UNIVERSITI MALAYSIA PERLIS
EKT 241/4:
ELECTROMAGNETIC
THEORY
CHAPTER 4 – MAGNETOSTATICS
PREPARED BY: NORDIANA MOHAMAD SAAID
dianams@unimap.edu.my
Chapter Outline
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Maxwell’s Equations
Magnetic Forces and Torques
The total electromagnetic force, known as Lorentz force
Biot- Savart’s law
Gauss’s law for magnetism
Ampere’s law for magnetism
Magnetic Field and Flux
Vector magnetic potential
Properties of 3 different types of material
Boundary conditions between two different media
Self inductance and mutual inductance
Magnetic energy
Maxwell’s equations
Maxwell’s equations:
  D  v
B
E  
t
B  0
D
H  H 
t
Where;
E = electric field intensity
D = electric flux density
ρv = electric charge density
per unit volume
H = magnetic field intensity
B = magnetic flux density
Maxwell’s equations
• For static case, ∂/∂t = 0.
• Maxwell’s equations is reduced to:
Electrostatics
Magnetostatics
  D  v
 B  0
 E  0
 H  J
Magnetic Force
B = Magnetic Flux Density
B
B
q
q
q
I
B
B
Fm  qu  B
N
Magnetic Torque on a CurrentCarrying Loop
• Applied force vector F and distance vector d are
used to generate a torque T
T = d× F (N·m)
• Rotation direction is governed by right-hand rule.
Magnetic Forces and Torques
• The electric force Fe per unit charge acting on a
test charge placed at a point in space with
electric field E.
• When a charged particle moving with a velocity
u passing through that point in space, the
magnetic force Fm is exerted on that charged
particle.
Fm  qu  B
N
where B = magnetic flux density (Cm/s or Tesla T)
Magnetic Forces and Torques
• If a charged particle is in the presence of both an
electric field E and magnetic field B, the total
electromagnetic force acting on it is:
F  Fe  Fm
Fe  qE;
Fm  qu  B
 F  Fe  Fm  qE  qu  B
 F  qE  u  B
(Lorentz force)
Magnetic Force on a CurrentCarrying Conductor
• For closed circuit of contour C carrying I , total
magnetic force Fm is:
I dl  B
Fm  qu  FBm N
N 
C
• In a uniform magnetic field, Fm is zero for a
closed circuit.
Magnetic Force on a CurrentCarrying Conductor
• On a line segment, Fm is proportional to the
vector between the end points.
Fm  I  B
Example 1
The semicircular conductor shown carries a current I.
The closed circuit is exposed to a uniform magnetic
field B  ŷB0 . Determine (a) the magnetic force F1 on the
straight section of the wire and (b) the force F2 on the curved
section.
Fm  I  dl  B
C
Fm  I  B
N 
Solution to Example 1
• a) Using Fm  I  B,
  xˆ 2r
B  yˆ B0
F1  xˆ 2Ir   yˆ B0  zˆ 2IrB0 N 
b) F2  ?
the productof dl  B is in the - ẑ direction:

B  yˆBo
F2  I dl  B

 0
 dl  (r )(sin )
0

 I r sin d  B0   ẑIrB0  cos0
 0
  ẑIrB0 ( 1  1)


 F2  ẑ 2 IrB0 N 
The Biot–Savart’s Law
The Biot–Savart law is used to compute the
magnetic field generated by a steady
current, i.e. a continual flow of charges, for
example through a wire
Biot–Savart’s law states that:
1 dl  R̂
dH 
4 R 2
 A/m 
where:
dH = differential magnetic field
dl = differential length
The Biot–Savart’s Law
• To determine the total H:
1 dl  R̂
dH 
4 R 2
 A/m 
1 dl  Rˆ
H
4 l R 2
A/m 
The Biot–Savart’s Law
• Biot–Savart’s law may be expressed in terms of
distributed current sources.
1
H 
4

1
H 
4

S
v
J s  Rˆ
ds
2
R
J  Rˆ
dv
2
R
for a surface current 
for a volume
current 
Example 2
Determine the magnetic field at the apex O of the
pie-shaped loop as shown. Ignore the
contributions to the field due to the current in the
small arcs near O.
O  A = dl
0
C  O = -dl
A C
?
1 dl  Rˆ
H
4 l R 2
• For segment AC, dl is in φ direction,
dl  R̂  ẑad
Ra
• Using Biot- Savart’s law: H  1
ẑad
4

a2
1
 H  ẑ

4a

1
4

A/m 
ẑad
a2
where is in radians
QUIZ!!
z
ε1=2ε0
ε2=8ε0
Ө2
E2
xy
E1
Ө1
Find E1 if E2 = 2x -3y +3z with s = 3.54 x 10-11(C/m2)
And find Ө1 and Ө2
Gauss’s Law for Magnetism
• Gauss’s law for magnetism states that:
  B  0 (different ial form)   B  ds  0 (integral form)
S
• Magnetic field lines always form continuous
closed loops.
  D  s
B  0
Ampere’s law for magnetism
• Ampere’s law states that:  H  dl  I
 Ampere' s
C
• true for an infinite length of conductor
C, +aø
H
dl
true for an infinite
length of conductor
I, +az
r
law
Magnetic Field of an infinite
length of conductor
From
terms of Hφ:
then re-arrange the equation in
H 
I
2r
Hence, the magnetic field vector, H:
I
H
a
2r
Example 3
• A toroidal coil with N turns carrying a current I ,
determine the magnetic field H in each of the
following three regions: r < a, a < r < b,and r > b,
all in the azimuthal plane of the toroid.
Solution to Example 3
• H = 0 for r < a as no current is flowing through
the surface of the contour
• H = 0 for r > b, as equal number of current coils
cross the surface in both directions.
• For a < r < b, we apply Ampere’s law:
2
 H  dl    φˆ H  φˆ rd  2rH   NI
C
0
NI
• Hence, H = NI/(2πr) . H  φˆ H  φˆ
2r
for a  r  b 
Magnetic Flux
• The amount of magnetic flux, φ in Webers from
magnetic field passing through a surface is
found in a manner analogous to finding electric
flux:
   B  dS
B  0 H
I
H
a
2r
I
 B  0
a
2r
Example 4
An infinite length coaxial cable with inner
conductor radius of 0.01m and outer conductor
radius of 0.05m carrying a current of 2.5A exists
along the z axis in the + az direction.
Find the flux passing through the region between
two conductors with height of 2 m in free space.
Solution to Example 4
Iaz=2.5A
z
1)
inner conductor radius = r1 0.01m
2)
outer conductor radius = r2 0.05m
3)
current of 2.5A (in the +az direction)
4)
Flux radius = 2m
   B  dS
I
B  0 H  0
a
2r
I


    0
a   dS
 2r 

aø
Flux,z
xy
r2
r1
Solution to Example 4
where dS is in the aø direction.
So, dS   drdza 
Therefore,

  B  dS
2


0.05

z  0 r  0.01
0 I
a  drdza
2r
2 0 I 0.05

ln
 1.61  10  6 Wb
2
0.01
Vector Magnetic Potential
• For any vector of vector magnetic potential A:
    A  0
• We are able to derive: B    A
Wb/m. 
2
• Vector Poisson’s equation is given as:
 2 A  J
where
 J
A
dv'

4 v ' R'
Wb/m 
Magnetic Properties of Materials
The behavior of magnetic dipole moments & magnetic susceptibility,
(  m ) of its atoms with an external magnetic field is used as a basis for
classifying magnetic materials.
Magnetic Properties of Materials
• Magnetization in a material is associated with
atomic current loops generated by two principal
mechanisms:
– Orbital motions of the electrons around the nucleus,
i.e orbital magnetic moment, mo
– Electron spin about its own axis, i.e spin magnetic
moment, ms
Magnetic Permeability
• Magnetization vector M is defined as
M  mH
where  m= magnetic susceptibility (dimensionless)
• Magnetic permeability is defined as:
B  H
  0 1   m  H/m 
where  0  4  10 7 H m
and to define the magnetic properties in term of
relative permeability is defined as:

r 
0
 1  m
Magnetic Materials Diamagnetic
metals have a very weak and negative
susceptibility (  m ) to magnetic field
slightly repelled by a magnetic field and the
material does not retain the magnetic properties
when the external field is removed
Most elements in the periodic table, including
copper, silver, and gold, are diamagnetic.
Magnetic Materials Paramagnetic
 Paramagnetic materials have a small and (  m )
positive susceptibilities to magnetic fields.
 slightly attracted by a magnetic field and the
material does not retain the magnetic properties
when the external field is removed.
 Paramagnetic materials include magnesium,
molybdenum, lithium, and tantalum.
Magnetic Materials –
Diamagnetic, Paramagnetic
• However, the absolute susceptibilities value of
both materials is in the order 10-5. Thus,  m can
be ignored. Hence, we have Magnetic
permeability:  r  1 or    0
• Diamagnetic and paramagnetic materials
include dielectric materials and most metals.
Magnetic Materials –
Ferromagnetic Materials
• Ferromagnetic materials is characterized by
magnetized domain - a microscopic region
within which the magnetic moments of all its
atoms are aligned parallel to each other.
• Hysteresis – “to lag behind”. It determines how
easy/hard for a magnetic material to be
magnetized and demagnetized.
Process of Magnetic Hysteresis
material is magnetized
and can serve as
permanent magnet!
B
material is demagnetize
Magnetic Hysteresis of
Ferromagnetic Materials
• Comparison of hysteresis curves for (a) a hard and (b) a
soft ferromagnetic material is shown.
Hard magnetic material- cannot be
easily magnetized & demagnetized by an
external magnetic field.
Soft magnetic material – easily
magnetized & demagnetized.
Magnetic Hysteresis of
Ferromagnetic Materials
• Properties of magnetic materials as follows:
Magnetic boundary conditions
• Boundary between medium 1 with μ1 and
medium 2 with μ2
Magnetic boundary conditions
• Boundary condition related to normal
components of the electric field;
 D  ds  Q
S
 D1n  D2n   S
• By analogy, application of Gauss’s law for
magnetism, we get first boundary condition:
 B  ds  0
S
 B1n  B2n
• Since B  H ,
• For linear, isotropic media, the first boundary
condition which is related to H; 1H1n  2 H 2n
z
xy
By applying
Ampere’s law
 H  dl  I
C
H 2t  H1t  J s
Magnetic boundary conditions
• The result is generalized to a vector form:
nˆ 2  H1  H 2   J s
• Where nˆ 2 is the normal vector pointing away from medium 2
• However, surface currents can exist only on the
surfaces of perfect conductors and perfect
superconductors (infinite conductivities).
• Hence, at the interface between media with finite
conductivities, Js=0. Thus:
H1t  H 2t
Example
xy (plane)
H1t  H 2t
  (  r )(0 )
• Solution:
1) H1t = H2t thus, H2t = 6ax + 2ay
2) Hn1 = 3az,
but, Hn2 = ??
  (  r )(0 )
μr1 = 6000 ; μr2 = 3000 ;
6000μ0(3az) = 3000 μ0(Hn2)
Hn2 = 6az
thus, H2 =6ax + 2ay + 6az
Inductance
• An inductor is the magnetic analogue of an
electrical capacitor.
• Capacitor can store electric energy in the electric
field present in the medium between its
conducting surfaces.
• Inductor can store magnetic energy in the
volume comprising the inductors.
INDUCTANCE
store magnetic
energy

L
I
12
L12 
(H )
I1
H 
Wm 1
wm 
 H 2
v
2
J/m 
3
Inductance
• Example of an inductor is a solenoid - a coil
consisting of multiple turns of wire wound in a
helical geometry around a cylindrical core.
Magnetic Field in a Solenoid
• For one cross section of
solenoid,
B  zˆ
nI
2
sin  2  sin 1 
• When l >a, θ1≈−90° and θ2≈90°,
For long solenoid with l / a  1
ẑNI
B  ẑnI 
l
Where, N=nl
=total number of turns
over the length l
Self Inductance
The self-inductance of a
circuit is used to describe
the reaction of the circuit
to a changing current in
the circuit,
(The ratio of the magnetic
flux to the current)
Self Inductance
• Self-inductance of any conducting structure is
the ratio of the magnetic flux linkage, Λ to the
current I flowing through the structure.

H 
L
I
• Magnetic flux linkage, Λ is the total magnetic
flux linking a given conducting structure.
 N2


  N   
IS  (Wb)
 l

Self Inductance
• Magnetic flux,  linking a surface S is given by:
   B  ds Wb 
S
• In a solenoid with uniform magnetic field, the flux
linking a single loop is:
 N 
  ẑ  I   ẑds
S  l 
N
  IS
l
where S  cross - sectional area of the loop

Self Inductance – magnetic flux in
solenoid
• Magnetic flux,    B  ds Wb 
S
ẑNI
B
l
 ds  S
S
 N
ẑ 
S  l
N
  IS
l



I   ẑds

Self Inductance
• Magnetic flux,  linking a surface S is given by:
   B  ds Wb 
S
• In a solenoid with uniform magnetic field, the flux
linking a single loop is:
 N 
  ẑ  I   ẑds
S  l 
N
  IS
l
where S  cross - sectional area of the loop

Self Inductance
• For a solenoid:
N2
L
S solenoid
l

• For two conductor
configuration:
  1
L     B  ds
I
I I S
Self Inductance for a solenoid

L
I
H 
  N 
  N  N    IS 

  l
Thus,

L
I
 N2 
  
IS  (Wb)
 l



 N2 

IS 
 l

 H 
 
I
N2
L  
S
l
Mutual Inductance
• Mutual inductance – produced by magnetic
coupling between two different conducting
structures.
Mutual Inductance
• Magnetic field B1 generated by current I1 results in a flux
Φ12 through loop 2:
12   B1  dS
S2
• If loop 2 consists of N2 turns all coupled by B1 in exactly
the same way, the total magnetic flux linkage through
loop 2 due to B1 is:
12  N 2 12  N 2  B1  dS
S2
Mutual Inductance
• Hence, the mutual inductance:
12 N 2
L12 

I1
I1
 B1  ds H 
s2
12  N 2 12  N 2  B1  dS
S2
12   B1  dS
S2
Magnetic Energy
• Consider an inductor with an inductance L
connected to a current source.
• The current I flowing through the inductor is
increased from zero to a final value I.
• The energy expended in building up the current
in the inductor:
Wm 

l
1 2
pdt  ivdt  L idi  LI
2


0
• i.e the magnetic energy stored in the inductor
Magnetic Energy
• Magnetic energy density (for solenoid):
Wm 1
wm 
 H 2
v
2
J/m3 
• i.e magnetic energy per unit volume
• Magnetic energy in magnetic field:
1
Wm 
H  Bdv
2

v
J/m3 
Example: Magnetic Energy in a
Coaxial Cable
• Derive an expression for the magnetic energy
stored in a coaxial cable of length l and inner
and outer a and b. The insulation material has
magnetic permeability μ.