Field exams mxq proplems engineering with solution

Concept of Magnetic Circuits
What are magnetic circuits ?
They are basically ferromagnetic
structures (mostly iron) with coils wound
around them.
Because of the material high permeability, most of the
magnetic flux is confined inside the
magnetic circuit.
Examples: Transformers,
Actuators, Electromagnets and
Electric machines.
1

2
It’s an approximate technique to compute magnetic
flux.
2
NI
H
a
π
=
2
NI
B
a
µ
π
=
2
( )
2
NI
b
a
µ
φ π
π
=
BA
φ =
2
2
( )
NI
a
b
φ
π
µ π
=

3
V
I
R
=
l
R
A
σ
=
2
2
( )
a
R
b
π
σ π
=
2
2
( )
V
I
a
b
π
σ π
=

4
2
2
( )
V
I
a
b
π
σ π
=
2
2
( )
NI
a
b
φ
π
µ π
=

5
I φ
( )
V EMF ( )
NI MMF
σ µ
l
R
A
σ
= mc
l
A
µ
ℜ =
Mean core
length
Reluctance
5

6
V IR
= NI φ
= ℜ
V
R
I
NI
ℜ
φ

7
DC Magnetic Circuits: the supply is DC, magnetic circuit
laws can be used directly.
1
DC
coil
V
i
R
= 2
DC
coil
V
i
R
=

8
DC Magnetic Circuits
1 2
i i
=
1 2
1 2
mc mc
c c
l l
A A
φ φ
µ µ
=
1 2
Ni Ni
=
1 1
2 2
φ µ
φ µ
=
1
Ni 1
ℜ
1
φ
2
Ni 2
ℜ
2
φ
1 1 2 2
φ φ
ℜ = ℜ

Example
Sheet (3), Prob. (1):
Find the value of the current that will produce a flux
of 0.0025 Wb.
5000
r
µ =
NI φ
= ℜ
mc
c
l
A
µ
ℜ =
4(30 2 2.5) 140
mc
l cm
= + × =
50
N =
10cm
2
5*10 50
c
A cm
= =
9

Example
Find the value of the current that will produce a flux
of 0.0025 Wb.
5000
r
µ =
7 4
1.4
(5000)(4 10 )(50 10 )
π − −
ℜ =
× ×
50
N =
10cm
1
44563.38 H−
ℜ =
NI φ
= ℜ
2.228
I A
=
10

11
AC Magnetic Circuits: the supply is AC, Faraday’s law
must be considered first.

12
AC Magnetic Circuits:
AC
v e zero
− =
AC
v e
=
(Neglecting Coil Resistance)
V zero
∑ =
d
e N
dt
φ
= −

13
AC Magnetic Circuits
AC
d
v N
dt
φ
=
sin( )
m
d
V t N
dt
φ
ω =
By Integration
cos( ) sin( )
2
m m
V V
t t
N N
π
φ ω ω
ω ω
=− = −
Ф is dependent on V and f

14
1 2
φ φ φ
= =
1 2
2 1
i
i
µ
µ
=
1
1
mc
c
l
Ni
A
φ
µ
= 2
2
mc
c
l
Ni
A
φ
µ
=
1 1 2 2
c c
mc mc
Ni A Ni A
l l
µ µ
=
1
Ni 1
ℜ
1
φ
2
Ni 2
ℜ
2
φ

15
To get the current
sin( )
2
mc m
c
l V
i t
N A N
π
ω
µ ω
 

 
= −
 
 

 
mc
c
l
i
N A
φ
µ
=
2
sin( )
2
m
c
mc
V
i t
N A
l
π
ω
µ
ω
= −
 

 
 
 

 

16
2
c
mc
V
I
N A
j
l
µ
ω
=
 

 
 
 

 
core
V
I
j L
ω
=
2
core
mc
c
N
L
l
A
µ
= mc
c
l
A
µ
ℜ =
2
N
L =
ℜ
Self Inductance

Example
Find the inductance of the coil.
5000
r
µ =
50
N =
10cm
1
44563.38 H−
ℜ =
2
N
L =
ℜ
0.056
L H
=
17

Magnetic Circuits: Electric Machines
18
Electrical machines have generally two basic parts named
"Stator" and "Rotor".
The stator is the stationary member. The rotor is the
rotating member.
A small air gap exists between the two member.

19

Salient Poles Stator
20
N
S

21
φ
φ
f
g
M
φ =
ℜ
2
2
f f
g
N I
φ =
ℜ
Neglecting the reluctance of the steel parts
g
o c
g
A
µ
ℜ =
f
o c
M
A
g
φ µ
= f
g o
M
B
g
µ
=
f f f
M N I
=

Faraday’s Law Applications
1
1. Inductors.
2. Transformers.
3. Electrical Generators.
4. Electrical Motors.
5. Transmission Lines.

Inductors
4
d
v N
dt
f
=
Ni
Â
f
Ni f
= Â
Ni
f =
Â
2
N di
v
dt
=
Â
di
v L
dt
=
2
N
L =
Â

Inductors
5
( )
V j L I
w
=
L
V jX I
=
sin( )
2
m
V
i t
L
p
w
w
= -
By Integration
V
I
j L
w
=

Inductors: Introduction to Inductor Design
6
Objective:
Design an inductor having a given inductance L, which
carries worst-case current Imax without saturating the core
(Bmax), and which has a given winding resistance R, or,
equivalently, exhibits a worst-case copper loss of Im
2R

Transformer
7
Single-phase transformer

Transformer
Single-phase transformer
8

Transformer
10
Ideal Transformer
- Rcoil = zero - µ → ∞
- No leakage flux - No core losses
1
1 1
d
v N
dt
f
= 2
2 2
d
v N
dt
f
=
1 2
f f
=
1 1
2 2
v N
v N
= 1 1
2 2
e N
e N
=
Since no leakage

Transformer
11
Ideal Transformer
1 1 2 2
N i N i f
- = Â
Since µ → ∞
zero
Â =
1 2
2 1
N i
N i
=
1 1
N i 2 2
N i
Â
f

Transformer
12
Ideal Transformer
1 1
2 2
v N
v N
= 1 1
2 2
e N
e N
=
2 1
1 2
i N
i N
=
1 1 2 2
v i v i
=

Transformer: Equivalent Circuit
14
Case A: Ideal Transformer
1 1
2 2
v N
v N
= 1 1
2 2
e N
e N
=
2 1
1 2
i N
i N
=
1 1 2 2
v i v i
=

15
Case B: - Rcoil = zero - µ ≠ ∞
- No leakage flux - No core losses
At No-load: (i2 = zero)
1
N im f
= Â
1
2
1
sin( )
2
m
V
i t
N
m
p
w
w
= Â -
1
1
sin( )
2
m
V
t
N
p
f w
w
= -
From Faraday’s Law
1
N im
Â
f

16
If the transformer is loaded:
core
X L
m w
=
1
sin( )
2
m
core
V
i t
L
m
p
w
w
= -
1
core
V
I
j L
m
w
=
1 1
N i 2 2
N i
Â
f
1 1 2 2
N i N i f
- = Â
1 1 2 2 1
N i N i N im
- =

17
2
1 2
1
N
i i i
N
m = -
2
' 2
2
1
N
i i
N
=
2
1 2
1
N
i i i
N
m
= +

18
Case C: - Rcoil = zero - µ ≠ ∞
- No leakage flux - core losses exist.
Core (iron) losses
1. Eddy Current Losses
2 2 2
( )
e m
P B f thickness
µ
2 2 2
( )
e e m
P k B f thickness
= W/m3

19
Case C:
2. Hysteresis Losses

20
Case C:
BH Curve (Magnetization Curve)

21
Case C:
B
H

22
Case C:
1 .
m
r
B
B
w H dB
-
= ò
2 .
r
m
B
B
w H dB
= ò
1 1 2 3
(A.m ) ( . . ) .
H B T J A m J m
- - - -
´ = =
Energy lost /unit volume/cycle = 2(w1 – w2)= area inside the loop

23
Case C:
n depends on the material (1.6 – 2.4)
Total core losses:
n
h
P B f
µ n
h h
P k B f
= W/m3
2 2 2
( ) n
c e h
P k B f thickness k B f
= +
2
c
P B
µ
W/m3
Since f is usually constant
2
c
P V
µ
2
c
P f
µ

24
Case C:
2
1
c
c
V
P
R
=

25
Case D: - Rcoil = zero - µ ≠ ∞
- Leakage flux exists. - core losses exist.

Case D:
1 1
N i 2 2
N i
Â
f
1
l
Â
2
l
Â
1
f 2
f
1
l
f 2
l
f
1 1
l
f f f
= +
2 2
l
f f f
= -
1 1 1 1
l l
N i f
= Â 2 2 2 2
l l
N i f
= Â
(1)
(2)
(3) (4)
From (1) & (3)
1 1
1 1 1
l
d d
d
N N N
dt dt dt
f f
f
= +
2
1 1 1
1 1
1
l
l
l
d N di
v N
dt R dt
f
= =
26

Case D:
1 1
N i 2 2
N i
Â
f
1
l
Â
2
l
Â
1
f 2
f
1
l
f 2
l
f
From (2) & (4)
2 2
2 2 2
l
d d
d
N N N
dt dt dt
f f
f
= -
2
2 2 2
2 2
2
l
l
l
d N di
v N
dt R dt
f
= =
2 2
l
f f f
= -
2 2 2 2
l l
N i f
= Â
(2)
(4)
27

Case D:
let
1 1
d
e N
dt
f
= 2 2
d
e N
dt
f
=
1 1 1
l
v e v
= + 2 2 2
l
v e v
= -
1
v
1
l
v
1
e 2
e
2
l
v
2
v
1 1
1 1 1
l
d d
d
N N N
dt dt dt
f f
f
= + 2 2
2 2 2
l
d d
d
N N N
dt dt dt
f f
f
= -
1
1 1
d
v N
dt
f
= 2
2 2
d
v N
dt
f
=
28

Case D:
2
1 1
1
1
l
l
N di
v
R dt
=
2
2 2
2
2
l
l
N di
v
R dt
=
1
1 1
l l
di
v L
dt
=
2
2 2
l l
di
v L
dt
=
1 1 1
l l
V j L I
w
= 2 2 2
l l
V j L I
w
=
1 1 1
l l
V jX I
= 2 2 2
l l
V jX I
=
1
v
1 1
l
jX I
1
e 2
e
2 2
l
jX I
2
v
29

Case D:
30

Case E: - Rcoil ≠ zero - µ ≠ ∞
- Leakage flux exists. - core losses exist.
Exact Equivalent Circuit of the Transformer
31

FARADAY’S LAW APPLICATIONS
1. Inductors.
2. Transformers.
3. Electrical Generators.
4. Electrical Motors.
5. Transmission Lines.
1

GENERATORS
 Faraday’s Disk Generator (1831)
2

GENERATORS
 Faraday’s Rotating Rectangle (1831)
3

GENERATORS
4
𝑒 = − න
𝑆
𝜕𝐵
𝜕𝑡
. 𝑑𝑆
+ ර
𝑐
(𝑣 × 𝐵) . 𝑑𝑙
𝐵 = 𝐵𝑜𝑢𝑦
𝑒 = ර
𝑐
(𝑣 × 𝐵) . 𝑑𝑙 𝑒 = න
1
2
+ න
2
3
+ න
3
4
+ න
4
1
= is the velocity in the direction normal to the loop
𝑣
dl = unit length along the extension of the loop
x
y
z

GENERATORS
5
𝑒12 = න
1=0
2=−ℎ
ӈ
𝑣 × 𝐵𝑜𝑢𝑦 . 𝑑𝑥 (−𝑢𝑥)
𝑎𝑛
𝑢𝑦
𝛼
𝑒12 = 𝑢𝐵𝑜ℎ sin 𝛼
𝑢𝑧
an normal to the loop
𝜈 = u 𝑎𝑛 = u cos(𝛼) 𝑢𝑦 − u sin(𝛼) 𝑢𝑧
ӈ
𝑣 × 𝐵𝑜𝑢𝑦 = u 𝐵𝑜 sin(𝛼) 𝑢𝑥
x
y
z

GENERATORS
6
𝑒23 = න
2=0
3=𝑤
(𝜈 × 𝐵𝑜𝑢𝑦 . 𝑑𝑦 (𝑢𝑦) = u 𝐵𝑜 sin(𝛼) 𝑢𝑥 . 𝑑𝑦 (𝑢𝑦) 𝑒23 = 𝑧𝑒𝑟𝑜
x
y
z

GENERATORS
7
𝑒34 = න
3=−ℎ
4=0
ӈ
𝑣 × 𝐵𝑜𝑢𝑦 . 𝑑𝑥 (𝑢𝑥)
𝑒34 = 𝑢𝐵𝑜ℎ sin 𝛼
x
y
z
𝑒34 = න
3=−ℎ
4=0
ӈ
𝑣 × 𝐵𝑜𝑢𝑦 . 𝑑𝑥 (𝑢𝑥)
ӈ
𝑣 × 𝐵𝑜𝑢𝑦 = u 𝐵𝑜 sin(𝛼) 𝑢𝑥

GENERATORS
8
𝑒41 = න
4
1
= 𝑧𝑒𝑟𝑜
𝑢 = 𝜔
𝑤
2
𝑒 = 2𝑢𝐵𝑜ℎ sin 𝛼
𝑒 = 𝜔𝐵𝑜(𝑤ℎ) sin 𝛼
𝛼 = 𝜔𝑡 + 𝜃𝑜
𝑒 = 𝜔𝐵𝑜(𝑤ℎ) sin( 𝜔𝑡 + 𝜃𝑜)
x
y
z
Radial speed w to linear speed u

GENERATORS
AC Generator (Alternator)
9
𝑒 = 𝜔𝐵𝑜𝐴 sin( 𝜔𝑡 + 𝜃𝑜)
𝑇 =
1
𝑓
𝑓 =
𝜔
2𝜋
𝐴 = ℎ𝑤

GENERATORS: AC GENERATOR
10
Field direction
The field lines has to perpendicularly cross
the loop
That is why there is a “sine” in the
expression
𝑒 = 𝜔𝐵𝑜𝐴 sin( 𝜔𝑡 + 𝜃𝑜)

11
Realistically the field lines are usually curved or more circular due to the use of
cylindrical magnetic poles to fit rotational motion

12
𝑒 = ර
𝑐
(𝑣 × 𝐵) . 𝑑𝑙
𝑒12 = න
1=0
2=ℎ
𝑢 𝑢𝜙 × 𝐵𝑜 𝑢𝑟 . 𝑑𝑧 (−𝑢𝑧) 𝑒12 = 𝑢𝐵𝑜ℎ
𝑢𝑟
𝑢𝜙
𝑢𝑧
𝑒 = න
1
2
+ න
2
3
+ න
3
4
+ න
4
1
N S
No field region

13
𝑒34 = න
3=ℎ
4=0
𝑢 𝑢𝜙 × 𝐵𝑜(−𝑢𝑟) . 𝑑𝑧 𝑢𝑧
𝑒34 = 𝑢𝐵𝑜ℎ
𝑒23 = 𝑧𝑒𝑟𝑜
𝑒41 = 𝑧𝑒𝑟𝑜
𝑢𝑟
𝑢𝜙
𝑢𝑧

14
𝑢 = 𝜔𝑟
𝑒 = 2𝑢𝐵𝑜ℎ
𝑒 = 2𝜔𝐵𝑜(𝑟ℎ)
Under the pole faces
𝑒 = 0
Beyond the pole edges B0=0
𝑒 =
2
𝜋
𝐴𝑝𝐵𝑜𝜔 𝑒 =
2
𝜋
𝜙𝜔 Ф = Total flux under one
pole
𝑢𝑟
𝑢𝜙
𝑢𝑧

15
𝑒 = 𝐾𝜙𝜔

GENERATORS: DC GENERATOR (DYNAMO)
16

GENERATORS: 3-PHASE AC GENERATOR
20

GENERATORS: 3-PHASE AC GENERATOR
21

Generators
2
 Faraday’s Disk Generator (1831)

Generators
3
 Faraday’s Disk Generator (1831)

Generators
4
 Faraday’s Rotating Rectangle (1831)

Generators
5
 Faraday’s Rotating Rectangle (1831)

Generators
6
.
.
S
c
B
e dS
t
v B dl
¶
=-
¶
+ ´
ò
ò

y
o
B B u
=
.
c
e v B dl
= ´
ò

2 3 4 1
1 2 3 4
e = + + +
ò ò ò ò

Generators
7
( )
2
12
1 0
( ) . ( )
h
n y x
o
e u a B u dx u
=
=
= - ´ -
ò
n
a
y
u
a
12 sin
o
e uB h a
=
(1)(1)sin ( )
n y x
a u u
a
´ =

Generators
8
( )
3
23
2 0
( ) . ( )
w
n y y
o
e u a B u dy u
=
=
= - ´
ò 23
e zero
=

Generators
9
( )
4
34
3 0
( ) . ( )
h
n y x
o
e u a B u dx u
=
=
= ´
ò
n
a
y
u
a
34 sin
o
e uB h a
=
(1)(1)sin ( )
n y x
a u u
a
´ =

Generators
10
1
41
4
e zero
= =
ò
2
w
u w
=
2 sin
o
e uB h a
=
( )sin
o
e B wh
w a
=
o
t
a w q
= +
( )sin( )
o o
e B wh t
w w q
= +

Generators
11
AC Generator (Alternator)
sin( )
o o
e B A t
w w q
= +
1
T
f
=
2
f
w
p
=
A hw
=

Generators: AC Generator
14
.
c
e v B dl
= ´
ò

( )
2
12
1 0
( ) ( ) . ( )
h
r z
o
e u u B u dz u
f
=
=
= ´ -
ò 12 o
e uB h
=
r
u
u f
z
u
2 3 4 1
1 2 3 4
e = + + +
ò ò ò ò

15
( )
4
34
3 0
( ) ( ) . ( )
h
r z
o
e u u B u dz u
f
=
=
= ´ -
ò 34 o
e uB h
=
23
e zero
=
41
e zero
=
r
u
u f
z
u

16
u r
w
=
2 o
e uB h
=
2 ( )
o
e B rh
w
=
0
e =
Beyond the pole edges
r
u
u f
z
u
2
p o
e A B w
p
=
2
e fw
p
=
Ф = Total flux under one
pole

17
2
e fw
p
=
2
e fw
p
=

Generators: DC Generator (Dynamo)
18

Generators: DC Generator
20
e K fw
=
2
e fw
p
=
0
e =

Generators: 3-phase AC Generator
22

23

24

MOTORS: DC MOTOR
3
𝑇 = 𝑚 × 𝐵
𝑚 = 𝐼𝑆
𝑚 = 𝐼(ℎ𝑤)𝑎𝑛
𝑃𝑚 = 𝑇𝜔 𝑃𝑚 = 𝜔𝐼𝐵𝑜𝐴 sin 𝛼
𝐴 = ℎ𝑤
𝑇 = 𝐼(ℎ𝑤)𝑎𝑛 × 𝐵𝑜𝑢𝑦
𝑇 = 𝐼𝐵𝑜𝐴 sin 𝛼 (𝑢𝑥)
𝑎𝑛 = cos(𝛼) 𝑢𝑦 − sin(𝛼) 𝑢𝑧

MOTORS: DC MOTOR
6
𝐹12 = 𝐼 න
1=0
2=−ℎ
𝑑𝑧(−𝑢𝑧) × 𝐵𝑜𝑢𝑟
𝐹12 = 𝐼𝐵𝑜ℎ(𝑢𝜙)
𝐹𝑚 = 𝐼 න 𝑑𝑙 × 𝐵
𝐵 = 𝐵𝑜𝑢𝑟
𝐹12 = 𝐼 න
1
2
𝑑𝑙 × 𝐵
𝑢𝑟
𝑢𝜙
𝑢𝑧
𝑒 = න
1
2
+ න
2
3
+ න
3
4
+ න
4
1
dl = in the direction of the current carrying conductor

MOTORS: DC MOTOR
7
𝐹34 = 𝐼 න
3=−ℎ
4=0
𝑑𝑧(𝑢𝑧) × 𝐵𝑜(−𝑢𝑟) 𝐹34 = 𝐼𝐵𝑜ℎ(𝑢𝜙)
𝐵 = 𝐵𝑜(−𝑢𝑟)
𝐹34 = 𝐼 න
3
4
𝑑𝑙 × 𝐵
𝑢𝑟
𝑢𝜙
𝑢𝑧

MOTORS: DC MOTOR
9
𝑇 = 𝑟 × 𝐹
𝑇 = 𝐹𝐿
𝑇 = (2𝑟)𝐼𝐵𝑜ℎ
𝑇 = 2𝐼𝐵𝑜(𝑟ℎ)
𝑇 = 0
𝑇 =
2
𝜋
𝐴𝑝𝐵𝑜𝐼 𝑇 =
2
𝜋
𝜙𝐼 Ф = Total flux under one
pole

MOTORS: AC MOTOR
10
𝐵 = 𝐵𝑜 sin 𝜔1 𝑡
𝜔1 = 2𝜋𝑓

MOTORS: AC MOTOR
11
𝑒 = − න
𝑆
𝜕𝐵
𝜕𝑡
. 𝑑𝑆
+ ර
𝑐
(𝑣 × 𝐵) . 𝑑𝑙
𝐵 = 𝐵𝑜 sin 𝜔1 𝑡𝑢𝑦
𝑒1 = − න
𝑆
𝜕𝐵
𝜕𝑡
. 𝑑𝑆
𝑒 = 𝑒1 + 𝑒2
𝑒1 = − න
𝑆=ℎ𝑤
𝜔1𝐵𝑜 cos( 𝜔1𝑡)𝑢𝑦. 𝑑𝑆 𝑎𝑛

MOTORS: AC MOTOR
12
𝑒1 = − න
𝑆=𝑤ℎ
𝜔1𝐵𝑜 cos( 𝜔1𝑡)𝑢𝑦. 𝑑𝑆 𝑎𝑛
𝑎𝑛 ⋅ 𝑢𝑦 = (1)(1) cos 𝛼
𝑒1 = −𝜔1𝐵𝑜(𝑤ℎ) cos( 𝜔1𝑡) cos( 𝛼)
𝑒1 = −𝜔1𝐵𝑜(𝑤ℎ) cos( 𝜔1𝑡) cos( 𝜔𝑡 + 𝜃𝑜)
𝑒1 = −𝜔1𝐵𝑜
𝐴
2
[cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜) + cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜)]

MOTORS: AC MOTOR
13
𝑒2 = 𝜔𝐵𝑜(𝑤ℎ) sin( 𝜔1𝑡) sin( 𝜔𝑡 + 𝜃𝑜)
𝑒2 = ර
𝑐
𝑣 × 𝐵. 𝑑𝑙
𝑒2 = 𝜔𝐵𝑜
(𝑤ℎ)
2
[cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜) − cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜)]

MOTORS: AC MOTOR
14
𝑒 = 𝑒1 + 𝑒2
𝑒 = −𝐵𝑜
𝐴
2
[(𝜔1 − 𝜔) cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜)
+(𝜔1 + 𝜔) cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜)]
𝑒1 = −𝜔1𝐵𝑜
𝐴
2
[cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜) + cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜)]
𝑒2 = 𝜔𝐵𝑜
𝐴
2
𝑒 = 𝑒𝑓 + 𝑒𝑏

MOTORS: AC MOTOR
15
𝑒𝑓
𝑒𝑏
𝑖𝑓
𝑖𝑏
𝑖𝑓 =
𝐵𝑜𝐴
2 𝑍𝑓
[(𝜔1 − 𝜔) cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓)]
𝑖𝑓
𝑖𝑏
𝑒𝑓
𝑍𝑓
𝑒𝑏
𝑍𝑏
𝑅 + 𝑗(𝜔1 − 𝜔)𝐿
𝑅 + 𝑗(𝜔1 + 𝜔)𝐿
𝑖𝑏 =
𝐵𝑜𝐴
2 𝑍𝑏
[(𝜔1 + 𝜔) cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜 − 𝜙𝑏)]

MOTORS: AC MOTOR
Torque
16
𝑖𝑓 = 𝐼𝑓𝑚 cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓)
𝑖𝑏 = 𝐼𝑏𝑚 cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜 − 𝜙𝑏)
𝑇𝑓 = 𝑚 × 𝐵 𝑚 = 𝑖𝑓𝑆
𝑇𝑓 = 𝐼𝑓𝑚 cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓)(𝐴𝑎𝑛) × 𝐵𝑜 sin 𝜔1 𝑡𝑢𝑦
𝐵 = 𝐵𝑜 sin 𝜔1 𝑡𝑢𝑦
𝑇𝑓 = 𝐼𝑓𝑚𝐵𝑜𝐴 cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓) sin 𝜔1 𝑡 sin( 𝜔𝑡 + 𝜃𝑜)
sin 𝜔1 𝑡 sin( 𝜔𝑡 + 𝜃𝑜) =
1
2

MOTORS: AC MOTOR
Torque
17
𝑇𝑓 =
𝐼𝑓𝑚𝐵𝑜𝐴
2
cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓)
∗ [cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜) − cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜)]
cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓) ∗ cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜) =
1
2
cos 𝜙𝑓 + cos[ 2((𝜔1 − 𝜔)𝑡 − 𝜃𝑜) − 𝜙𝑓)]
cos( (𝜔1 − 𝜔)𝑡 − 𝜃𝑜 − 𝜙𝑓) ∗ cos( (𝜔1 + 𝜔)𝑡 + 𝜃𝑜) =
1
2
cos[ 2(𝜔𝑡 + 𝜃𝑜) + 𝜙𝑓) cos( 2𝜔1 − 𝜙𝑓)]

MOTORS: AC MOTOR
Torque
18
𝑇𝑓 =
4
∗
cos 𝜙𝑓 + cos[ 2((𝜔1 − 𝜔)𝑡 − 𝜃𝑜) − 𝜙𝑓)]
− cos( 2𝜔1 − 𝜙𝑓) − cos[ 2(𝜔𝑡 + 𝜃𝑜) + 𝜙𝑓)]
𝑇𝑓𝑎𝑣𝑔 =
4
cos 𝜙𝑓 Similarly 𝑇𝑏𝑎𝑣𝑔 = −
𝐼𝑏𝑚𝐵𝑜𝐴
4
cos 𝜙𝑏
𝑇𝑛𝑒𝑡 =
𝐵𝑜𝐴
4
𝐼𝑓𝑚 cos 𝜙𝑓 − 𝐼𝑏𝑚 cos 𝜙𝑏

MOTORS: AC MOTOR
19
For ω = 0
𝑇𝑛𝑒𝑡 =
𝐵𝑜𝐴
4
𝐼𝑓𝑚 cos 𝜙𝑓 − 𝐼𝑏𝑚 cos 𝜙𝑏
𝜙𝑓 = 𝜙𝑏
𝐼𝑓𝑚 = 𝐼𝑏𝑚
𝑇𝑛𝑒𝑡 = 𝑧𝑒𝑟𝑜
This motor cannot start from rest by its own developed torque
For any speed ω
𝐼𝑓𝑚 ≠ 𝐼𝑏𝑚 𝜙𝑓 ≠ 𝜙𝑏
𝑇𝑛𝑒𝑡 ≠ 𝑧𝑒𝑟𝑜

MOTORS: AC MOTOR
The motor needs an external method to start-up
It is called Induction Motor
Single Phase
Induction Motor
20

Motors: DC Motor
4
o
B B
=
DC
N S

Motors: DC Motor
6
2
12
1 0
( )
h
x y
o
F I dx u B u
=
=
= - ´
ò 12 ( )
z
o
F IB h u
= -
m
F I dl B
= ´
ò
y
o
B B u
=
2
12
1
F I dl B
= ´
ò
12
F

Motors: DC Motor
7
4
34
3 0
( )
h
x y
o
F I dx u B u
=
=
= ´
ò 12 ( )
z
o
F IB h u
=
4
34
3
F I dl B
= ´
ò
12
F
34
F

Motors: DC Motor
8
w
34
F
12
F
w
T r F
= ´
T FL
=
L
sin
L w a
=
( )sin
o
T IB hw a
=
o
F IB h
=
( )sin( )
o
T IB hw t
w
=
a

Motors: DC Motor
10
T m B
= ´
m I S
=
( ) n
m I hw a
=
m
P T w
= sin
m o
P IB A
w a
=
A hw
=
(1)(1)sin ( )
n y x
a u u
a
´ =
( ) n y
o
T I hw a B u
= ´
sin ( )
x
o
T IB A u
a
=

Motors: DC Motor
12
2
12
1 0
( )
h
z r
o
F I dz u B u
=
=
= ´
ò 12 ( )
o
F IB h uf
=
m
F I dl B
= ´
ò
r
o
B B u
=
2
12
1
F I dl B
= ´
ò
r
u
u f
z
u

Motors: DC Motor
13
4
34
3 0
( ) ( )
h
z r
o
F I dz u B u
=
=
= - ´ -
ò 34 ( )
o
F IB h uf
=
( )
r
o
B B u
= -
4
34
3
F I dl B
= ´
ò
r
u
u f
z
u

Motors: DC Motor
15
T r F
= ´
T FL
=
(2 ) o
T r IB h
=
2 ( )
o
T IB rh
=
0
T =
2
p o
T A B I
p
=
2
T I
f
p
=
Ф = Total flux under one
pole

Motors: AC Motor
16
1
sin
o
B B t
w
=
1 2 f
w p
=

Motors: AC Motor
17
.
.
S
c
B
e dS
t
v B dl
¶
=-
¶
+ ´
ò
ò

1
sin y
o
B B tu
w
=
1 .
S
B
e dS
t
¶
=-
¶
ò
1 2
e e e
= +
1 1 1
cos( ) .
y n
o
S hw
e B t u dS a
w w
=
=- ò

Motors: AC Motor
18
1 1 1
( )cos( )cos( )
o o
e B wh t t
w w w q
=- +
1 1 1 1
[cos(( ) ) cos(( ) )]
2
o o o
A
e B t t
w w w q w w q
=- - - + + +
1 1 1
cos( ) .
y n
o
S wh
e B t u dS a
w w
=
=- ò
(1)(1)cos
n y
a u a
⋅ =
1 1 1
( )cos( )cos( )
o
e B wh t
w w a
=-

Motors: AC Motor
19
2 1
( )sin( )sin( )
o o
e B wh t t
w w w q
= +
2 .
c
e v B dl
= ´
ò

2 1 1
[cos(( ) ) cos(( ) )]
2
o o o
A
e B t t
w w w q w w q
= - - - + +

Motors: AC Motor
20
1 2
e e e
= +
1 1
1 1
[( )cos(( ) )
2
( )cos(( ) )]
o o
o
A
e B t
t
w w w w q
w w w w q
=- - - -
+ + + +
1 1 1 1
[cos(( ) ) cos(( ) )]
2
o o o
A
e B t t
w w w q w w q
=- - - + + +
2 1 1
[cos(( ) ) cos(( ) )]
2
o o o
A
e B t t
w w w q w w q
= - - - + +
f b
e e e
= +

Motors: AC Motor
21
f
e
b
e
f
i
b
i
1 1
[( ) cos(( ) )]
2
o
f o f
f
B A
i t
Z
w w w w q f
= - - - -
f
i
b
i
f
f
e
Z
b
b
e
Z
1 1
[( ) cos(( ) )]
2
o
b o b
b
B A
i t
Z
w w w w q f
= + + + -
1
( )
R j L
w w
+ -
1
( )
R j L
w w
+ +

Motors: AC Motor
22
Torque
1
cos(( ) )
f fm o f
i I t
w w q f
= - - -
1
cos(( ) )
b bm o b
i I t
w w q f
= + + -
f
T m B
= ´ f
m i S
=
1 1
cos(( ) )( ) sin
n y
f fm o f o
T I t Aa B tu
w w q f w
= - - - ´
1
sin y
o
B B tu
w
=
1 1
cos(( ) )sin sin( )
f fm o o f o
T I B A t t t
w w q f w w q
= - - - +
1 1 1
1
sin sin( ) [cos(( ) ) cos(( ) )]
2
o o o
t t t t
w w q w w q w w q
+ = - - - + +

Motors: AC Motor
23
Torque
1
1 1
cos(( ) )
2
*[cos(( ) ) cos(( ) )]
fm o
f o f
o o
I B A
T t
t t
w w q f
w w q w w q
= - - -
- - - + +
[ ]
1 1
1
cos(( ) )*cos(( ) )
1
cos cos[2(( ) ) )]
2
o f o
f o f
t t
t
w w q f w w q
f w w q f
- - - - - =
+ - - -
1 1
1
cos(( ) )*cos(( ) )
1
cos[2( ) )] cos(2 )
2
o f o
o f f
t t
t t
w w q f w w q
w q f w f
- - - + + =
é ù
+ + + -
ê ú
ë û

Motors: AC Motor
24
Torque
1
1
cos cos[2(( ) ) )]
*
cos(2 ) cos[2( ) )]
4
f o f
fm o
f
f o f
t
I B A
T
t t
f w w q f
w f w q f
é ù
+ - - -
ê ú
= ê ú
- - - + +
ê ú
ë û
cos
4
fm o
favg f
I B A
T f
= Similarly cos
4
bm o
bavg b
I B A
T f
=-
[ ]
cos cos
4
o
net fm f bm b
B A
T I I
f f
= -

Motors: AC Motor
25
For ω = 0
[ ]
cos cos
4
o
net fm f bm b
B A
T I I
f f
= -
f b
f f
=
fm bm
I I
=
net
T zero
=
This motor cannot start from rest by its own developed torque
For any speed ω
fm bm
I I
¹ f b
f f
¹
net
T zero
¹

Motors: AC Motor
26
The motor needs an external method to start-up
It is called Induction Motor
Single Phase
Induction Motor

Maxwell’s equations
1
Static Fields (ρ & J are constants)
Electrostatics Magnetostatics
Steady Current Flow
E zero
´ =
.D r
 =
H J
´ =
.B zero
 =
D E
e
= B H
m
=
E zero
´ =
.J zero
 = J E
s
=

2
Time varying fields: Faraday’s Law
for a single turn
d
e
dt
f
=-
.
S
B dS
f = ò
.
c
emf E dl
= ò

. .
c S
d
E dl B dS
dt
=-
ò ò


3
Time varying fields: Faraday’s Law
for a stationary loop
Using Stoke’s theorem
B
E
t
¶
´ =-
¶
. .
c S
B
E dl dS
t
¶
=-
¶
ò ò

. .
cons non cons
E E E -
= +
S c
( A).dS A.dl
´ =
ò ò


4
Time varying fields
. .
cons non cons
D D D -
= +
.
. cons
D r
 =
.
. non cons
D zero
-
 =
.D r
 =

5
Time varying fields: Law of conservation of charge
net
dQ
I
dt
=-
.
S
dQ
J dS
dt
=-
ò
 v
Q dv
r
= ò
.
S v
d
J dS dv
dt
r
=-
ò ò


6
Time varying fields: Law of conservation of charge
Using Divergence theorem
.
S v
J dS dv
t
r
¶
=-
¶
ò ò

.J
t
r
¶
 =-
¶ V S
( .A)dv A.dS
 =
ò ò


7
Time varying fields: Modified Ampere’s Law
Contradicts with the law of conservation of charge
.J zero
t
r
¶
 + =
¶
H J
´ =
.( )
H zero
 ´ = .J zero
 =
.D r
 =

8
Time varying fields: Modified Ampere’s Law
( . )
.
D
J zero
t
¶ 
 + =
¶
.( )
H zero
 ´ =
.( ) .( )
D
H J
t
¶
 ´ =  +
¶
D
H J
t
¶
´ = +
¶

9
B, H , D, E are field quantities
ρ , J are source quantities
D
H J
t
¶
´ = +
¶
.D r
 =
B
E
t
¶
´ =-
¶
.B zero
 =

10
Integral form (using Divergence and Stoke’s)
. .
C S
B
E dl dS
t
¶
=-
¶
ò ò

. ( ).
C S
D
H dl J dS
t
¶
= +
¶
ò ò

.
S V
D dS dv
r
=
ò ò

.
S
B dS zero
=
ò


11
Explanatory example:
For path 1
For path 2
Applying Ampere’s law will give B = zero since there is no
conduction current passing in the capacitor.
. en
c
H dl I
=
ò

2
o c
i
B u
r
f
m
p
=

12
For path 2
Applying modified Ampere’s law
For parallel plate capacitor:
. ( ).
c S
D
H dl J dS
t
¶
= +
¶
ò ò

J zero
= o
D E
e
=
V
E
d
=
sin
m
V V t
w
=
sin
m
o
V
D t
d
e w
=

13
For path 2
. .
c S
D
H dl dS
t
¶
=
¶
ò ò

cos
m
o
V
D
t
t d
e w w
¶
=
¶
( cos )
2
m
o o
V
A t
d
B
r
m e w w
p
= 2
o d
i
B u
r
f
m
p
=

14
For path 2
Displacement Current
From circuits equations:
cos
m
d o
V
i A t
d
e w w
=
c
dV
i C
dt
= cos
c m
i CV t
w w
= o A
C
d
e
=
cos
m
c o
V
i A t
d
e w w
= c d
i i
=
dE
dt
J

15
Boundary Conditions
1 2
2 ( ) 0
n
u E E
  
1 2
t t
E E

1 2
2 ( )
n s
u D D 
  
1 2
2 ( )
n s
u H H J
  
1 2
2.( ) 0
n
u B B
 
1 2
n n s
D D 
 
1 2 s
t t
H H J
 
1 2
n n
B B


Maxwell’s equations: Time-Harmonic
16
Time Harmonic Fields
A time-harmonic field is one that varies periodically or sinusoidally
with time. When the source quantities (ρ , J) vary sinusoidally, all the
field quantities will vary sinusoidally
cos( )
m
v V t
 
 
2 2
j
m m
V V
V e 

  
( )
( , ) Re[ ( )e ]
j t
E r t E r  


( )
( ) Re[ e ]
j t
m
v t V  


( , ) ( )cos( )
E r t E r t
 
 
( ) ( )e ( )
j
h
E r E r E r


  
( )
( , ) Re[ ( )e ]
j t
h
E r t E r 


17
( )
( , ) Re[ ( )e ]
j t
h
B r t B r 

( ) ( )
( )(Re[e ]) ( ) Re[e ]
j t j t
h h
E r B r
t
w w
¶
´ =-
¶
( )
(Re[e ])
j t
h
E w
´ ( )
(Re[e ])
j t
h
j B w
w
=-
h h
E j B
w
´ =-
( )
( , ) Re[ ( )e ]
j t
h
E r t E r 

B
E
t
¶
´ =-
¶

18
h h
h
H J j D
w
´ = +
. h h
D r
 =
. h
B zero
 =
h h
E j B
w
´ =-

Poynting’s Theorem
19
V/m * A/m=VA/m2
S E H
= ´
. .( ) . .
S E H H E E H
 =  ´ = ´ - ´
. .( ) .( )
B D
S H E J
t t
¶ ¶
 = - - +
¶ ¶
2 2 2
1 1
. ( ) ( )
2 2
S E E H
t t
s e m
¶ ¶
 =- - -
¶ ¶
2 2 2
1 1
. ( )
2 2
S v v
S dS E dv E H dv
t
s e m
¶
=- - +
¶
ò ò ò

2
d df ( x )
f ( x ) 2 f ( x )
dt dt
=

20
For time harmonic fields
complex Poynting vector
. ( )
d e m
S v v
S dS P dv w w dv
t
¶
- = + +
¶
ò ò ò

2
d
P E
s
=
2
1
2
e
w E
e
=
2
1
2
m
w H
m
=
2 2 2
1 1 1
. ( )
2 2 2
h h h
h
S v v
S dS E dv j H E dv
s w m e
- = + -
ò ò ò

*
1
2
h h
h
S E H
= ´

21
For time harmonic fields
. ( )
h davg mavg eavg
S v v
S dS P dv j w w dv
w
- = + -
ò ò ò

.
h
S
S dS P jQ
- = +
ò


11/15/2022
1
Time Varying Fields
Applications
Maxwell’s Equations
By:
Emad Fathy Yassin
Faculty of Engineering
Cairo University
Maxwell’s Equations “ Static Fields”
Magnetostatics
Electrostatics
Steady Current Flow
• Static Fields and
1
2

11/15/2022
2
Maxwell’s Equations “ Time Varying Fields”
𝑒𝑚𝑓 = −
𝜕𝜙
𝜕𝑡
As 𝑒𝑚𝑓 = ∮ 𝐸. 𝑑𝑙 , and 𝜙 = ∫ 𝐵. 𝑑𝑠
Then:
𝐸. 𝑑𝑙 = −
𝜕
𝜕𝑡
𝐵. 𝑑𝑠
= −
𝜕𝐵
𝜕𝑡
. 𝑑𝑠
• Time varying Fields: Faraday’s Law
For a single stationary turn
Stokes ‘ Theorem
𝐴̅. 𝑑𝑙 = ∇ × 𝐴̅. 𝑑𝑠
𝑎𝑟𝑒𝑎 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
𝑡ℎ𝑒 𝑝𝑎𝑡ℎ
𝑐𝑙𝑜𝑠𝑒𝑑 𝑝𝑎𝑡ℎ
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− −(𝟏)
𝐸. 𝑑𝑙 = ∇ × 𝐸. 𝑑𝑠
𝐷. 𝑑𝑠 = 𝑄
= 𝜌 𝑑𝑣
Then:
𝐷. 𝑑𝑠 = ∇. 𝐷 𝑑𝑣
= 𝜌 𝑑𝑣
• Time varying Fields: Gaussian’s Law
Divergence ‘ Theorem
𝐴̅. 𝑑𝑠 = ∇. 𝐴̅ 𝑑𝑣
𝑣𝑜𝑙𝑢𝑚𝑒
𝑐𝑙𝑜𝑠𝑒𝑑 𝑎𝑟𝑒𝑎
𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟐)
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11/15/2022
3
𝐼 = −
𝑑𝑄
𝑑𝑡
= −
𝝏
𝝏𝑡
𝜌 𝑑𝑣
𝐼 = 𝐽̅. 𝑑𝑠 = ∇. 𝐽̅ 𝑑𝑣
• Time varying Fields: Law of Conservation of Charge
𝐴̅. 𝑑𝑠 = ∇. 𝐴̅ 𝑑𝑣
𝜵. 𝑱̅ = −
𝝏𝝆𝒗
𝝏𝒕
This is one of the fundamental postulates of physics, that
electric changes may not be created or destroyed.
𝑸
𝐻. 𝑑𝑙 = 𝐼
= 𝐽̅. 𝑑𝑠
𝐻. 𝑑𝑙 = ∇ × 𝐻. 𝑑𝑠 = 𝐽. 𝑑𝑠
• Time varying Fields: Ampere's Law
𝜵 × 𝑯 = 𝑱̅
Stokes ‘ Theorem
𝐴̅. 𝑑𝑙 = ∇ × 𝐴̅. 𝑑𝑠
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅
Null Identity
𝜵. 𝜵 × 𝑨 = 𝟎
𝜵 × 𝜵𝑽 = 𝟎
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅ = 𝟎
Take divergence
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6

11/15/2022
4
• Time varying Fields: Ampere's Law
𝜵 × 𝑯 = 𝑱̅ Stokes ‘ Theorem
𝐴̅. 𝑑𝑙 = ∇ × 𝐴̅. 𝑑𝑠
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅
Null Identity
𝜵. 𝜵 × 𝑨 = 𝟎
𝜵 × 𝜵𝑽 = 𝟎
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅ = 𝟎
Take divergence
This contradicts with the continuity equations
𝜵. 𝑱̅ = −
𝝏𝝆𝒗
𝝏𝒕
Then ampere's law has to be modified
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅ +
𝝏𝝆𝒗
𝝏𝒕
= 𝟎
• Time varying Fields: Modified Ampere's Law
Then ampere's law has to be modified
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅ +
𝝏𝝆𝒗
𝝏𝒕
= 𝟎
From gaussian law sub. with 𝝆𝒗 = 𝜵. 𝑫
𝜵. 𝜵 × 𝑯 = 𝜵. 𝑱̅ +
𝝏
𝝏𝒕
𝜵. 𝑫 = 𝟎
𝜵. 𝜵 × 𝑯 = 𝜵. (𝑱̅ +
𝝏
𝝏𝒕
𝑫) = 𝟎
𝜵. 𝜵 × 𝑯 = 𝜵. (𝑱̅ +
𝝏
𝝏𝒕
𝑫) = 𝟎 𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟑)
Conduction
Current
density
Displacement
Current
density
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8

11/15/2022
5
𝐵. 𝑑𝑠 = 0
∇. 𝐵 𝑑𝑣 = 0
• Time varying Fields: The Law of conservation of Magnetic Fields
Divergence Theorem
𝐴̅. 𝑑𝑠 = ∇. 𝐴̅ 𝑑𝑣
𝜵. 𝑩 = 𝟎 −− −(𝟒)
This equation is one of the fundamental postulates in the magnetic
field. This means that the magnetic flux lines always close upon
themselves and there are no isolated magnetic charges.
Differential form
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
Integral form
𝑬. 𝒅𝒍 = −
𝝏𝑩
𝝏𝒕
. 𝒅𝒔 −− − 𝟏
𝑯. 𝒅𝒍 = (𝑱̅ +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔 −− −(𝟐)
𝑫. 𝒅𝒔 = 𝑸𝒆𝒏 −− −(𝟑)
𝑩. 𝒅𝒔 = 𝟎 −− −(𝟒)
𝑩, 𝑯, 𝑫, 𝑬 are field quantities
𝝆𝒗 , 𝑱 are source quantities
Substitutive relations
𝑫 = 𝝐𝑬
𝑩 = 𝝁 𝑯
𝑱̅ = 𝝈 𝑬
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10

11/15/2022
6
Maxwell’s Equations “ Explanatory Example ”
Maxwell’s 2nd law
𝑯. 𝒅𝒍 = (𝑱̅ +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝒊𝒄
𝒔𝟏
𝒔𝟐
𝒗𝒄(𝒕)
𝑯. 𝒅𝒍 = (𝑱𝒄 +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝑺𝟏
= (𝑱𝒄 +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝑺𝟐
(𝑱𝒄 +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝑺𝟏
= (𝑱𝒄 + 𝟎). 𝒅𝒔
𝑺𝟏
= 𝒊𝒄
Since there is no dielectric over the surface 𝑺𝟏
(𝑱𝒄 +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝑺𝟐
= (𝟎 +
𝝏𝑫
𝝏𝒕
). 𝒅𝒔
𝑺𝟐
= 𝒊𝒅
Since there is no free charges in the dielectric
over the surface 𝑺𝟐, then
𝒊𝒄 = 𝒊𝒅
𝒊𝒅
Maxwell’s Equations “Explanatory Example ”
𝒊𝒄
𝒔𝟏
𝒔𝟐
𝒗𝒄(𝒕)
𝒊𝒄 = 𝒊𝒅
𝒊𝒅
𝑫 = 𝝐𝑬
𝑫 = 𝝐
𝒗𝒄
𝒅
𝒊𝒅 =
𝝏𝑫
𝝏𝒕
. 𝒅𝒔 =
𝝐
𝒅
𝝏𝒗𝒄
𝝏𝒕
. 𝒅𝒔 =
𝝐 𝑨
𝒅
𝝏𝒗𝒄
𝝏𝒕
= 𝑪
𝝏𝒗𝒄
𝝏𝒕
−−− −(𝟐)
𝑨
𝒅
In another method, the capacitance of this capacitor can be given by:
𝑪 =
𝝐 𝑨
𝒅
𝒊𝒄 = 𝑪
𝒅𝒗𝒄
𝒅𝒕
=
𝝐 𝑨
𝒅
𝒅𝒗𝒄
𝒅𝒕
−−− −(𝟏)
Then the capacitor current can be calculated from:
The flux density in the dielectric is given by:
Where 𝐸 =
𝒗𝒄
𝒅
Then,
From (1) and (2)
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11/15/2022
7
Maxwell’s Equations “ Time Harmonic (phasor) Form”
with time. When the source quantities (𝜌 , 𝐽) vary sinusoidally, all the
𝑣 𝑡 = 𝑉 cos(𝜔𝑡 + 𝜃) in instantaneous value
𝑉 =
𝑉
2
∠𝜃 =
𝑉
2
𝑒
in phasor value
𝒎
𝒋𝜽
with time. When the source quantities (𝜌 , 𝐽) vary sinusoidally, all the
𝐸 𝑟, 𝑡 = 𝐸 𝑟 cos(𝜔𝑡 + 𝜃)
= ℜ 𝐸 𝑟 e
= ℜ 𝐸 𝑟 e e
= ℜ 𝐸 e 𝑬𝒉 = 𝑬 𝒓 𝒆𝒋𝜽
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11/15/2022
8
Differential form
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
𝐸 𝑟, 𝑡 = 𝐸 𝑟 cos(𝜔𝑡 + 𝜃)
= ℜ 𝐸 e 𝐸 = 𝐸 𝑟 e
𝜕
𝜕𝑡
𝐸 𝑟, 𝑡 =
𝜕
𝜕𝑡
ℜ 𝐸 𝑒
= 𝑗𝜔 ℜ 𝐸 𝑒
𝝏
𝝏𝒕
==⇒ 𝒋𝝎
Time harmonic form
𝜵 × 𝑬𝒉 = −𝒋𝝎𝝁𝑯𝒉 −− − 𝟏
𝜵 × 𝑯𝒉 = 𝑱̅ + 𝒋𝝎 𝝐 𝑬𝒉 −− −(𝟐)
𝜵. 𝑫𝒉 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩𝒉 = 𝟎 −− −(𝟒)
𝝏𝟐
𝝏𝒕𝟐
==⇒ 𝒋𝝎 𝟐
Poynting Theorem
Poynting vector S is used to calculate the power losses in a medium using
the electric and magnetic field quantities
𝑺 = 𝑬 × 𝑯 𝑉𝐴/𝑚
𝛁. 𝑺 = 𝛁. (𝑬 × 𝑯 )
= 𝑯 . 𝛁 × 𝑬 − 𝑬. 𝛁 × 𝑯
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
= 𝑯 . (−
𝝏𝑩
𝝏𝒕
) − 𝑬. (𝑱̅ +
𝝏𝑫
𝝏𝒕
)
Substitutive relations
𝑫 = 𝝐𝑬
𝑩 = 𝝁 𝑯
𝑱̅ = 𝝈 𝑬
= −𝝁𝑯 .
𝝏𝑯
𝝏𝒕
− 𝑬. (𝝈𝑬 + 𝝐
𝝏𝑬
𝝏𝒕
)
= −𝝁𝑯 .
𝝏𝑯
𝝏𝒕
− 𝝈𝑬𝟐
− 𝝐 𝑬.
𝝏𝑬
𝝏𝒕
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11/15/2022
9
Poynting Theorem
𝑺 = 𝑬 × 𝑯 𝑉𝐴/𝑚
𝛁. 𝑺 = 𝛁. (𝑬 × 𝑯 )
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
= −𝝁𝑯 .
𝝏𝑯
𝝏𝒕
− 𝝈𝑬𝟐
− 𝝐 𝑬.
𝝏𝑬
𝝏𝒕
= −𝝈𝑬𝟐
−
𝟏
𝟐
𝝁
𝝏𝑯𝟐
𝝏𝒕
−
𝟏
𝟐
𝝐
𝝏𝑬𝟐
𝝏𝒕
𝛁. 𝑺 = 𝝈𝑬𝟐
+
𝝏
𝝏𝒕
𝟏
𝟐
𝝁𝑯𝟐
+
𝝏
𝝏𝒕
𝟏
𝟐
𝝐𝑬𝟐
− 𝑆̅. 𝑑𝑠 = 𝝈𝑬𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝁𝑯𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝐𝑬𝟐
𝑑𝑣
𝐴̅. 𝑑𝑠 = ∇. 𝐴̅ 𝑑𝑣
− 𝛁. 𝑺 𝑑𝑣 = 𝝈𝑬𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝁𝑯𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝐𝑬𝟐
𝑑𝑣
Poynting Theorem
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
− 𝑆̅. 𝑑𝑠 = 𝝈𝑬𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝁𝑯𝟐
𝑑𝑣 +
𝝏
𝝏𝒕
𝟏
𝟐
𝝐𝑬𝟐
𝑑𝑣
− 𝑆̅. 𝑑𝑠 = 𝑝 𝑑𝑣 +
𝝏
𝝏𝒕
𝒘𝒎 𝑑𝑣 +
𝝏
𝝏𝒕
𝒘𝒆 𝑑𝑣
𝒑𝒅 = 𝝈𝑬𝟐
, 𝒘𝒎 =
𝟏
𝟐
𝝁𝑯𝟐
, 𝒂𝒏𝒅 𝒘𝒆 =
𝟏
𝟐
𝝐𝑬𝟐
𝑺𝒉 =
1
2
𝑬𝒉 × 𝑯∗
𝒉
− 𝑆̅ . 𝑑𝑠 = 𝝈𝑬𝟐
𝑑𝑣 + 𝒋𝝎
𝟏
𝟐
𝝁𝑯𝟐
𝑑𝑣 + 𝒋𝝎
𝟏
𝟐
𝝐𝑬𝟐
𝑑𝑣
− 𝑆̅ . 𝑑𝑠 = 𝑃 + 𝒋𝑸
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12/7/2022
1
Time Varying Fields
Applications
Wave Equations
By:
Emad Fathy Yassin
Cairo University
Wave Equations
Differential form
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
𝜵 × 𝜵 × 𝑬 = −
𝝏
𝝏𝒕
𝜵 × 𝑩
Take curl for maxwell’s 1st equation
𝜵 × 𝜵 × 𝑬 = −
𝝏
𝝏𝒕
𝝁 (𝑱̅ +
𝝏𝑫
𝝏𝒕
)
Sub. using maxwell’s 2nd equation
Substitutive quantities
𝑫 = 𝝐𝑬
𝑩 = 𝝁 𝑯
𝑱̅ = 𝝈 𝑬
𝜵𝜵. 𝑬 − 𝜵𝟐
𝑬 = −
𝝏
𝝏𝒕
𝝁 (𝝈 𝑬 + 𝝐
𝝏𝑬
𝝏𝒕
)
Sub. using maxwell’s 3rd equation
−𝝁𝝈
𝝏
𝝏𝒕
𝑬 − 𝝁𝝐
𝝏𝟐
𝑬
𝝏𝒕𝟐
𝜵
𝝆
𝝐
− 𝜵𝟐
𝑬 =
𝜵𝟐
𝑬 − 𝝁𝝈
𝝏
𝝏𝒕
𝑬 − 𝝁𝝐
𝝏𝟐
𝑬
𝝏𝒕𝟐
=
𝜵𝝆
𝝐
−− −(𝟏)
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2
Wave Equations
Differential form
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
𝜵 × 𝜵 × 𝑯 = 𝜵 × (𝑱̅ +
𝝏𝑫
𝝏𝒕
)
Take curl for maxwell’s 2nd equation
Sub. using maxwell’s 1st and 3rd equations
Substitutive quantities
𝑫 = 𝝐𝑬
𝑩 = 𝝁 𝑯
𝑱̅ = 𝝈 𝑬
𝜵𝜵. 𝑯 − 𝜵𝟐
𝑯 = 𝝈 𝜵 × 𝑬 + 𝝐
𝝏
𝝏𝒕
𝜵 × 𝑬
𝟎 − 𝜵𝟐
𝑯 = −𝝁𝝈
𝝏
𝝏𝒕
𝑯 − 𝝁𝝐
𝝏𝟐
𝑯
𝝏𝒕𝟐
𝜵𝟐
𝑯 − 𝝁𝝈
𝝏
𝝏𝒕
𝑯 − 𝝁𝝐
𝝏𝟐
𝑯
𝝏𝒕𝟐
= 𝟎 −− −(𝟐)
𝜵 × 𝜵 × 𝑯 = 𝜵 × (𝝈 𝑬 + 𝝐
𝝏 𝑬
𝝏𝒕
)
Wave Equations
Max. Eqns. (Differential form)
𝜵 × 𝑬 = −
𝝏𝑩
𝝏𝒕
−− − 𝟏
𝜵 × 𝑯 = 𝑱̅ +
𝝏𝑫
𝝏𝒕
−− −(𝟐)
𝜵. 𝑫 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩 = 𝟎 −− −(𝟒)
𝜵𝟐
𝑯 − 𝝁𝝈
𝝏
𝝏𝒕
𝑯 − 𝝁𝝐
𝝏𝟐
𝑯
𝝏𝒕𝟐
= 𝟎 −− −(𝟐)
𝜵𝟐
𝑬 − 𝝁𝝈
𝝏
𝝏𝒕
𝑬 − 𝝁𝝐
𝝏𝟐
𝑬
𝝏𝒕𝟐
=
𝜵𝝆
𝝐
−− −(𝟏)
Wave equations in differential form
Wave equations in time harmonic (phasor) form
𝜵𝟐
𝑯 − 𝒋𝝎𝝁𝝈𝑯 + 𝝎𝟐
𝝁𝝐𝑯 = 𝟎 −− −(𝟐)
𝜵𝟐
𝑬 − 𝒋𝝎𝝁𝝈𝑬 + 𝝎𝟐
𝝁𝝐𝑬 =
𝜵𝝆
𝝐
−− −(𝟏)
Max. Eqns. (Time harmonic form)
𝜵 × 𝑬𝒉 = −𝒋𝝎𝝁𝑯𝒉 −− − 𝟏
𝜵 × 𝑯𝒉 = 𝑱̅ + 𝒋𝝎 𝝐 𝑬𝒉 −− −(𝟐)
𝜵. 𝑫𝒉 = 𝝆𝒗 −− −(𝟑)
𝜵. 𝑩𝒉 = 𝟎 −− −(𝟒)
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3
Wave Equations “Different Media”
Solve the wave equations for different medias as follows:
 Free space, [𝝈 = 𝟎, 𝝁𝒐, 𝝐𝒐] (vacuum, space, dry air)
 Lossless dielectric media , [𝝈 = 𝟎, 𝝁𝒐𝝁𝒓, 𝝐𝒐𝝐𝒓] (dry wood, plastic)
 Lossy dielectric media , [𝝈 ≠ 𝟎, 𝝁𝒐𝝁𝒓, 𝝐𝒐𝝐𝒓] (wet soil, sea water)
 Good conductor, [𝝈 𝒗𝒆𝒓𝒚 𝒉𝒊𝒈𝒉, 𝝁𝒐𝝁𝒓, 𝝐𝒐𝝐𝒓] (Cupper, Aluminum)
𝜵𝟐
𝝁𝝐𝑯 = 𝟎 −− −(𝟐)
𝜵𝟐
𝝁𝝐𝑬 =
𝜵𝝆
𝝐
−− −(𝟏)
Wave Equations “Free space ”
Wifi/WLAN antenna
𝜵𝟐
𝑯 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑯 = 𝟎 − −(𝟐)
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 − −(𝟏)
TV antenna
Satellite antenna
Military antenna
==⇒ 𝒌𝒐
𝟐 = 𝝎𝟐 𝝁𝒐𝝐𝒐
1) Free space, [𝝈 = 𝟎, 𝝁𝒐, 𝝐𝒐] assume source free 𝝆 = 𝟎
𝜵𝟐
𝝁𝝐𝑯 = 𝟎 − −(𝟐)
𝜵𝟐
𝝁𝝐𝑬 =
𝜵𝝆
𝝐
− − (𝟏)
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4
𝜵𝟐
𝑯 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑯 = 𝟎 −− −(𝟐)
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝝏𝟐
𝝏𝒙𝟐
+
𝝏𝟐
𝝏𝒚𝟐
+
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝒌𝒐
𝟐
𝑬 = 𝟎
𝒖𝒙
𝒖𝒛
𝒖𝒚
Planner waves
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝝏𝟐
𝝏𝒙𝟐
+
𝝏𝟐
𝝏𝒚𝟐
+
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝒌𝒐
𝟐
𝑬 = 𝟎
𝝏𝟐
𝝏𝒛𝟐 𝑬 + 𝒌𝒐
𝟐
𝑬 = 𝟎  ODE
𝑬 𝒛 = 𝑬𝒐 𝒆 𝒋𝒌𝒐𝒛
𝑭𝒐𝒓𝒘𝒂𝒓𝒅 𝒘𝒂𝒗𝒆
+ 𝑬𝒐 𝒆𝒋𝒌𝒐𝒛
𝑩𝒂𝒄𝒌𝒘𝒂𝒓𝒅 𝒘𝒂𝒗𝒆
How to solve ODE
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝒌𝒐
𝟐
𝑬 = 𝟎
Let 𝐌 =
𝝏
𝝏𝒛
⇒ 𝑴𝟐
𝑬 + 𝒌𝒐
𝟐
𝑬 = 𝟎
𝑴𝟐
= −𝒌𝒐
𝟐
⇒ 𝑴 = ±𝒋𝒌𝒐
𝑬 𝒛 = 𝑨𝒆 𝒋𝒌𝒐𝒛
+ 𝑪𝒆 𝒋𝒌𝒐𝒛
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𝒖𝒙
𝒖𝒛
𝒖𝒚
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
+ 𝑬𝒐 𝒆𝒋𝒌𝒐𝒛
Direction of propagation
Direction
of oscillation
𝒖𝒙
From Maxwells 1st equation
𝜵 × 𝑬𝒉 = −𝒋𝝎𝝁𝑯𝒉
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁𝒐
𝜵 × 𝑬𝒉
𝒖𝒙
𝒖𝒛
𝒖𝒚
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝒖𝒙 Direction of propagation
Direction
of oscillation
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁𝒐
𝜵 × 𝑬𝒉
𝐻 =
1
−𝑗𝜔𝜇
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐸 𝐸 𝐸
𝐻 =
1
−𝑗𝜔𝜇
𝑢 𝑢 𝑢
0 0
𝜕
𝜕𝑧
𝐸 0 0
=
1
−𝑗𝜔𝜇
𝜕𝐸
𝜕𝑧
𝑢
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𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝒖𝒙
𝒖𝒙
𝒖𝒛
𝒖𝒚
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁𝒐
𝜵 × 𝑬𝒉
𝑯 𝒛 =
1
−𝑗𝜔𝜇
𝑢 𝑢 𝑢
0 0
𝜕
𝜕𝑧
𝐸 0 0
=
1
−𝑗𝜔𝜇
𝜕𝐸
𝜕𝑧
𝑢
=
−𝒋𝒌𝒐 𝑬𝒐
−𝑗𝜔𝜇
𝒆 𝒋𝒌𝒐𝒛
𝑢
𝒌𝒐
𝟐
= 𝝎𝟐
𝝁𝒐𝝐𝒐
𝒌𝒐 = 𝝎 𝝁𝒐𝝐𝒐
𝒌𝒐 =
𝝎
𝑪
Speed of light 𝑪 =
𝟏
𝝁𝒐𝝐𝒐
=
𝝎 𝝁𝒐𝝐𝒐 𝑬𝒐
𝜔𝜇
𝑢
Wave
Number
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝒖𝒙
𝒖𝒙
𝒖𝒛
𝒖𝒚
𝑯 𝒛 =
𝟏
−𝒋𝝎𝝁𝒐
𝜵 × 𝑬𝒉
𝒌𝒐
𝟐
= 𝝎𝟐
𝝁𝒐𝝐𝒐
𝒌𝒐 =
𝝎
𝑪
𝟏
𝝁𝒐𝝐𝒐
=
𝝎 𝝁𝒐𝝐𝒐𝑬𝒐
𝜔𝜇
𝑢
=
𝝐𝒐
𝝁𝒐
𝑬𝒐 𝒆 𝒋𝒌𝒐𝒛
𝑢
=
𝟏
𝜼𝒐
𝑢
𝜼𝒐 =
𝝁𝒐
𝝐𝒐
= 𝟏𝟐𝟎𝝅 𝛀
Intrinsic impedance
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𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎 −− −(𝟏)
𝒖𝒙
𝒖𝒙
𝒖𝒛
𝒖𝒚
𝝀 =
𝒖
𝒇
=
𝟏
𝒇
𝝎
𝒌𝒐
=
𝟐𝝅
𝒌𝒐
𝑯 𝒛 =
𝟏
𝜼𝒐
𝒖𝒚
𝒌𝒐
𝟐
= 𝝎𝟐
𝝁𝒐𝝐𝒐
𝒌𝒐 =
𝝎
𝑪
𝟏
𝝁𝒐𝝐𝒐
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆 𝒋𝒌𝒐𝒛
e 𝒖𝒙
= 𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒐𝒛) 𝒖𝒙
𝑯 𝒛, 𝒕 =
𝟏
𝜼𝒐
𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒐𝒛) 𝒖𝒚
𝝎𝒕 − 𝒌𝒐𝒛 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝝎 − 𝒌𝒐
𝜕𝒛
𝜕𝑡
= 𝟎
𝒖 =
𝝎
𝒌𝒐
𝒖
𝒖 = 𝑪
Wave speed (speed
of light in free space)
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝒐𝝐𝒐𝑬 = 𝟎
𝑬 𝒛, 𝒕 = 𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒐𝒛) 𝒖𝒙
𝑯 𝒛, 𝒕 =
𝟏
𝜼𝒐
𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒐𝒛) 𝒖𝒚
EMANIM: Interactive animation of
electromagnetic waves (szialab.org) File:EM-Wave.gif - Wikimedia Commons
𝒖𝒚 𝒖𝒛
𝒖𝒙
Electromagnetic EM wave
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8
Wave Equations “Lossless dielectric Medium”
2) Lossless dielectric Medium, [𝝈 = 𝟎, 𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝑯 + 𝝎𝟐
𝝁𝝐𝑯 = 𝟎 −− −(𝟐)
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝝐 𝑬 = 𝟎 −− −(𝟏)
𝝏𝟐
𝝏𝒙𝟐
+
𝝏𝟐
𝝏𝒚𝟐
+
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝒌𝟐
𝑬 = 𝟎
𝒖𝒙
𝒖𝒛
𝒖𝒚
Planner waves
𝒌𝟐
= 𝝎𝟐
𝝁𝝐
𝝏𝟐
𝝏𝒛𝟐 𝑬 + 𝒌𝟐
𝑬 = 𝟎  ODE
𝑬 𝒛 = 𝑬𝒐 𝒆 𝒋𝒌𝒛
+ 𝑬𝒐 𝒆𝒋𝒌𝒛
𝒖𝒙
From Maxwells 1st equation
𝜵 × 𝑬𝒉 = −𝒋𝝎𝝁𝑯𝒉
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁
𝜵 × 𝑬𝒉
𝒖𝒙
𝒖𝒛
𝒖𝒚
Direction of propagation
Direction
of oscillation
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝝐 𝑬 = 𝟎 −− −(𝟏)
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𝒖𝒙
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁
𝜵 × 𝑬𝒉
𝐻 =
𝟏
−𝒋𝝎𝝁
𝑢 𝑢 𝑢
0 0
𝜕
𝜕𝑧
𝐸 0 0
=
𝟏
−𝒋𝝎𝝁
𝜕𝐸
𝜕𝑧
𝑢
=
−𝒋𝒌𝑬𝒐
−𝑗𝜔𝝁
𝒆 𝒋𝒌𝒛
𝑢
𝒌𝟐
= 𝝎𝟐
𝝁𝝐
𝒌 = 𝝎 𝝁𝝐
𝒌 =
𝝎
𝒖
𝒖 =
𝟏
𝝁𝝐
=
𝑪
𝝁𝒓𝝐𝒓
=
𝝎 𝝁𝝐𝑬𝒐
𝜔𝝁
𝒆 𝒋𝒌𝒛
𝑢
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝝐 𝑬 = 𝟎 −− −(𝟏)
𝒖𝒙
𝒖𝒙
𝒖𝒛
𝒖𝒚
𝑯 𝒛 =
𝟏
−𝒋𝝎𝝁
𝜵 × 𝑬𝒉
=
𝟏
𝜼
𝑬𝒐 𝒆 𝒋𝒌𝒛
𝑢 𝜼 =
𝝁𝒐𝝁𝒓
𝝐𝒓𝝐𝒐
= 𝟏𝟐𝟎𝝅
𝝁𝒓
𝝐𝒓
𝛀
𝜵𝟐
𝑬 + 𝝎𝟐
𝝁𝝐 𝑬 = 𝟎 −− −(𝟏)
𝑬 𝒛, 𝒕 = 𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒛) 𝒖𝒙
𝑯 𝒛, 𝒕 =
𝟏
𝜼
𝑬𝒐 𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒛) 𝒖𝒚
𝝀 =
𝟐𝝅
𝒌
17
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Wave Equations “Lossy dielectric Medium”
3) Lossy dielectric Medium, [𝝈 ≠ 𝟎, 𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝝁𝝐 𝑬 = 𝟎
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝒌𝒄
𝟐
𝑬 = 𝟎
𝜵𝟐
𝑬 + (−𝒋𝝎𝝁𝝈 + 𝝎𝟐
𝝁𝝐) 𝑬 = 𝟎
𝒌𝒄
𝟐
= 𝝎𝟐
𝝁𝝐 −𝒋𝝎𝝁𝝈
𝒌𝒄
𝟐
= 𝝎𝟐
𝝁𝝐(𝟏 −
𝒋𝝈
𝝎𝝐
)
𝒌𝒄 = 𝝎𝟐𝝁𝝐 𝟏 −
𝒋𝝈
𝝎𝝐
𝟏
𝟐
𝑬 𝒛 = 𝑬𝒐 𝒆 𝒋𝒌𝒄𝒛
𝒖𝒙
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁
𝜵 × 𝑬𝒉 =
𝟏
−𝒋𝝎𝝁
𝑢 𝑢 𝑢
0 0
𝜕
𝜕𝑧
𝐸 0 0
=
𝟏
−𝒋𝝎𝝁
𝜕𝐸
𝜕𝑧
𝑢 =
𝒌𝒄𝑬𝒐
𝜔𝝁
𝒆 𝒋𝒌𝒄𝒛
𝑢
𝒌𝒄 = 𝒌 𝟏 −
𝒋𝝈
𝝎𝝐
𝟏
𝟐
𝟏 + 𝒙 𝒏 = 𝟏 +
𝒏𝒙
𝟏!
+
𝒏 𝒏 − 𝟏 𝒙𝟐
𝟐!
+ ⋯
Binomial Theory
𝒌𝒄 ≃ 𝑘 (𝟏 −
𝒋𝝈
𝟐𝝎𝝐
)
𝒌 = 𝝎 𝝁𝝐
3) Lossy dielectric Medium, [𝝈 ≠ 𝟎, 𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝝁𝝐 𝑬 = 𝟎 𝒌𝒄
𝟐
= 𝝎𝟐
𝒖𝒙
𝑯𝒉 𝒛 =
𝒌𝒄𝑬𝒐
𝝎𝝁
𝒖𝒚
𝒌𝒄 ≃ 𝑘 (𝟏 −
𝒋𝝈
𝟐𝝎𝝐
)
𝒌𝒄 ≃ 𝒌 − 𝒋𝜶
𝑯𝒉 𝒛 =
(𝒌 − 𝒋𝜶)𝑬𝒐
𝝎𝝁
𝒖𝒚
𝛼 =
𝑘𝜎
2𝜔𝜖
𝛼 =
𝜎 𝜔 𝜇𝜖
2𝜔𝜖
𝛼 =
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆 𝒋𝒌𝒄𝒛
e 𝒖𝒙
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆 𝜶𝒛
𝒆 𝒋𝒌𝒛
e 𝒖𝒙
= 𝑬𝒐 𝒆 𝜶𝒛
𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒛) 𝒖𝒙
= ℜ 𝑬𝒐 𝒆 𝒋(𝒌 𝒋𝜶)𝒛
e 𝒖𝒙
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3) Lossy dielectric Medium, [𝝈 ≠ 𝟎,𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝝁𝝐 𝑬 = 𝟎 𝒌𝒄
𝟐
= 𝝎𝟐
𝒖𝒙
𝒌𝒄 ≃ 𝒌 − 𝒋𝜶
𝛼 =
𝑘𝜎
2𝜔𝜖
𝛼 =
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆 𝜶𝒛
𝒆 𝒋𝒌𝒛
e 𝒖𝒙
= 𝑬𝒐 𝒆 𝜶𝒛
𝒄𝒐𝒔 (𝝎𝒕 − 𝒌𝒛) 𝒖𝒙
𝒆 𝒋𝜶𝒛
: Attenuation factor
Wave Equations “Good conductor ”
4) Good conductor , [𝝈 ≠ 𝟎, 𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝝁𝝐 𝑬 = 𝟎
𝝏𝟐
𝝏𝒛𝟐
𝑬 + 𝜸𝟐
𝑬 = 𝟎
𝜵𝟐
𝑬 + (−𝒋𝝎𝝁𝝈 + 𝝎𝟐
𝝁𝝐) 𝑬 = 𝟎
𝜸𝟐 = 𝝎𝟐𝝁𝝐 −𝒋𝝎𝝁𝝈
𝑬 𝒛 = 𝑬𝒐 𝒆 𝒋𝜸𝒛
𝒖𝒙
𝑯𝒉 =
𝟏
−𝒋𝝎𝝁
𝜵 × 𝑬𝒉 =
𝟏
−𝒋𝝎𝝁
𝑢 𝑢 𝑢
0 0
𝜕
𝜕𝑧
𝐸 0 0
=
𝟏
−𝒋𝝎𝝁
𝜕𝐸
𝜕𝑧
𝑢 =
𝜸 𝑬𝒐
𝜔𝝁
𝒆 𝒋𝜸 𝒛
𝑢
𝛾 = −𝑗𝜔𝜇𝜎
𝛾 = −𝑗 𝜔𝜇𝜎
𝑒 = cos
𝜋
2
+
𝑗 sin
𝜋
2
= 0 + 𝑗
𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = cos
𝜋
4
−𝑗 sin
𝜋
4
−𝑗 =
1 − 𝑗
2
𝛾 =
1 − 𝑗
2
𝜔𝜇𝜎
𝛾 = 1 − 𝑗
𝜔𝜇𝜎
2
𝛾 =
1 − 𝑗
𝛿 𝛿 =
2
𝜔𝜇𝜎
Skin depth
Penetration
depth
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12
4) Good conductor , [𝝈 ≠ 𝟎, 𝝁 = 𝝁𝒐𝝁𝒓, 𝝐 = 𝝐𝒐 𝝐𝒓] assume source free 𝝆 = 𝟎
𝜵𝟐
𝑬 + 𝜸𝟐
𝑬 = 𝟎
𝑬 𝒛 = 𝑬𝒐 𝒆 𝒋𝜸𝒛
𝒖𝒙
𝑯𝒉 𝒛 =
𝜸𝑬𝒐
𝝎𝝁
𝒆 𝒋𝜸𝒛
𝒖𝒚
𝑯𝒉 𝒛 =
(
1 − 𝑗
𝛿
)𝑬𝒐
𝝎𝝁
𝒆 𝒋𝜸𝒛
𝒖𝒚
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆 𝒋𝜸𝒛
e 𝒖𝒙
𝑬 𝒛, 𝒕 = ℜ 𝑬𝒐 𝒆
𝒛
𝜹 𝒆
𝒋𝒛
𝜹 e 𝒖𝒙
= ℜ 𝑬𝒐 𝒆
𝒋( )𝒛
e 𝒖𝒙
𝛾 =
1 − 𝑗
𝛿 𝛿 =
2
𝜔𝜇𝜎
= 𝑬𝒐 𝒆
𝒛
𝜹 𝒄𝒐𝒔 (𝝎𝒕 −
𝒛
𝜹
) 𝒖𝒙
23

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1
Time Varying Fields
Applications
Magnetic Diffusion Equation
Applications
“Induction Heating”
By:
Emad Fathy Yassin
Cairo University
∇ 𝐸 − 𝑗𝜔𝜇𝜎𝐸 + 𝜔 𝜇𝜖𝐸 =
∇𝜌
𝜖
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Diffusion Equation
∇ 𝐻 + −𝑗𝜔𝜇𝜎𝐻 = 0
∇ 𝐻 − 𝑗𝜔𝜇𝜎𝐻 + 𝜔 𝜇𝜖𝐻 = 0
Wave Equations
∇ × 𝐸 = −𝑗𝜔𝜇𝐻̅
∇ × 𝐻 = 𝐽̅+𝑗𝜔𝜖𝐸
∇. 𝐷 = 𝜌
∇. 𝐵 = 0
Maxwell’s Equations
∇ 𝐸 − 𝑗𝜔𝜇𝜎𝐸 + 0 =
∇𝜌
𝜖
∇ 𝐻 − 𝑗𝜔𝜇𝜎𝐻 + 0 = 0
Wave Equations
∇ × 𝐸 = −𝑗𝜔𝜇𝐻̅
∇ × 𝐻 = 𝐽̅+ 0
∇. 𝐷 = 𝜌
∇. 𝐵 = 0
Maxwell’s Equations in good conductors
Diffusion Equations
∇ 𝐻 + 𝛾 𝐻 = 0
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2
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
How to Solve Diffusion Equation?
∇ 𝐻 + 𝛾 𝐻 = 0
𝑒 = cos
𝜋
2
+ 𝑗 sin
𝜋
2
= 0 + 𝑗
𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = cos
𝜋
4
− 𝑗 sin
𝜋
4
−𝑗 =
1 − 𝑗
2
𝛾 =
1 − 𝑗
2
𝜔𝜇𝜎
𝛾 = 1 − 𝑗
𝜔𝜇𝜎
2
𝛾 =
1 − 𝑗
𝛿
𝛿 =
2
𝜔𝜇𝜎
Skin depth
‘Penetration
depth’
𝜕
𝜕𝑢
𝐻 + 𝛾 𝐻 = 0
Assume inf. extension in 2 direction depends on
the problem dimensions
𝜕
𝜕𝑢
−−−→ 𝑀
𝑀 𝐻 + 𝛾 𝐻 = 0
𝑀 + 𝛾 = 0
𝑀 = ± 𝑗𝛾
𝑀 = ±𝛽
𝛽 = 𝑗𝛾 =
1 + 𝑗
𝛿
𝐻 = 𝐴 𝑒 + 𝐶 𝑒
How to Solve Diffusion Equation?
∇ 𝐻 + 𝛾 𝐻 = 0
𝑒 = cos
𝜋
2
+ 𝑗 sin
𝜋
2
= 0 + 𝑗
𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = 𝑒
−𝑗 = cos
𝜋
4
− 𝑗 sin
𝜋
4
−𝑗 =
1 − 𝑗
2
𝛾 =
1 − 𝑗
2
𝜔𝜇𝜎
𝛾 = 1 − 𝑗
𝜔𝜇𝜎
2
𝛾 =
1 − 𝑗
𝛿
𝛿 =
2
𝜔𝜇𝜎 Skin depth
Penetration depth
𝛽 = 𝑗𝛾 =
1 + 𝑗
𝛿
𝐻 = 𝐴 𝑒 + 𝐶 𝑒
Using Ampere's low to get H outside material and
direction of H
𝐻 = 𝐴 𝑒 + 𝐶 𝑒
To get 𝐴 𝑎𝑛𝑑 𝐶 Using BCs
Special cases
• Semi inf. extension
• Even symmetry
• Odd symmetry
𝐽̅ = ∇ × 𝐻
𝐸 =
𝐽̅
𝜎
=
1
𝜎
∇ × 𝐻
𝑆̅ =
1
2
𝐸 × 𝐻∗
− 𝑆̅. 𝑑𝑠 = 𝑃 + 𝑗𝑄
𝐻 − 𝐻 = 𝑘
𝐵 = 𝐵
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3
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Assume inf. extension in height == 𝑥 → ∞ and
diameter is very high z → ∞ 𝑧
𝑦
𝑥
𝑘 = 𝑘 cos 𝜔𝑡
(1) Induction Heating
𝑎
𝒙
𝒚
𝒛
Conducting Slab
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Assume inf. Extension in x and z direction
𝜕
𝜕𝑦
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑦 = 𝐴 𝑒 + 𝐶 𝑒
Assume semi-inf. extension in +𝑣𝑒 y direction
𝑦 = ∞ 𝐻 ≠ ∞ 𝑡ℎ𝑒𝑛 𝐶 𝑚𝑢𝑠𝑡 𝑒𝑞𝑢𝑎𝑙 𝑧𝑒𝑟𝑜
𝐻 𝑦 = 𝐴 𝑒
𝐻 𝑦 = 𝐴 𝑒 (−𝑢 )
𝑧
𝑦
𝑥
𝑘 = 𝑘
1
4 3
2
+ + + = 𝐼
𝐻 ∗ 𝑙 + 0 + 𝐻 ∗ 𝑙 + 0 = 𝑘 𝑙
From ampere's low
𝐻
𝑙
=
𝑘
2
𝐻 =
𝑘
2
𝑢𝑠𝑖𝑛𝑔 𝐵𝐶𝑠 @ 𝑦 = 0
𝐻 =
𝑘
2
𝑎
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𝐻 𝑦 = 𝐴 𝑒 (−𝑢 )
𝑧
𝑦
𝑥
𝑘 = 𝑘
1
4 3
2
+ + + = 𝐼
𝐻 ∗ 𝑙 + 0 + 𝐻 ∗ 𝑙 + 0 = 𝑘 𝑙
𝐻 =
𝑘
2
From ampere's low
𝐻
𝑙
=
𝑘
2
𝐻=
𝑘
2
𝑢𝑠𝑖𝑛𝑔 𝐵𝐶𝑠 @ (𝑦 = 0)
𝐻 = 𝐻
1
2
𝑘
2
−𝑢 = 𝐴 (−𝑢 )
𝐻 𝑦 =
𝑘
2
𝑒 (−𝑢 )
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
0
𝜕
𝜕𝑦
0
𝐻 0 0
= −
𝜕
𝜕𝑦
𝐻 𝑢̅
= −
𝑘 𝛽
2
𝑒 𝑢̅
𝑎
𝑧
𝑦
𝑥
𝑘 = 𝑘
1
4 3
2
+ + + = 𝐼
𝐻 ∗ 𝑙 + 0 + 𝐻 ∗ 𝑙 + 0 = 𝑘 𝑙
𝐻 =
𝑘
2
From ampere's low
𝐻
𝑙
=
𝑘
2
𝐻=
𝑘
2
1
2
𝐻 𝑦 =
𝑘
2
𝑒 (−𝑢 )
𝐽̅ = −
𝑘 𝛽
2
𝑒 𝑢
𝐸 =
𝐽̅
𝜎
= −
𝑘 𝛽
2𝜎
𝑒 𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝑘 𝛽
2𝜎
(−𝑢 ) ×
𝑘
2
(−𝑢 )
𝑆̅|@ =
𝑘 𝛽
8𝜎
𝑢
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𝑧
𝑦
𝑥
𝑘 = 𝑘
1
4 3
2
+ + + = 𝐼
𝐻 ∗ 𝑙 + 0 + 𝐻 ∗ 𝑙 + 0 = 𝑘 𝑙
𝐻 =
𝑘
2
From ampere's low
𝐻
𝑙
=
𝑘
2
𝐻=
𝑘
2
1
2
𝐻 𝑦 =
𝑘
2
𝑒 (−𝑢 ) 𝐸 =
𝐽̅
𝜎
= −
𝑘 𝛽
2𝜎
𝑒 𝑢
𝑆̅|@ =
𝑘 𝛽
8𝜎
𝑢
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠 =
− 𝑆̅. 𝑑𝑠 =
𝑘 𝛽
8𝜎
∗ 1 ∗ 1
𝑃 + 𝑗𝑄 =
𝑘 𝛽
8𝜎
𝛽 =
1 + 𝑗
𝛿
=
𝑘
8𝜎
1 + 𝑗
𝛿
𝑃 =
𝑘
8𝜎𝛿
𝑤/𝑚 𝑄 =
𝑘
8𝜎𝛿
𝑉𝐴𝑅/𝑚
𝑏𝑜𝑡𝑡𝑜𝑚 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
( @ 𝑦 = 0)
𝑘 𝛽
8𝜎
𝑢 . dxd𝑧 (−𝑢 )
9

12/21/2022
1
EPMN203: Time Varying
Fields Applications
Magnetic Diffusion Equation Applications
“Transformer Eddy Current Losses”
By:
Emad Fathy Yassin
Cairo University
Transformer Eddy Current Losses
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
1
2

12/21/2022
2
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Assume inf. extension in height == 𝑧 → ∞ and
2ℎ ≫ 2𝑑 −−→ y → ∞
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = 𝐴 𝑒 + 𝐶 𝑒
+ + + = 𝐼
From ampere's low
𝐻. 𝑑𝑙 = 𝐼 1
4
3
2
𝐻 ∗ 𝑙 + 0 + 0 + 0 = 𝑁𝐼
𝐻 = 𝑛𝐼
𝐻
𝐻
𝐻 𝑥 = 𝑑 = 𝐻 𝑥 = −𝑑
Using BCs @ 𝑥 = 𝑑
𝐻 = 𝐻
1
2
𝐻 = 𝐻 𝑥 = 𝑑
= 𝐴 cosh 𝛽𝑑
𝑒𝑣𝑒𝑛 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦 𝑡ℎ𝑒𝑛 𝐴 = 𝐶
𝐻 𝑥 = 𝐴 cosh 𝛽𝑥
𝐴 =
𝐻
cosh 𝛽𝑑
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
∇ 𝐻 + 𝛾 𝐻 = 0
+ + + = 𝐼
From ampere's low
𝐻. 𝑑𝑙 = 𝐼 1
4
3
2
𝐻 ∗ 𝑙 + 0 + 0 + 0 = 𝑁𝐼
𝐻 = 𝑛𝐼
𝐻
𝐻
1
2
𝐻 𝑥 = 𝐴 cosh 𝛽𝑥
𝐴 =
𝐻
cosh 𝛽𝑑
𝐻 𝑥 = 𝐻
cosh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
0 0
0 0 𝐻
= −
𝜕
𝜕𝑥
𝐻 𝑢̅
= −𝐻 𝛽
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
3
4

12/21/2022
3
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
1
4
3
2
𝐻 = 𝑛𝐼
𝐻
𝐻
1
2
𝐻 𝑥 = 𝐻
cosh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝐽̅ = −𝐻 𝛽
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝐸 =
𝐽̅
𝜎
= −
𝐻 𝛽
𝜎
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(−𝑢 ) × 𝐻 (𝑢 )
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(−𝑢 )
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
1
4
3
2
𝐻 = 𝑛𝐼
𝐻
𝐻
1
2
𝐻 𝑥 = 𝐻
cosh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝐽̅ = −𝐻 𝛽
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝐸 =
𝐽̅
𝜎
= −
𝐻 𝛽
𝜎
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(𝑢 ) × 𝐻 (𝑢 )
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(𝑢 )
5
6

12/21/2022
4
𝑧
𝒚
𝑥
𝑥
𝒚
𝑥
𝑦
2𝑑
2ℎ
𝑥
𝑦
2ℎ
2𝑑
1
4
3
2
𝐻 = 𝑛𝐼
𝐻
𝐻
1
2
𝐻 𝑥 = 𝐻
cosh 𝛽𝑥
cosh 𝛽𝑑
𝑢 𝐸 =
𝐽̅
𝜎
= −
𝐻 𝛽
𝜎
sinh 𝛽𝑥
cosh 𝛽𝑑
𝑢
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(𝑢 )
𝑆̅|@ =
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(−𝑢 )
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠
@
− 𝑆̅. 𝑑𝑠
@
= −
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
(−𝑢 ). dyd𝑧 (𝑢 )
−
1
2
𝐻 𝛽
𝜎
sinh𝛽𝑑
cosh 𝛽𝑑
(𝑢 ). dyd𝑧 (−𝑢 )
= 2 ∗
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
∗ 2ℎ ∗ 1 = 𝑃 + 𝑗𝑄
𝑃 + 𝑗𝑄 = 2 ∗
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
∗ 2ℎ ∗ 1
sinh 𝛽𝑑
cosh 𝛽𝑑
cosh (𝑎 + 𝑗𝑏) = cosh 𝑎 cos 𝑏 + 𝑗 sinh 𝑎 sin 𝑏
sinh (𝑎 + 𝑗𝑏) = sinh 𝑎 cos 𝑏 + 𝑗 cosh 𝑎 sin 𝑏
=
sinh
1 + 𝑗
𝛿
𝑑
cosh
1 + 𝑗
𝛿
𝑑
=
sinh
𝑑
𝛿
cos
𝑑
𝛿
+ 𝑗 cosh
𝑑
𝛿
sin
𝑑
𝛿
cosh
𝑑
𝛿
cos
𝑑
𝛿
+ 𝑗 sinh
𝑑
𝛿
sin
𝑑
𝛿
∗
cosh
𝑑
𝛿
cos
𝑑
𝛿
− 𝑗 sinh
𝑑
𝛿
sin
𝑑
𝛿
cosh
𝑑
𝛿
cos
𝑑
𝛿
− 𝑗 sinh
𝑑
𝛿
sin
𝑑
𝛿
=
cosh
𝑑
𝛿
cos
𝑑
𝛿
+ sinh
𝑑
𝛿
sin
𝑑
𝛿
sinh
𝑑
𝛿
cosh
𝑑
𝛿
cos
𝑑
𝛿
+ sin
𝑑
𝛿
+𝑗 cos
𝑑
𝛿
sin
𝑑
𝛿
cosh
𝑑
𝛿
− sinh
𝑑
𝛿
=
sinh
𝑑
𝛿
cosh
𝑑
𝛿
+ 𝑗 cos
𝑑
𝛿
sin
𝑑
𝛿
cosh
𝑑
𝛿
cos
𝑑
𝛿
+ sinh
𝑑
𝛿
sin
𝑑
𝛿
7
8

12/21/2022
5
𝑃 + 𝑗𝑄 = 2 ∗
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
∗ 2ℎ ∗ 1
sinh 𝛽𝑑
cosh 𝛽𝑑
=
sinh
𝑑
𝛿
cosh
𝑑
𝛿
+ 𝑗 cos
𝑑
𝛿
sin
𝑑
𝛿
cosh
𝑑
𝛿
cos
𝑑
𝛿
+ sinh
𝑑
𝛿
sin
𝑑
𝛿
sinh 𝑎 cosh 𝑎 =
1
2
sinh 2𝑎
sin 𝑎 cos 𝑎 =
1
2
sin 2𝑎
cosh 𝑎 =
1
2
1 + cosh 2𝑎
cos 𝑎 =
1
2
1 + cos 2𝑎
sinh a = −1 + cosh 2𝑎
sin 𝑎 =
1
2
1 − cos 2𝑎
=
1
2
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
1
4
1 + cosh
2𝑑
𝛿
1 + cos
2𝑑
𝛿
+ −1 + cosh
2𝑑
𝛿
1 − cos
2𝑑
𝛿
=
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
1
2
1 + cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
+ cosh
2𝑑
𝛿
cos
2𝑑
𝛿
−1 + cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
− cosh
2𝑑
𝛿
cos
2𝑑
𝛿
=
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
1
2
2 cosh
2𝑑
𝛿
+ 2 cos
2𝑑
𝛿
=
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑃 + 𝑗𝑄 = 2 ∗
1
2
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
∗ 2ℎ ∗ 1
sinh 𝛽𝑑
cosh 𝛽𝑑
sinh 𝑎 cosh 𝑎 =
1
2
sinh 2𝑎
sin 𝑎 cos 𝑎 =
1
2
sin 2𝑎
cosh 𝑎 =
1
2
cosh 2𝑎 + 1
cos 𝑎 =
1
2
cos 2𝑎 + 1
sinh a = cosh 2𝑎 − 1
sin 𝑎 =
1
2
− cos 2𝑎 + 1
=
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑃 + 𝑗𝑄 =
𝐻 𝛽
𝜎
sinh 𝛽𝑑
cosh 𝛽𝑑
∗ 2ℎ ∗ 1
𝑃 + 𝑗𝑄 =
𝐻
𝜎
1 + 𝑗
𝛿
sinh
2𝑑
𝛿
+ 𝑗 sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
∗ 2ℎ ∗ 1
𝑃 =
𝐻
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
− sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑤/𝑚
𝑄 =
𝐻
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
+ sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑉𝐴𝑅 /𝑚
9
10

12/21/2022
6
𝑃 =
(𝑛𝐼 )
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
− sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑤/𝑚
𝛿 =
2
𝜔𝜇𝜎
=
2
2𝜋 ∗ 50 ∗ 𝜇 ∗ 5000 ∗ 0.17 ∗ 10 ∗ 10
= 2.4413 ∗ 10
𝑃 =
100 ∗ 8
0.17 ∗ 10 ∗ 10
∗
0.45 ∗ 2
2.4413 ∗ 10
∗
sinh
0.2 ∗ 10
2.4413 ∗ 10
− sin
0.2 ∗ 10
2.4413 ∗ 10
cosh
0.2 ∗ 10
2.4413 ∗ 10
+ cos
0.2 ∗ 10
2.4413 ∗ 10
= 12.49 𝑤/𝑚
From Tailor Series Expansion
sin 𝑥 =
𝑥
1!
−
𝑥
3!
+
𝑥
5!
… … . .
𝑃 =
𝐻
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
− sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑤/𝑚
cos 𝑥 = 1 −
𝑥
2!
+
𝑥
4!
… … . .
sinℎ 𝑥 =
𝑥
1!
+
𝑥
3!
+
𝑥
5!
… … . .
cosh 𝑥 = 1 +
𝑥
2!
+
𝑥
4!
… … . .
𝑃 =
𝐻
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
− sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
∗
1
2𝑑 ∗ 2ℎ
𝑤/𝑚
𝑥
𝑦
2ℎ
2𝑑
𝑃 =
𝐻
𝜎
1
2𝑑 ∗ 𝛿
2𝑑
𝛿
+
1
3!
(
2𝑑
𝛿
) −[
2𝑑
𝛿
−
1
3!
(
2𝑑
𝛿
) ]
1 +
1
2!
(
2𝑑
𝛿
) +[1 −
1
2!
(
2𝑑
𝛿
) ]
𝑤/𝑚
𝑃 =
𝐻
𝜎
1
2𝑑 ∗ 𝛿
2 ∗
1
3!
(
2𝑑
𝛿
)
2
𝑤/𝑚
𝑃 =
𝐻
𝜎
1
2𝑑 ∗ 𝛿
1
3!
(
2𝑑
𝛿
) 𝑤/𝑚
11
12

12/21/2022
7
From Tailor Series Expansion
sin 𝑥 =
𝑥
1!
−
𝑥
3!
+
𝑥
5!
… … . .
𝑃 =
𝐻
𝜎
2ℎ
𝛿
sinh
2𝑑
𝛿
− sin
2𝑑
𝛿
cosh
2𝑑
𝛿
+ cos
2𝑑
𝛿
𝑤/𝑚
cos 𝑥 = 1 −
𝑥
2!
+
𝑥
4!
… … . .
sinℎ 𝑥 =
𝑥
1!
+
𝑥
3!
+
𝑥
5!
… … . .
cosh 𝑥 = 1 +
𝑥
2!
+
𝑥
4!
… … . .
𝑥
𝑦
2ℎ
2𝑑
𝑃 =
𝐻
𝜎
1
2𝑑 ∗ 𝛿
1
3!
(
2𝑑
𝛿
) 𝑤/𝑚
𝑃 =
𝐻
𝜎
1
𝛿
1
3!
(2𝑑) 𝑤/𝑚
𝛿 =
2
𝜔𝜇𝜎
𝑃 =
1
3!
𝐻
𝜎
1
4
∗ 𝜔 𝜇 𝜎 (2𝑑) 𝑤/𝑚
𝑃 =
1
3!
𝐻 𝜇
4
𝜎 ∗ (2𝜋𝐹) (2𝑑) 𝑤/𝑚
𝑃 =
1
3!
𝐵 𝜎 ∗ (𝜋𝐹) (2𝑑) =
𝜋 𝜎
6
𝐹 𝐵 (2𝑑) 𝑤/𝑚
𝑃 = 𝐾 𝐹 𝐵 ∗ 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑤/𝑚
13

12/28/2022
1
EPMN203: Time Varying
Fields Applications
“Bus Bar and Transmission Line”
By:
Emad Fathy Yassin
Cairo University
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
1
2

12/28/2022
2
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Assume inf. extension in length == 𝑧 → ∞ and
2ℎ ≫ 2𝑑 −−→ y → ∞
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
𝐻 𝑥 = 𝑑 = −𝐻 𝑥 = −𝑑
𝐻 = 𝐻
𝐼
4ℎ
= 𝐻 𝑥 = 𝑑
= 𝐶 sinh 𝛽𝑑
𝑜𝑑𝑑 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦 𝑡ℎ𝑒𝑛 𝐴 = −𝐶
𝐻 𝑥 = 𝐶 sinh 𝛽𝑥
𝐶 =
𝐼
4ℎ
1
sinh 𝛽𝑑
1
2
𝐻
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
∇ 𝐻 + 𝛾 𝐻 = 0
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
𝐻 𝑥 =
𝐼
4ℎ
sinh 𝛽𝑥
sinh 𝛽𝑑
𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
0 0
0 𝐻 0
=
𝜕
𝜕𝑥
𝐻 𝑢̅
= 𝛽
𝐼
4ℎ
cosh 𝛽𝑥
sinh 𝛽𝑑
𝑢̅
𝐸 =
𝐽̅
𝜎
=
𝐼
4ℎ
𝛽
𝜎
cosh 𝛽𝑥
sinh 𝛽𝑑
𝑢
𝐽̅ = 𝛽
𝐼
4ℎ
cosh 𝛽𝑥
sinh 𝛽𝑑
𝑢̅
Eddy currents are induced inside the
main conductor due to main AC current.
This causes current redistribution
3
4

12/28/2022
3
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
𝐻 𝑥 =
𝐼
4ℎ
sinh 𝛽𝑥
sinh 𝛽𝑑
𝑢 𝐸 =
𝐼
4ℎ
𝛽
𝜎
cosh 𝛽𝑥
sinh 𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝐼
4ℎ
𝛽
𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
𝑢 ×
𝐼
4ℎ
𝑢
𝑆̅|@ =
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
(−𝑢 )
𝑆̅|@ =
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
(𝑢 )
𝑆̅|@ =
1
2
𝐼
4ℎ
𝛽
𝜎
cosh 𝛽𝑑
sinh𝛽𝑑
𝑢 ×
𝐼
4ℎ
(−𝑢 )
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
𝑆̅|@ =
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
(−𝑢 )
𝑆̅|@ =
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
(𝑢 )
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠
@
− 𝑆̅. 𝑑𝑠
@
= −
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh𝛽𝑑
sinh𝛽𝑑
(−𝑢 ). dyd𝑧 (𝑢 )
−
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh𝛽𝑑
sinh 𝛽𝑑
(𝑢 ). dyd𝑧 (−𝑢 )
= 2 ∗
1
2
𝐼 𝛽
16 ℎ 𝜎
cosh 𝛽𝑑
sinh𝛽𝑑
∗ 2ℎ ∗ 1
=
𝐼 𝛽
8ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
= 𝑃 + 𝑗𝑄
5
6

12/28/2022
4
Bus Bar
𝑥
𝑦
2ℎ
2𝑑
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
− 𝑆̅. 𝑑𝑠 =
𝐼 𝛽
8ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
= 𝑃 + 𝑗𝑄
=
1
2
𝐼 (𝑅 + 𝑗𝑋)
𝑅 + 𝑗𝑋 =
𝛽
4ℎ 𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
𝑅 = ℜ
𝛽
4ℎ 𝜎
cosh 𝛽𝑑
sinh𝛽𝑑
=
1
4ℎ 𝛿 𝜎
sinh
2𝑑
𝛿
+ sin
2𝑑
𝛿
cosh
2𝑑
𝛿
− cos
2𝑑
𝛿
Ω/𝑚
𝑅 =
𝐿
𝜎 𝐴
𝑅
𝑅
=
=
1
𝜎 ∗ 2ℎ ∗ 2𝑑
=
1
4ℎ𝑑 𝜎
Ω/𝑚
𝑑
𝛿
sinh
2𝑑
𝛿
+ sin
2𝑑
𝛿
cosh
2𝑑
𝛿
− cos
2𝑑
𝛿
=
𝛼
2
sinh𝛼 + sin 𝛼
cosh 𝛼 − cos 𝛼
Transmission Line
2ℎ
2𝑑
2ℎ
2𝑑
𝐼
X
𝐼
7
8

12/28/2022
5
Transmission Line
2ℎ
2𝑑
2ℎ
2𝑑
𝐼
X
𝐼
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ 𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
2ℎ
2𝑑
𝐼
2ℎ
2𝑑
X
𝐼
𝐻 =
𝐼
2ℎ
𝐻 = 0
𝐻 = 0
𝑥
𝑦
2ℎ
2𝑑
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝐻 = 0
𝑥
𝑦
Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
2ℎ ≫ 2𝑑 −−→ y → ∞
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = 𝑑 ≠ 𝐻 𝑥 = −𝑑
Using BCs @ 𝑥 = −𝑑
𝐻 = 𝐻
0 = 𝐻 𝑥 = −𝑑
0 = 𝐴 cosh 0 + 𝐶 sinh 0
𝑁𝑜𝑡 𝑒𝑣𝑒𝑛 , 𝑁𝑜𝑡 𝑜𝑑𝑑
𝐻 𝑥 = 𝐴 cosh 𝛽(𝑥 + 𝑑) + 𝐶 sinh 𝛽(𝑥 + 𝑑)
𝐴 = 0
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
𝐻 = 𝐻
𝐼
2ℎ
= 𝐻 𝑥 = 𝑑
= 𝐶 sinh 2𝛽𝑑
𝐶 =
𝐼
2ℎ
1
sinh2𝛽𝑑
𝐻 𝑥 = 𝐶 sinh 𝛽(𝑥 + 𝑑)
𝐻 𝑥 =
𝐼
2ℎ
sinh 𝛽(𝑥 + 𝑑)
sinh 2𝛽𝑑
𝑢
Solution 1
9
10

12/28/2022
6
Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 =
𝐼
2ℎ
sinh 2𝛽𝑑
𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
0 0
0 𝐻 0
=
𝜕
𝜕𝑥
𝐻 𝑢̅
= 𝛽
𝐼
2ℎ
cosh 𝛽(𝑥 + 𝑑)
sinh 2𝛽𝑑
𝑢̅
𝐸 =
𝐽̅
𝜎
=
𝐼
2ℎ
𝛽
𝜎
sinh 2𝛽𝑑
𝑢
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
Solution 1
Transmission Line
𝐻 𝑥 =
𝐼
2ℎ
sinh 2𝛽𝑑
𝑢 𝐸 =
𝐼
2ℎ
𝛽
𝜎
sinh 2𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝐼
2ℎ
𝛽
𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
𝑢 ×
𝐼
2ℎ
𝑢
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 )
𝑆̅|@ = 0
𝑆̅|@ =
1
2
𝐼
4ℎ
𝛽
𝜎
cosh 𝛽𝑑
sinh 𝛽𝑑
𝑢 × 0
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
Solution 1
11
12

12/28/2022
7
Transmission Line
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 )
𝑆̅|@ = 0
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠
@
− 𝑆̅. 𝑑𝑠
@
= −
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 ). dyd𝑧 (𝑢 )
=
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh2𝛽𝑑
sinh 2𝛽𝑑
∗ 2ℎ ∗ 1
=
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
Solution 1
Transmission Line
− 𝑆̅. 𝑑𝑠 =
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
=
1
2
𝑅 + 𝑗𝑋 =
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh2𝛽𝑑
𝑅 = ℜ
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
+ sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
𝑋 = Ι𝑚
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
− sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
Solution 1
13
14

12/28/2022
8
Transmission Line
2ℎ
2𝑑
2ℎ
2𝑑
𝐼
X
𝐼
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ 𝐻 =
𝐼
4ℎ
𝐻 =
𝐼
4ℎ
2ℎ
2𝑑
𝐼
2ℎ
2𝑑
X
𝐼
𝐻 =
𝐼
2ℎ
𝐻 = 0
𝐻 = 0
𝑥
𝑦
2ℎ
2𝑑
+ + + = 𝐼
From ampere's low
4
3
𝐻 ∗ 2ℎ + 0 + 𝐻 ∗ 2ℎ + 0 = 𝐼
𝐻 =
𝐼
4ℎ
𝐼
𝐻
1
2
𝐻
𝑥
𝑦
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
2ℎ ≫ 2𝑑 −−→ y → ∞
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = 𝑑 ≠ 𝐻 𝑥 = −𝑑
𝐻 = 𝐻
0 = 𝐻 𝑥 = 𝑑
𝐻 𝑥 = 𝐴 cosh 𝛽(𝑥 − 𝑑) + 𝐶 sinh 𝛽(𝑥 − 𝑑)
𝐴 = 0
Using BCs @ 𝑥 = −𝑑
𝐻 = 𝐻
𝐼
2ℎ
= 𝐻 𝑥 = −𝑑
= −𝐶 sinh2𝛽𝑑
𝐶 = −
𝐼
2ℎ
1
sinh 2𝛽𝑑
𝐻 𝑥 = 𝐶 sinh 𝛽(𝑥 − 𝑑)
𝐻 𝑥 = −
𝐼
2ℎ
sinh 𝛽(𝑥 − 𝑑)
sinh 2𝛽𝑑
𝑢
𝑥
𝑦
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
15
16

12/28/2022
9
Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = −
𝐼
2ℎ
sinh 2𝛽𝑑
𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
0 0
0 𝐻 0
=
𝜕
𝜕𝑥
𝐻 𝑢̅
= −𝛽
𝐼
2ℎ
cosh 𝛽(𝑥 − 𝑑)
sinh 2𝛽𝑑
𝑢̅
𝐸 =
𝐽̅
𝜎
= −
𝐼
2ℎ
𝛽
𝜎
sinh 2𝛽𝑑
𝑢
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
Transmission Line
𝐻 𝑥 = −
𝐼
2ℎ
sinh 2𝛽𝑑
𝑢 𝐸 = −
𝐼
2ℎ
𝛽
𝜎
sinh 2𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ = −
1
2
𝐼
2ℎ
𝛽
𝜎
cosh 2𝛽𝑑
sinh2𝛽𝑑
𝑢 ×
𝐼
2ℎ
(𝑢 )
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(𝑢 )
𝑆̅|@ = 0
𝑆̅|@ =
1
2
𝐼
4ℎ
𝛽
𝜎
cosh 0
sinh2𝛽𝑑
𝑢 × 0
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
17
18

12/28/2022
10
Transmission Line
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(𝑢 )
𝑆̅|@ = 0
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠
@
− 𝑆̅. 𝑑𝑠
@
= −
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(𝑢 ). dyd𝑧 (−𝑢 )
=
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh2𝛽𝑑
sinh 2𝛽𝑑
∗ 2ℎ ∗ 1
=
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
Transmission Line
− 𝑆̅. 𝑑𝑠 =
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
=
1
2
𝑅 + 𝑗𝑋 =
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh2𝛽𝑑
𝑅 = ℜ
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
+ sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
𝑋 = Ι𝑚
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
− sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
𝐻 =
𝐼
2ℎ
2ℎ
2𝑑
X
𝐻 = 0
Solution 2
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Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
2ℎ ≫ 2𝑑 −−→ y → ∞
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = 𝑑 ≠ 𝐻 𝑥 = −𝑑
Using BCs @ 𝑥 = 0
𝐻 = 𝐻
0 = 𝐻 𝑥 = 0
𝐻 𝑥 = 𝐴 cosh 𝛽𝑥 + 𝐶 sinh 𝛽𝑥
𝐴 = 0
Using BCs @ 𝑥 = 2𝑑
𝐻 = 𝐻
𝐼
2ℎ
= 𝐻 𝑥 = 2𝑑
= 𝐶 sinh 2𝛽𝑑
𝐶 =
𝐼
2ℎ
1
sinh2𝛽𝑑
𝐻 𝑥 = 𝐶 sinh 𝛽𝑥
𝐻 𝑥 =
𝐼
2ℎ
sinh 𝛽𝑥
sinh 2𝛽𝑑
𝑢
2ℎ
2𝑑
𝐼
𝐻 =
𝐼
2ℎ
𝑥
𝑦
𝐻 = 0
Solution 3
Transmission Line
∇ 𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 =
𝐼
2ℎ
sinh 𝛽𝑥
sinh 2𝛽𝑑
𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝐻 𝐻 𝐻
=
𝑢 𝑢 𝑢
𝜕
𝜕𝑥
0 0
0 𝐻 0
=
𝜕
𝜕𝑥
𝐻 𝑢̅
= 𝛽
𝐼
2ℎ
cosh 𝛽𝑥
sinh 2𝛽𝑑
𝑢̅
𝐸 =
𝐽̅
𝜎
=
𝐼
2ℎ
𝛽
𝜎
cosh 𝛽𝑥
sinh 2𝛽𝑑
𝑢
2ℎ
2𝑑
𝐼
𝑥
𝑦
𝐻 = 0
Solution 3
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12
Transmission Line
𝐻 𝑥 =
𝐼
2ℎ
sinh 𝛽𝑥
sinh 2𝛽𝑑
𝑢 𝐸 =
𝐼
2ℎ
𝛽
𝜎
cosh 𝛽𝑥
sinh 2𝛽𝑑
𝑢
𝑆̅ =
1
2
𝐸 × 𝐻∗
𝑆̅|@ = 𝐸| × 𝐻|
∗
𝑆̅|@ =
1
2
𝐼
2ℎ
𝛽
𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
𝑢 ×
𝐼
2ℎ
(𝑢 )
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 )
𝑆̅|@ = 0
𝑆̅|@ =
1
2
𝐼
4ℎ
𝛽
𝜎
cosh 0
sinh 2𝛽𝑑
𝑢 × 0
2ℎ
2𝑑
𝐼
𝑥
𝑦
𝐻 = 0
Solution 3
Transmission Line
𝑆̅|@ =
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 )
𝑆̅|@ = 0
− 𝑆̅. 𝑑𝑠 = − 𝑆̅. 𝑑𝑠
@
− 𝑆̅. 𝑑𝑠
@
= −
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
(−𝑢 ). dyd𝑧 (𝑢 )
=
1
2
𝐼 𝛽
4 ℎ 𝜎
cosh2𝛽𝑑
sinh 2𝛽𝑑
∗ 2ℎ ∗ 1
=
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
2ℎ
2𝑑
𝐼
𝑥
𝑦
𝐻 = 0
Solution 3
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13
Transmission Line
− 𝑆̅. 𝑑𝑠 =
𝐼 𝛽
4ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
= 𝑃 + 𝑗𝑄
=
1
2
𝑅 + 𝑗𝑋 =
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh2𝛽𝑑
𝑅 = ℜ
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
+ sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
𝑋 = Ι𝑚
𝛽
2ℎ 𝜎
cosh 2𝛽𝑑
sinh 2𝛽𝑑
=
1
2ℎ 𝛿 𝜎
sinh
4𝑑
𝛿
− sin
4𝑑
𝛿
cosh
4𝑑
𝛿
− cos
4𝑑
𝛿
Ω/𝑚
2ℎ
2𝑑
𝐼
𝑥
𝑦
𝐻 = 0
Solution 3
Proximity Effect
Proximity effect is the current redistribution mechanism due to changing magnetic fields from currents
in other nearby conductors
Litz Wire, HF-Litz, High Frequency Litz Wire, Litz
Wire for High Efficiency | ELEKTRISOLA
25
26

12/27/2022
1
EPE2020: Time Varying
Fields Applications
“Electromagnetic Shielding”
By:
Emad Fathy Yassin
Cairo University
Electromagnetic Shielding
1
2

12/27/2022
2
∇ 𝐻 + 𝛾 𝐻 = 0
(
𝜕
𝜕𝑥
+
𝜕
𝜕𝑦
+
𝜕
𝜕𝑧
)𝐻 + 𝛾 𝐻 = 0
Assume inf. extension in height == 𝑧 → ∞ and
h >> t then y → ∞
𝒙
𝒚
𝒛
𝜕
𝜕𝑥
𝐻 + 𝛾 𝐻 = 0
𝐻 𝑥 = 𝐴 𝑒 + 𝐶 𝑒 𝑢
𝐻 𝑥 = 𝑎 ≠ 𝐻 𝑥 = 𝑎 + 𝑡
𝑁𝑜𝑡 𝑒𝑣𝑒𝑛 , 𝑁𝑜𝑡 𝑜𝑑𝑑 𝐻
𝐻
𝐻
Using BCs @ 𝑥 = 𝑎
𝐻 = 𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
Using BCs @ 𝑥 = 𝑎 + 𝑡
𝐻 = 𝐻
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
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3
𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
𝐻 𝑥 = 𝐴 𝑒 + 𝐶 𝑒 𝑢
𝐽̅ = ∇ × 𝐻
𝐽̅ =
𝑢̅ 𝑢̅ 𝑢̅
𝜕
𝜕𝑥
0 0
0 𝐻 0
=
𝜕
𝜕𝑥
𝐻 𝑢̅
𝐽̅ 𝑥 = 𝛽 −𝐴 𝑒 + 𝐶 𝑒 𝑢
𝐸 𝑥 = −𝐴 𝑒 + 𝐶 𝑒 𝑢
𝐹𝑟𝑜𝑚 𝐹𝑎𝑟𝑎𝑑𝑎𝑦 𝑠 𝑙𝑜𝑤
𝑒𝑚𝑓 = −
𝑑𝜙
𝑑𝑡
𝑒𝑚𝑓 = −𝑗𝜔 𝐵 ∗ 𝐴𝑟𝑒𝑎
𝑒𝑚𝑓 = −𝑗𝜔 𝜇 𝐻 ∗ 𝐴𝑟𝑒𝑎
𝐻
𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
𝐸 𝑥 = −𝐴 𝑒 + 𝐶 𝑒 𝑢
𝐹𝑟𝑜𝑚 𝐹𝑎𝑟𝑎𝑑𝑎𝑦 𝑠 𝑙𝑜𝑤
𝑒𝑚𝑓 = −𝑗𝜔 𝐵 ∗ 𝐴𝑟𝑒𝑎
𝑒𝑚𝑓 = −𝑗𝜔 𝜇 𝐻 ∗ 𝐴𝑟𝑒𝑎
𝐻
𝐸. 𝑑𝑙 = −𝑗𝜔 𝜇 𝐻 ∗ 𝐴𝑟𝑒𝑎
𝐸(𝑥 = 𝑎). 𝑑𝑙 ≈ −𝑗𝜔 𝜇 𝐻 ∗ 2𝑎 ∗ 𝐿
𝐸 𝑥 = 𝑎 ∗ 2𝐿 ≈ −𝑗𝜔 𝜇 𝐻 ∗ 2𝑎 ∗ 𝐿
𝛽
𝜎
−𝐴 𝑒 + 𝐶 𝑒 ≈ −𝑗𝜔 𝜇 𝐻 ∗ 𝑎
𝛽 −𝐴 𝑒 + 𝐶 𝑒 ≈ −𝑗𝜔 𝜇 𝜎 𝐻 ∗ 𝑎
𝛾 = −𝑗𝜔 𝜇 𝜎
𝛽 = 𝑗 𝛾
𝛽 = −𝛾
𝛽 −𝐴 𝑒 + 𝐶 𝑒 ≈ −𝛽 𝐻 ∗ 𝑎
−𝛽 𝐻 𝑎 = −𝐴 𝑒 + 𝐶 𝑒 (3)
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𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
𝐻
−𝛽 𝐻 𝑎 = −𝐴 𝑒 + 𝐶 𝑒 (3)
𝑒𝑞𝑛. 1 + 𝑒𝑞𝑛. (3)
𝐻 (1 − 𝛽𝑎) = 2𝐶 𝑒
𝐶 =
𝐻 1 − 𝛽𝑎
2
∗ 𝑒
𝑠𝑢𝑏. 𝑖𝑛 𝑒𝑞𝑛. 1
𝐴 =
𝐻 1 + 𝛽𝑎
2
∗ 𝑒
𝑠𝑢𝑏. 𝑖𝑛 𝑒𝑞𝑛. 2 𝑡𝑜 𝑔𝑒𝑡 𝑡
𝐻 =
𝐻 1 + 𝛽𝑎
2
∗ 𝑒 𝑒 ( ) +
𝐻 1 − 𝛽𝑎
2
∗ 𝑒 𝑒 ( )
𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
𝐻
−𝛽 𝐻 𝑎 = −𝐴 𝑒 + 𝐶 𝑒 (3)
𝐶 =
𝐻 1 − 𝛽𝑎
2
∗ 𝑒
𝐴 =
𝐻 1 + 𝛽𝑎
2
∗ 𝑒
𝑠𝑢𝑏. 𝑖𝑛 𝑒𝑞𝑛. 2 𝑡𝑜 𝑔𝑒𝑡 𝑡
𝐻 =
𝐻 1 + 𝛽𝑎
2
∗ 𝑒 𝑒 ( ) +
𝐻 1 − 𝛽𝑎
2
∗ 𝑒 𝑒 ( )
𝐻
𝐻
=
1 + 𝛽𝑎
2
𝑒 +
1 − 𝛽𝑎
2
𝑒
𝐻
𝐻
=
𝑒 + 𝑒
2
− 𝛽𝑎
𝑒 − 𝑒
2
𝐻
𝐻
= cosh 𝛽𝑡 − 𝛽𝑎 sinh 𝛽𝑡
cosh 𝑥 = 1 +
𝑥
2!
+
𝑥
4!
… … …
sinh 𝑥 =
𝑥
1!
+
𝑥
3!
+
𝑥
5!
… … …
𝐻
𝐻
≈ 1 − 𝛽𝑎 ∗ 𝛽𝑡
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12/27/2022
5
𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻 = 𝐴 𝑒 + 𝐶 𝑒 (1)
𝐻 = 𝐴 𝑒 ( ) + 𝐶 𝑒 ( ) (2)
𝐻
−𝛽 𝐻 𝑎 = −𝐴 𝑒 + 𝐶 𝑒 (3)
𝐶 =
𝐻 1 − 𝛽𝑎
2
∗ 𝑒
𝐴 =
𝐻 1 + 𝛽𝑎
2
∗ 𝑒
cosh 𝑥 = 1 +
𝑥
2!
+
𝑥
4!
sinh𝑥 =
𝑥
1!
+
𝑥
3!
+
𝑥
5!
𝐻
𝐻
≈ 1 − 𝛽 𝑎𝑡
𝐻
𝐻
≈ 1 − 𝑗𝜔 𝜇 𝜎𝑎𝑡
𝛾 = −𝑗𝜔 𝜇 𝜎
𝛽 = −𝛾
𝐻
𝐻
≈ 1 + 𝜔 𝜇 𝜎𝑎𝑡
𝑡 ≈
1
𝜔 𝜇 𝜎𝑎
𝐻
𝐻
− 1
𝒙
𝒚
𝒛
𝐻
𝐻
𝐻
𝐻
𝑡 ≈
1
𝜔 𝜇 𝜎𝑎
𝐻
𝐻
− 1
𝑡 ≈
1
2𝜋 ∗ 50 ∗ 𝜇 ∗ 10 ∗ 3
4
1
− 1 ≈ 0.327 mm
𝑡 ≈
1
2𝜋 ∗ 50 ∗ 𝜇 ∗ 10 ∗ 3
8
1
− 1 ≈ 0.67 mm
𝑐𝑎𝑠𝑒 (𝑎)
𝑐𝑎𝑠𝑒 (𝑏)
9
10

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