 
a
3/ 2 
4 r 2  z 2


 a


11
Vector magnetic potential for other
current distributions

For current sheet
0 KdS
A  S
4 R

For current volume
A   vol
0 Jdv
4 R
12
Magnetic force

Force on a moving charge
F  qv  B

N
Force on a differential current element
d F  dQv  B
13
Hall effect

Hall effect is the voltage exerted from the separation of electrons
and positive ions influenced by the magnetic force in the
conductor. This Hall voltage is perpendicular to both magnetic
field and the charge velocity.
-
+
B
-
+
-
+
-
+
Fq
-
 qv(a x )  Ba z
+
+
F q  qv  B
+
+
+
 qvBa y
N
+
I
14
Magnetic force on the current
carrying conductor (1)

For the current carrying conductor, consider the magnetic force
on the whole conductor not on the charges.
From
J  rv v
and
dQ = rvdv
then
d F  rv dv(v  B)
d F  J  Bdv.
15
Magnetic force on the current
carrying conductor (2)
From
Jdv  KdS  Id L
we can write
d F  Id L  B
then
F   Id L  B.
 For a straight conductor in a uniform magnetic field
(still maintains the closed circuit),
F  I L B
or F = ILBsin
 Force between differential current elements determine
the force on the conductor influenced by the other nearby. 16
Ex2 Determine the force action on circuit
2 by circuit 1.
z
I1
I2
F21
F12
y
x
d
17
Force and torque on a closed circuit
(1)
dT  R  d F
Nm
z
where T = torque (Nm)
R = distance from the origin (m)
T
O
y
R
d F = Force (N)
F
x
If the current is uniform, T  R  F.
18
Force and torque on a closed circuit
(2)
 For a current loop, we can express torque as
dT  Id S  B.
 If B is constant or uniform, we can express torque as
T  I S  B.
 Define magnetic dipole moment Id S  dm
where m = magnetic dipole moment (Am2).
Therefore, torque can be shown as T  m  B.
19
Ex3 To illustrate some force and torque calculations,
consider the rectangular loop shown. Calculate the
total force and torque contribution on each side. Let
the current I flow in the loop lied in the uniform
magnetic field B  Bx a x  By a y  Bz a z tesla.
y
1
4
R1
I
b2
x
2
z
3
b1
20
21
Ex4 A 2.5 m length conductor is located at z =
0, x =4m and has a uniform current of 12 A in
the direction a y . DetermineB in this area if
the force acting on the conductor is 1.210-2 N
in the direction(a x  a z ) / 2 .
22
Ex5 The rectangular conductor loop is
located in the field that has B  0.05 a x  a y T.
2
Determine the torque around z axis if the
current in the loop is 0.5 A.
z
0.08m
I=0.5A
y
-0.04m
x
23
The nature of magnetic materials


Combine our knowledge of the action of a magnetic field on a
current loop with a simple model of an atom and obtain some
appreciation of the difference in behavior of various types of
materials in magnetic fields.
The magnetic properties of the materials depend on ‘magnetic
moment’. Three types of magnetic moment are
1. The circular orbiting of electrons around the positive nucleus
results in the current and then the magnetic field m = IdS.
2. Electron spinning around its own axis and thus generates a
magnetic dipole moment.
3. Nuclear spin, this factor provides a negligible effect on the
overall magnetic properties of materials.
24
Types of magnetic material (1)



diamagnetic The small magnetic filed produced by the
motion of the electrons in their orbits and those
produced by the electron spin combine to produce a
net field of zero or we can say the permanent magnetic
moment m0 = 0.
The external field would produce an internal magnetic
field.
Some examples of materials that has diamagnetic effect
are Metallic bismuth, hydrogen, helium, the other ‘inert’
gases, sodium chloride, copper, gold silicon,
germanium, graphite, and sulfur.
25
Types of magnetic material (2)

paramagnetic The net magnetic moment of each atom is
not zero but the average over the volume is, due to
random orientation of the atoms. The material shows
no magnetic effects in the absence of the external field.

Whenever there is an external field and the alignment
of magnetic moments acts to increase the value of B ,
the material is called ‘paramagnetic’ but if it acts to
decrease the value of B , it is still called diamagnetic.

For example, Potassium, Oxygen, Tungsten, and some
rare earth elements.
26
Types of magnetic material (3)

Ferromagnetic each atom has a relatively large dipole moment due
to uncompensated electron spin moments. These moments are
forced to line up in parallel fashion over region containing a large
number of atoms, these regions are called ‘domains’. The domain
moments vary in direction from domain to domain. The overall
effect is therefore one of cancellation, and the material as a
whole has no magnetic moment.

When the external field is applied, those domains which are in
the direction of the applied field increase their size at the
expense of their neighbors, and the internal field increases greatly
over that of the external field alone. When the external field is
removed, a completely random domain alignment is not usually
attained, and a residual dipole field remains in the macroscopic
structure.
27
Types of magnetic material (4)

The magnetic state of material is a function of its magnetic
history or ‘hysteresis’. For example, Iron, Nickel, and Cobalt.

Antiferromagnetic The forces between adjacent atoms cause the
atomic moments to line up in anti parallel fashion. The net
magnetic moment is zero. The antiferromagnetic materials are
affected slightly by the presence of and external magnetic field.

For example, nickel oxide (NiO), ferrous sulfide (FeS), and
cobalt chloride (CoCl2).
28
Types of magnetic material (5)

Ferrimagnetic Substances show an antiparallel alignment of
adjacent atomic moments, but the moments are not equal. A
large response to an external magnetic field therefore occurs.

For example, the ferrites, the iron oxide magnetite (Fe3O4), a
nickel-zinc ferrite, and a nickel ferrite.
29
Types of magnetic material (6)

Superparamagnetic materials are composed of an assembly of
ferromagnetic particles in a nonferromagnetic matrix. The
domain walls cannot penetrate the intervening matrix material to
the adjacent particles.

For example, the magnetic tape.
30
Magnetization and permeability


how do the magnetic dipoles act as a distributed source for the
magnetic field? The result will look like Ampere’s circuital law,
the current will be the movement of bound charges, and the field,
which has the dimension of H will be called the magnetization M .
From
m  Ib d S
the total magnetic dipole
nv
mtotal   mi
i 1

where n = number of dipole moment per unit volume.
Define the magnetization
1 nv
M  lim
 mi
A/m
v0 v i 1
31
Bounded current

Consider the alignment of the dipole when there is an external
magnetic field,
m=IdS

m
m
m
dL
aL
in a volume of d S d L , there will be n dipoles aligning. Each
dipole has the current I, therefore there will be the increase in the
total current in that small volume which can be shown as
dIb  nIb d S d L  M d L
or
Ib   M d L.
32
The relationship between B and H in
mediums other than free space (1)
From Ampère’s circuital law,

B
0
d L  I total
It  Ib  I f
where If is free current, then
I f  It  Ib   (
B
0
 M ) d L.
33
The relationship between B and H in
mediums other than free space (2)
B
Define
H
or
B  0 (M  H )
therefore, we can write
0
M
I f   H d L.
Note: this H is the magnetic field in any medium.
34
The relationship between B and H in
mediums other than free space (3)
For a linear, isotropic media
M  m H
where m is the magnetic susceptibility,
therefore
B  0 ( H  m H )
 0 H (1  m )
 0 r H
which shows that
r  1   m .
35
The relationship between B and H in
mediums other than free space (4)
Therefore we can write
B  H
where
 = r0.
If r  1, diamagnetic material
If r  1, paramagnetic material
If r 1, ferromagnetic material
 The magnetic susceptibility of some materials
hydrogen = -210-5
 copper = -0.910-5
 germanium = -0.810-5
 silicon = -0.310-5
36