Download Chapter 4: Magnetostatics

UNIVERSITI MALAYSIA PERLIS
EKT 241/4:
ELECTROMAGNETIC
THEORY
CHAPTER 4 – MAGNETOSTATICS
PREPARED BY: NORDIANA MOHAMAD SAAID
[email protected]
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 for magnetostatics:
 B  0
 H  J
Where;
J = current density
H = magnetic field intensity
B = magnetic flux density
•Relationship between B and H:
B  H
unit: Tesla or Weber/m2
Where: μ = magnetic permeability
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  qE  qu  B  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:
Fm  I  dl  B
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.
Solution to Example 1
• a) Using Fm  I  B,
  xˆ 2r
B  yˆ B0
F1  xˆ 2Ir   yˆ B0  zˆ 2IrB0 N 
• b) the product of dl  B is in the - zˆ direction :


 0
 0
F2  I  dl  B  zˆ I  rB0 sin d  zˆ 2 IrB0 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.
The Biot–Savart’s Law
Biot–Savart’s law states that:
ˆ
1 dI  R
dH 
4 R 2
A/m 
where:
dH = differential magnetic field
dI = differential current element
The Biot–Savart’s Law
• To determine the total H:
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.
Solution o Example 2
• For segment OA and OC, the magnetic field at O
is zero since dl is parallel and anti-parallel to R̂ .
• For segment AC, dl is in φ direction,
ˆ  zˆ dl  zˆ ad
dl   R

• Using Biot- Savart’s law:
1
H
4
zˆ ad
1
 a 2  zˆ 4a 
where  is in radians
Magnetic Force between Two
Parallel Conductors
• Force per unit length on
parallel current-carrying
conductors is:
 0 I1 I 2
F'1  yˆ
2d
where F’1 = -F’2 (attract each
other with equal force)
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.
Ampere’s law for magnetism
• Ampere’s law states that:
 H  dl  I Ampere' s
law 
C
• The directional path of current C follows the
right-hand rule.
Magnetic Field of an infinite
length of conductor
• Consider a conductor lying on
the z axis, carrying current I in
+az direction.
• Using Ampere’s law:
 H  dl  I
Ampere' s
law 
C
• The path to evaluate is along the
aφ direction, hence use dLφ.
az
I
a
Magnetic Field of an infinite
length of conductor
Using Ampere’s law:
 H  dL  Ienc
Where;
H  H  a
dL   rda 
Thus,
 H  dL  I
2
enc

 H  a  rda  I
0
Magnetic Field of an infinite
length of conductor
Integrating and then re-arrange the equation in
terms of Hφ:
I
H 
2r
Hence, the magnetic field vector, H:
H
I
2r
a
Note: this equation is true for an infinite length of
conductor
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
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
The relation between B and H is:
B  0H  0
I
2r
a
To find magnetic flux crossing the region, we use:
   B  dS
where dS is in the aφ direction.
unit: Weber
Solution to Example 4
So, dS   drdza 
Therefore,
   B  dS
0 I
  
a   drdza 
2r
z  0 r  0.01
2 0 I 0.05

ln
 1.61  10 6 Wb
2
0.01
2
0.05
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
• Magnetic behavior of a material is due to the
interaction of magnetic dipole moments of its
atoms with an external magnetic field.
• This behavior is used as a basis for classifying
magnetic materials.
• 3 types of magnetic materials: diamagnetic,
paramagnetic, and ferromagnetic.
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:
  0 1   m  H/m 
where  0  4  10 7 H m
and relative permeability is defined as

r   1   m
0
Magnetic Materials
• Diamagnetic materials have negative
susceptibilities.
• Paramagnetic materials have positive
susceptibilities.
• However, the absolute susceptibilities value of
both materials is in the order 10-5. Thus,  m can
be ignored. Hence, we have  r  1 or    0
• Diamagnetic and paramagnetic materials
include dielectric materials and most metals.
Magnetic Hysteresis of
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.
• Hard magnetic material- cannot be easily
demagnetized by an external magnetic field.
• Soft magnetic material – easily magnetized &
demagnetized.
Magnetic Hysteresis of
Ferromagnetic Materials
• Properties of magnetic materials as follows:
Magnetic Hysteresis of
Ferromagnetic Materials
• Comparison of hysteresis curves for (a) a hard
and (b) a soft ferromagnetic material is shown.
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
• i.e the normal component of B is continuous
across the boundary between two adjacent
media
Magnetic boundary conditions
• Since B  H ,
• For linear, isotropic media, the first boundary
condition which is related to H;
1H1n  2 H 2 n
• Reversal concept: whereas the normal
component of B is continuous across the
boundary, the normal component of D (electric
flux density) may not be continuous (unless
ρs=0)
Magnetic boundary conditions
• A similar reversal concept applies to tangential
components of the electric field E and magnetic
field H.
• Reversal concept related to tangential
components:
– Whereas the tangential component of E is continuous
across the boundary, the tangential component of H
may not be continuous (unless Js=0).
• By applying Ampere’s law and using the same
method of derivation as for electric field E:
H 2t  H 1t  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
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
• 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
zˆ NI
ˆ
B  znI 
l
Where, N=nl
=total number of turns
over the length 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 
   zˆ   I   zˆ ds
S
 l 
N
  IS
l
where S  cross - sectional area of the loop
Self Inductance
• Magnetic flux linkage, Λ is the total magnetic
flux linking a given conducting structure.
 N2

  N   
IS  (Wb)
 l

• Self-inductance of any conducting structure is
the ratio of the magnetic flux linkage, Λ to the
current I flowing through the structure.

L
I
H 
Self Inductance
• For a solenoid:
N2
L
S solenoid 
l
• For two conductor
configuration:
  1
L     B  ds
I
I I S
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 

B1  ds H 

I1
I 1 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:
l
1 2
Wm   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/m 
3
• i.e magnetic energy per unit volume