Transcript Fundamentals of Applied Electromagnetics

Anayet Karim
Chapter Objectives
 Magnetic force
 The total electromagnetic force, known as Lorentz
force
 Biot–Savart law
 Gauss’s law for magnetism
 Ampere’s law
 Vector magnetic potential
 3 different types of material
 Boundary between two different media
 Magnetic energy density
Chapter Outline
4-1)
4-2)
4-3)
4-4)
4-5)
4-6)
4-7)
4-8)
4-9)
Magnetic Forces and Torques
The Biot–Savart Law
Magnetic Force between Two Parallel Conductors
Maxwell’s Magnetostatic Equations
Vector Magnetic Potential
Magnetic Properties of Materials
Magnetic Boundary Conditions
Inductance
Magnetic Energy
Today We Will Focus the Following Topics
Magnetic Forces
Apart between Fe & Fm
Magnetic Torques
Few Examples
4-1 Magnetic Forces and Torques
•
When charged particle moving with a velocity u,
magnetic force Fm is produced.
Fm  qu  B N
where B = magnetic flux density (C.m/s or Tesla T)
•
When charged particle has E (electric field) and B
(Magnetic field), total electromagnetic force is
F  Fe  Fm  qE  qu  B  qE  u  B (Lorentzforce)
Apart Fe & Fm
•
•
•
•
•
•
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Fe always in the direction of electric field
Fm always perpendicular to magnetic field
Fe acts on a charged particles, which is static
Fm acts on a charged particles, which is moving
or in motion
E & D (Electric field & Density)
H & B (magnetic field & Density)
Fe=qE for “Force on charge q)
Fm = quXB “Force on charge q)
Stationary charges (Electrostatics) & Steady
currents (Magnetostatics)
4-1.1 Magnetic Force on a Current-Carrying Conductor
•
For closed circuit of contour C carrying I , total
magnetic force Fm is
Fm  I  dl  B
N 
C
•
Fm is zero for a closed circuit. On a line segment,
it is proportional to the vector between the end
point.
4-1.1 Magnetic Force on a Current-Carrying Conductor
•
The total magnetic force Fm on any closed current
loop in a uniform magnetic field is zero
Fm  I  dl  B
C
N   0
Example 4.1 Force on a Semicircular Conductor
The semicircular conductor shown lies in the x–y plane
and carries a current I . The closed circuit is exposed to
a uniform magnetic field B  yˆ B0 . Determine (a) the
magnetic force F1 on the straight section of the wire
and (b) the force F2 on the curved section.
Solution
a) F1  xˆ2Ir yˆB0  zˆ2IrB0 N


 0
 0
b) F2  I  dl  B   zˆI  rB0 sin d  zˆ 2 IrB0 N 
4-1.2 Magnetic Torque on a Current-Carrying 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 Field in the Plane of the Loop
• F1 and F3 generates a torque
in clockwise direction.
4-1.2 Magnetic Torque on a Current-Carrying Loop
Magnetic Field Perpendicular to the Axis of a Loop
•
When loop consists of N turns, the total torque is
T  N I A B0 sin
where N I A  magnetic moment,m
•
The vector m with normal vector is expressed as

m  nˆN I A
•
A  m 
2
T can be written as T  m  B
N  m 
What we learn today
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Electric force Fe
Magnetic force Fm
Electromagnetic force F
Lorentz force
Apart between Fe & Fm (5 types)
•
Examples are very Important
Test Yourself
•
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What is Force
Apart between Force & Torque
What is 2D axis
What is 3D axis
Q&A
•
Feel Free to ASK ME
•
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Next Class would be various lawsDo some study in advance
THANK YOU
TERIMA KASIH
4-2 The Biot–Savart Law
•
Biot–Savart law states that
1 dl  Rˆ
dH 
4 R 2
A/m
where dH = differential magnetic field
dl = differential length
•
To determine the total H we have
1 dl  Rˆ
H 
4 l R2
A/m 
4-2.1 Magnetic Field due to Surface and Volume Current Distributions
•
Biot–Savart law may be expressed in terms of
distributed current sources.
1
H
4
J s  Rˆ
S R 2 ds
for a surface current
1
H
4
J  Rˆ
v R 2 dv
for a volumecurrent
Example 4.3 Magnetic Field of a Pie-Shaped Loop
Determine the magnetic field at the apex O of the pieshaped loop as shown. Ignore the contributions to the
field due to the current in the small arcs near O.
Solution
For segment AC,
dl  Rˆ  zˆdl  zˆad
Consequently,
H
1
4
zˆad
1
ˆ

z
 where  is in radians
 a2
4a
4-2.2 Magnetic Field of a Magnetic Dipole
•
To find H in a spherical coordinate system, we
have
H

m ˆ
ˆ sin θ'
R
2
cos
θ'

θ
4R'3
where R’ >> a

A/m 
4-3 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)
4-4 Maxwell’s Magnetostatic Equation
•
There are 2 important properties: Gauss’s and
Ampere’s Law.
4-4.1 Gauss’s Law for Magnetism
•
Gauss’s law for magnetism states that

  B  0 (different ial form)  Bds  0 (integral form)
S
•
Net electric magnetic flux
through a closed surface
is zero.
4-4.2 Ampere’s Law
•
Ampere’s law states that
 Hdl  I Ampere's law 
C
•
The directional path of current C follows the righthand rule.
Example 4.6 Magnetic Field inside a Toroidal Coil
A toroidal coil (also called a torus or toroid) is a
doughnut-shaped structure (called its core) with closely
spaced turns of wire wrapped around it as shown. For a
toroid with N turns carrying a current I , determine the
magnetic field H in each of the following three regions:
r < a, a < r < b, andr > b, all in the azimuthal plane of
the toroid.
Solution 4.6 Magnetic Field inside a Toroidal Coil
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.
Application of Ampere’s law then gives
 H .dl 
C
  ˆH ˆrd  2rH   NI
2
0
Hence, H = NI/(2πr) and
NI
ˆ
ˆ
for a  r  b 
H  H  
2r
4-5 Vector Magnetic Potential
•
For any vector of vector magnetic potential A,
•
We are able to derive
•
Vector Poisson’s equation is given as
    A  0
 A  J
2
B   A
Wb/m .
 J
where A 
dv'

4 v ' R'
2
Wb/m
4-6 Magnetic Properties of Materials
•
•
Magnetic behavior is due to the interaction of
dipole and field.
3 types of magnetic materials: diamagnetic,
paramagnetic, and ferromagnetic.
4-6.1 Orbital and Spin Magnetic Moments
•
Electron generates around the nucleus and spins
about its own axis.
4-6.2 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
and relative permeability is defined as

r 
 1  m
0
4-6.3 Magnetic Hysteresis of Ferromagnetic Materials
•
•
Ferromagnetic materials is described by
magnetized domains.
Properties of magnetic materials are shown below.
4-6.3 Magnetic Hysteresis of Ferromagnetic Materials
•
Comparison of hysteresis curves for (a) a hard
and (b) a soft ferromagnetic material is shown.
4-7 Magnetic Boundary Conditions
•
For 2 different media when applying Gauss’s law,
we have
D1n  D2 n   s
B1n  B2 n
•
•
Boundary condition for H is 1H1n  2 H 2n .
Vector defined by the right-hand rule is
nˆ2  H1  H 2   J s
•
At interface between media with finite
conductivities, Js = 0 and H1t  H 2t .
4-8 Inductance
•
•
An inductor is a magnetic capacitor.
An example is a solenoid as shown below.
4-8.1 Magnetic Field in a Solenoid
•
For a cross section of solenoid,
B  zˆ
•
nI
2
sin 2  sin 1 
When l >a, θ1≈−90° and θ≈90°,
zˆNl
B  zˆnI 
l
long solenoid
with l / a  1
4-8.2 Self-Inductance
•
Magnetic flux  is given by
   Bds
Wb 
S
•
To compute the inductance we need area S.
4-8.2 Self-Inductance
•
The self-inductance L of conducting structure is
defined as

H 
L
I
where  = total magnetic flux (magnetic flux linkage)
•
For a solenoid,
N2
L
S solenoid
l
•
For two-conductor configurations,
  1
L     Bds
I
I I S
4-8.3 Mutual Inductance
•
•
Magnetic field lines are generated by I1 and S2 of
loop 2.
The mutual inductance is
12 N 2
L12 

B1ds H

I1
I1 s2
•
Transformer uses torodial coil with 2 windings.
4-9 Magnetic Energy
•
Total energy (Joules) expended in building up the
current in inductor is
l
1 2
Wm   pdt   ivdt  L  idi  LI
2
0
•
•
Some of the energy is stored in the inductor,
called magnetic energy, Wm.
The magnetic energy density wm is defined as
Wm 1
wm 
 H 2
v
2
J/m 
3
Example 4.9 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 radii a
and b. The insulation material has permeability µ.
Solution
The magnitude of the magnetic field is
B
1
H 
 2r
Magnetic energy stored in the coaxial cable is given by
2
1

I
1
2
Wm   H dv  2  2 dv
2V
8 V r
I 2 1
I 2  b 
 2  2  2rldr   2 ln  J 
8 V r
8
a