30100347 Electromagnetic Theory

S u m m a r y C h a p t e r 1 V e c t o r A l g e b r a 1 Scalars and Vectors Unit Vector Vector Addition, Subtraction Vector Multiplication C h a p t e r 2 C o o r d i n a t e S y s t e m s 5 Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates C h a p t e r 3 V e c t o r C a l c u l u s 7 Differential Length, Area and Volume Del Operator Gradient of a Scalar Divergence of a Vector Curl of a Vector Laplacian of a Scalar Classification of Vector Fields C h a p t e r 4 E l e c t r o s t a t i c F i e l d s 1 5 Coulomb’s Law and Field Intensity Electric Flux Density Gauss’s Law Electric Potential Electrostatic Energy C h a p t e r 5 E l e c t r i c F i e l d s i n M a t e r i a l s 3 3 Convection and Conduction Current Conductors Dielectrics Boundary Conditions C h a p t e r 6 P o i s s o n ’ s & L a p l a c e ’ s E q u a t i o n s – C a p a c i t a n c e 4 3 Poisson’s & Laplace’s Equations Capacitance C h a p t e r 7 M a g n e t o s t a t i c F i e l d s 4 5 Biot-Savart’s Law and Magnetic Flux Intensity Ampere’s Circuit Law Magnetic Flux Density Magnetic Vector Potential

C h a p t e r 8 M a g n e t i c F o r c e s a n d M a t e r i a l s – I n d u c t a n c e 5 3 Magnetic Forces Magnetization Magnetic Materials Boundary Conditions Inductance Magnetic Energy C h a p t e r 9 M a x w e l l ’ s E q u a t i o n s 6 3 Faraday’s law Displacement Current Maxwell’s Equations Time Varying Potentials Time Harmonic Fields C h a p t e r 1 0 E l e c t r o m a g n e t i c P l a n e W a v e 7 1 Waves Equations in Source-Free Region Electromagnetic Power and Poynting Theorem Solution of Uniform, Time-Harmonic Plane Wave Motion T u t o r i a l s Tutorial 1 Tutorial 2 Tutorial 3 Tutorial 4 Tutorial 5 Tutorial 6 Tutorial 7 Tutorial 8 Tutorial 9

Chapter 1

V E C T O R A L G E B R A I. Scalars and Vectors

Scalar – a quantity that has only magnitude (exe: time (t), mass (m), distance (d), temperature (T)) Vector – a quantity that has both magnitude and direction (exe: velocity ( vr ), force ( F

r), displacement ( d

r) )

Field – a function that specifies a particular quantity everywhere in a region. If the quantity is a scalar (or vector), the field is said to be a scalar field (or vector field).

II. Unit Vector

A unit vector Au along a vector Ar

is defined as a vector whose magnitude is unity and its

direction is along Ar

;

AA

AAuA

r

r

r

== where AA =r

: magnitude of Ar

Note that 1=Au .

We may write : AuAA .=r

In Cartesian (rectangular) coordinates: zzyyxx uAuAuAA ++=r

Where Ax , Ay , Az : the components of Ar

in the x, y, z directions respectively.

zyx uuu ,, : unit vectors in the x, y, z directions respectively.

We may write 222zyx AAAAA ++==

r

222

zyx

zzyyxxA

AAA

uAuAuAu

++

++=

Ax

z

Au

Ay

Az

Ar

z

xu yu

zu

x

y y

x

1

III. Vector Addition, Subtraction

Given 2 vectors ; zzyyxx

zzyyxx

uBuBuBB

uAuAuAA

++=

++=r

r

Addition :

zzzyyyxxx uBAuBAuBA

BAC

)()()( +++++=

+=rrr

Subtraction :

zzzyyyxxx uBAuBAuBA

BAD

)()()( −+−+−=

−=rrr

3 basic laws of algebra obeyed by any given vectors CBA

rrr,, :

LAW Addition Multiplication

Commutative ABBArrrr

+=+ kAAkrr

=

Associative ( ) ( ) CBACBArrrrrr

++=++ AkAkr

lrl )()( =

Distributive ( ) BkAkBAkrrrr

+=+

IV. Position and Distance Vector

The position vector Pr (radius vector) of point P is the directed distance from the origin O to P. (Cartesian :) zyxP uzuyuxOPr ++==

The distance vector (or separation vector) in Cartesian of P andP to Q; QyPQxPQPQPQ zuyyuxxrrr ()()( +−+−=−=

Pr

P

z P

O

y

x

2

Q is th

Pz−

Pr

PQr

e displacem

zu)

Qr

ent from

Q

V. Vector Multiplication V.1. Scalar (or dot) product

ABBABA θcos... =rr

ABθ : the smaller angle between A

r and B

r

(In Cartesian) zzyyxx BABABABA .... ++=

rr

Scalar product obeys the following : • Commutative : ABBA

rrrr.. =

• Distributive : ( ) CABACBArrrrrrr

... +=+

• 22. AAAA ==

rrr

• 0. =⇒⊥ BABArrrr

(Cartesian :) ⎪⎩

⎪⎨⎧

==

1.0.

nn

mn

uuuu with

⎩⎨⎧

≠∈

mnzyxmn ,,,

V.2. Cross product

nAB uBABABA .sin.. θ=∧=×rrrr

ABθ : the smaller angle between A

r and B

r

nu : unit vector, normal to the plane containing Ar

and Br

. The direction of

nu is defined using the right hand rule. Remark : (Lagrange’s Identity) ( )2222 .vuvuvu rrrrrr

−=∧

Br

Ar

BArr

∧

θAB

BArr

∧

Ar

Br

θAB

3

( ) ( ) ( ) zxyyxyzxxzxyzzy

z

y

x

z

y

x

uBABAuBABAuBABABBB

AAA

BA ... −+−+−=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛∧

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=∧

rr

or,

( ) ( ) ( ) zxyyxyzxxzxyzzy

zyx

zyx

zyx

uBABAuBABAuBABABBBAAAuuu

BA ... −+−+−==∧rr

Basic properties: i) It is not commutative: ABBA

rrrr∧≠∧

It is anticommutative: ( )ABBArrrr

∧−=∧

ii) It is not associative: ( ) ( ) CBACBArrrrrr

∧∧≠∧∧

iii) It is distributive: ( ) CABACBArrrrrrr

∧+∧=+∧

iv) Ar

parallel to 0rrrr

=∧⇒ BAB V.3. Scalar Triple Product

( ) ( ) ( )BACACBCBArrrrrrrrr

∧=∧=∧ V.4. Vector Triple Product

( ) ( ) ( )BACCABCBArrrrrrrrr

.. −=∧∧

Remark : ( ) ( )CBACBArrrrrr

.. ≠ , but ( ) ( )BACCBArrrrrr

.. =

4

Chapter 2

S Y S T E M S O F C O O R D I N A T E We shall restrict ourselves to 3 coordinate systems: Cartesian, Cylindrical and Spherical coordinate systems. I. Cartesian coordinate system

A point P can be represented as (x,y,z) and the ranges of the coordinate variables x, y, z are:

+∞<<∞−+∞<<∞−+∞<<∞−

zyx

A vector A

r in Cartesian can be written as:

zzyyxx uAuAuAA ++=r

where zyx uuu ,, are unit vectors along x, y and z directions.

II. Circular cylindrical coordinates (ρ, φ, z)

The ranges of variables are:

+∞<<∞−<≤+∞<≤

zπφ

ρ20

0

Given a vector A

r;

zz uAuAuAA ++= φφρρ

r

5

III. Spherical coordinates (r, θ, φ)

The ranges of variables are:

πφπθ20

00

<≤<≤+∞<≤ r

Given a vector A

r;

zzrr uAuAuAuAA +++= φφθθ

r

Note that the direction of φu is the same as the direction of φu in cylindrical coordinate system.

6

Chapter 3

V E C T O R C A L C U L U S I. Differential length, area and volume

I.1. Cartesian coordinates

i) Differential displacement : zyx udzudyudxd ... ++=l

ii) Differential normal area :

z

y

x

udydx

udzdx

udzdydS

..

..

..=

Remark : nn umudSdS ... l== l

iii) Differential volume : dzdydxdv ..= Remark : dv is a scalar

dz

dx

x

O

zu

z

ddy

z

x

y O

dy

7

dS m

un

y

dx

dx

xu

yu dz

y dz

I.2. Cylindrical coordinates

i) Differential displacement : zudzududd .... ++= φρ φρρl

ii) Differential normal area :

zudd

udzd

udzddS

...

..

...

ρφρ

ρ

φρ

φ

ρ=

iii) Differential volume : dzdddv ... φρρ=

z

dρ

dz

ρ.dφ

x

y

I.3. Spherical coordinates

i) Differential displacement : φθ φθθ udrudrudrd r ..sin... ++=l

ii) Differential normal area :

φ

θ

θ

φθ

φθθ

uddrr

uddrr

uddrdS r

...

...sin

...sin2=

iii) Differential volume : φθθ dddrrdv ...sin.2=

z

dr

r.dθ

ρ. dφ=r.sinθ.dφ

y

x

8

I.4. Line, surface and volume integrals Given a vector field A

r and a curve L (curve, or line, or contour), we define the line

integral of Ar

around L as :

∫ ∫=L

b

adAdA l

rl

r.cos. θ

If the path of integration is a closed curve (or patintegral becomes a closed contour integral ; and we as :

∫L dA lr.

CIRCULATION of Ar

around Given a vector field A

r, continuous in a region co

define the surface integral (or the flux) of Ar

through

∫∫∫∫∫∫

=

=

=

S

S n

S

dSA

dSuA

dSA

.cos

..

.

θ

ψ

r

r

r

Remark : for a closed surface (defining a volume),

∫∫=S

dSA.r

ψ

We define the volume integral of the scalar ρv over th ∫∫∫v v dv.ρ

b

c

θ

Ar

9

h) such as a-b-c-a (see figure), the define the circulation of A

r around L

L

ntaining the smooth surface S, we S as :

we ha

e volu

path L a

ld

Ar

u

ve :

me v, as :

surface S dS

n

θ

II. Del operator The del operator, written∇

r, is the vector differential operator ;

Cartesian : zyx uz

uy

ux

⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇r

Vector differential operator / Gradient operator

Cylindrical : zuz

uu ⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∇ φρ φρρ1r

Spherical : φθ φθθu

ru

ru

r r ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇sin11r

III. Gradient of a scalar

The gradient of a scalar V is a vector that represents both the magnitude and the direction of

the maximum space rate of increase of V. [ also noted: V∇r

or )(Vgrad ]

Cartesian : zyx uzVu

yVu

xVV ⎟

⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇r

Cylindrical : zuzVuVuVV ⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∇ φρ φρρ1r

Spherical : φθ φθθuV

ruV

ru

rVV r ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇sin11r

Properties: • ( ) UVUV ∇+∇=+∇

rrr

• ( ) VUUVVU ∇+∇=∇rrr

• 2UUVVU

UV ∇−∇

=⎟⎠⎞

⎜⎝⎛∇

rrr

• VnVV nn ∇=∇ −rr

1 U,V : scalar n : integer

Remarks : • The magnitude of V∇

r equals the maximum rate of change in V per unit distance.

• V∇r

points the direction of the maximum rate of change. • V∇

r at any point is perpendicular to the constant V surface which passes through that

point. • If VA ∇=

rr, V is said to be the scalar potential of A

r.

10

IV. Divergence of a vector – DIVERGENCE THEOREM The divergence of vector A

r is the net outward flow per unit volume over a closed

incremental surface, or, the divergence of Ar

at a given point P is the outward flux per unit volume as the volume shrinks about P.

Cartesian : z

Ay

Ax

AA zyx

∂∂

+∂

∂+

∂∂

=∇rr

Cylindrical : ( )z

AAAA z

∂∂

+∂

∂+

∂∂

=∇φρ

ρρρ

φρ

1.1rr

Spherical : ( ) ( )φθ

θθθ

φθ ∂

∂+

∂∂

+∂∂

=∇A

rA

rAr

rrA r sin

1sinsin1.1 2

2

rr

Properties: • The divergence of a vector is a scalar. • ( ) BABA

rrrrrrr∇+∇=+∇

• There is no divergence of a scalar V. D I V E R G E N C E T H E O R E M (or GAUSS-OSTROGRADSKY THEOREM) The total outward flux of a vector field A

r through the closed surface S is the same as the volume integral of

the divergence of Ar

.

Where, ∫∫∫∫∫ ∇=vS

dvAdSA ...rrr

11

V. Curl (Rotational) of a vector – STOKE’S THEOREM The curl (or rotational) of a vector field A

r is an axial (or rotational) vector whose magnitude

is the maximum circulation of Ar

per unit area as the area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum. Cartesian :

zxy

yzx

xyz

z

y

x

uyA

xA

uyA

zA

uz

AyA

AAA

z

y

x

AArotAcurl

⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥

⎦

⎤⎢⎣

⎡∂∂

−∂∂

+⎥⎦

⎤⎢⎣

⎡∂

∂−

∂∂

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛∧

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

∂∂∂∂∂∂

=

∧∇==rrrr

)()(

Cylindrical :

( )z

zz uAA

uA

zA

uz

AAA ⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂

∂+⎥

⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂∂

=∧∇φρρ

ρρρφρ

ρφφ

ρρ

φ 111rr

Spherical :

( ) ( ) ( )φ

θθ

φθφ

θφθφθθ

θu

Ar

rAr

ur

rAAr

uAA

rA rr

r ⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂∂

+⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂=∧∇

1sin

11sinsin1rr

Properties: ( ) BABA

rrrrrrr∧∇+∧∇=+∧∇•

• ( ) ( ) ( ) ( ) ( )BAABABBABArrrrrrrrrrrrrrr

∇−∇+∇−∇=∧∧∇ ....

• ( ) 0. =∧∇∇ Arrr

(the divergence of the rotational of a vector field vanishes)

• 0rrr

=∇∧∇ V (the rotational of the gradient of a scalar field vanishes)

S T O K E ’ S T H E O R E M (or ROTATIONAL THEOREM) The circulation of A

r around a closed path L is equal to the surface integral of the curl of A

r over the open

surface S bounded by L. ( Ar

and Arr

∧∇ are continuous on S)

Where, ( )∫∫∫ ∧∇=SL

dSAdA ..rr

lr

dSd ,l : the direction obtained using right-hand rule.

12

VI. Laplacian of a scalar The Laplacian of the scalar field V, is the divergence of the gradient of V.

Cartesian : 2

2

2

2

2

22

zV

yV

xVV

∂∂

+∂∂

+∂∂

=∇r

Cylindrical : 2

2

2

2

22 11

zVVVV

∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∇φρρ

ρρρ

r

Spherical :

2

2

2222

22

sin1sin

sin11

φθθθ

θθ ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇V

rV

rrVr

rrV

r

Properties: • A scalar field V is said to be harmonic in a given region if its

laplacian vanishes in that region, or: 02 =∇ V

r (Laplace’s Equation)

• The laplacian of a vector A

r :

( ) AAArrrrrrrr

∧∇∧∇−∇∇=∇ .2

(Cartesian :) zzyyxx uAuAuAA 2222 ∇+∇+∇=∇rrrrr

VII. Classification of vector fields A vector field is uniquely characterized by its divergence and curl. • A divergenceless field is solenoidal : 0. =∇ A

rr

• A curl-free field (the circulation of the vector field around a closed path is zero) is

irrotational :

0rrr

=∧∇ A A solenoidal (or divergenceless) field has neither source nor sink of flux. An irrotational field is also known as a conservative field (the line integral of an irrotational field is independent of the chosen path)

13

We may classify vector fields as : 1. Solenoidal and irrotational.

(exe : a static electric field in a charge-free region)

2. Solenoidal but not irrotational. (exe : a steady magnetic field in a current-carrying conductor)

3. Irrotational but not solenoidal. (exe : a static electric field in a charged-region)

4. Neither solenoidal nor irrotational. (exe : an electric field in charged medium with a time-varying field)

H E M H O L T Z ’ S T H E O R E M

Any vector Ar

satisfying the equations : ⎭⎬⎫

⎩

⎧

=∧∇=∇

S

v

AA

ρρ

rr

rr.

⎨ (where vρ can be regarded as the source density

of Ar

and Sρ its circulation density) with both vρ and Sρ vanishing at infinity can be written as the sum of 2 vectors: one irrotational, the other solenoidal. Where, BVA

rrrr∧∇+∇−=

If we let VAi ∇−=r

and BAS

rr∧∇= it is evident that 0

rr=∧∇ iA and 0. =∇ SA

r.

( iA is irrotational and SA is solenoidal)

14

Chapter 4

E L E C T R O S T A T I C F I E L D S We begin with fundamental concepts that are applicable to static electric fields in free-space (vacumm). An electrostatic field is produced by a static charge distribution. We investigate 2 fundamental laws governing electrostatic fields :

• Coulomb’s law • Gauss’s Law

I. Coulomb’s Law and field intensity

Point charge : a charge that is located on a body whose dimensions are much smaller than

other relevant dimensions. Charge : Symbol, q. Unit, coulombs (C). 1 electron charge : e = -1.6019 x 10-19 C. 1 coulomb equivalent to 6 x 1018 electrons. C O U L O M B ’ S L A W The force between 2 charged bodies, q1 and q2, that are very small compared with the distance of separation, R12, is proportional to the product of the charges and inversely proportional to the square of the distance, the direction of the force being along the line connecting the charges. Unlike charges attract and like charges repel each other. Using vector notation, Coulomb’s Law can be written mathematically as:

1212 2

12

21

02

12

2112 4

1RR u

Rqq

uR

qqkF

πε== k : the proportionality constant

In this case, we say 12F is the force exerted on point charge q2 due to q1

with Fmk /1094

1 9

0

×≈=πε

and 0ε : permittivity of free space, where

mF

mF

/10361

/10854.8

9

120

+

−

≈

×≈

π

ε

We may write, ( )

3

12

12

0

21123

0

212

120

2112 444 12

rr

rrqqR

Rqq

uRqq

F R

−

−===

πεπεπε

RR

u

RRR

rrR

R12

2112

1212

12=

==

−=

1r

2r q1

O (Origin)

12F q2

21F

15

Remarks :

• Since 2112 RR uu −= , we have : 2112 FF −=

• The distance R between q1 and q2 must be large compared with the linear dimensions of the bodies.

• q1 and q2 must be static (at rest). • The signs of q1 and q2 must be taken into account. For N cha ges (qr 1,q2, …. qN) located respectively at points with position vectors

( )Nrrr ,...., 21 , the resultant force on a charge q located at point rr is the vector sum of the forces exerted on q by each of the charges q1,q2, …. qN.

( )∑= −

−=

N

kk

kkq

rr

rrqqF1

304 r

r

πε

ELECTRIC FIELD INTENSITY

The force exerted to a charge is due to the electric field created by another charge. By definition, the electric field intensity is the force per unit charge that a very small stationary test charge experiences when it is placed where an electric field exists.

qFEr

r= , EqF

rr=

exe: Given 2 point charges q1 and q2 , placed respectively at points 1 and 2.

We define, 21F : force exerted on q1 due to q2.

21121 EqF =

where 21E : electric field intensity created by q2 at point 1.

By identification, we may write, 212

210

221 4 Ru

Rq

Eπε

=

In general, the electric field intensity created by a point charge q at a point of observation P can be written as:

RuR

qE 204πε

=r

V/m

For N point charges (superposition): (∑

= −

−=

N

k

k

rr

rqE

1041

r

rr

πε

q r

P

Er

16

R

)k

kr3

II. Electric fields due to continuous charge distributions We denote ; ρl (C/m) : line charge density

ρS (C/m2) : surface charge density ρv (C/m3) : volume charge density

We start from the expression RuR

qE 204πε

=r

. By replacing q with the charge element

ρl.dl, ρS.dS and ρv.dv and integrating, we get :

RuRd

E ∫= 204.

περ lrl : electric field at a point given due to a line charge

RS u

RdS

E ∫∫= 204.

περr

: electric field at a point given due to a surface charge

Rv u

Rdv

E ∫∫∫= 204.

περr

: electric field at a point given due to a volume charge

II.1. A line charge

Consider a finite line charge AB placed on z-axis, as shown in figure. The density of the line charge is ρl (C/m). If dl is the element (small portion) of the line AB, the total charge carried by dl is dq, where dq = ρl. dl. Hence, the total charge carried by the line charge AB is q, where

with dl = dz’ ∫∫ == B

A

z

z

B

Adzdq '.. ll l ρρ

Remark : We denote the field point by (x, y, z) and the source point (of the line

charge AB) by (x’, y’, z’).

α2

Rr

R

Pz),,( φρ α

dE zdE

α1α

z

ρdE

B

ldz )',','( φρ

z’

Ox

u

A

17

y

We have : RR

dzu

Rd

E R

rlrll ∫∫ == 3

02

0 4'.

4.

περ

περ

where :

αα

ρ

αα

ραρ

αρ

ρ ρ

ddz

dzzd

zz

R

uzzuR z

2

2

cos'

cos)'(

tan)'(cos

)'(

−=

=−

=−

=

−+=r

Hence,

[ ][ ]

( )( ) ( )[ ]z

z

z

z

z

uu

udud

udud

udud

udzz

udE

12210

0

0

0

23

3

023

3

0

coscossinsin4

..sin..cos4

..sin..cos4

..cos.tan..cos4

.cos

cos4

)'(.

coscos

4.

2

1

ααααρπε

ρ

ααααρπε

ρ

ααααρπε

ρ

αααααρπε

ρ

αα

ρρ

απε

ρα

αρ

ρα

περρ

ρ

α

α ρ

ρ

ρ

ρ

−+−=

⎥⎦⎤

⎢⎣⎡ +−=

+−=

+−=

⎟⎠⎞

⎜⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟

⎠⎞

⎜⎝⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

∫

∫∫

∫∫

∫∫

l

l

l

l

llr

(electrostatic field at point P due a finite line charge AB) Remark : For an infinite line charge (A at (0, 0, -∞) and B at (0, 0, +∞)), α1 will tend to π/2, and α1 will tend to - π/2, where the electric field at point P becomes :

ρρπερ

uE02l

r=

18

II.2. A surface charge Consider a finite surface charge of density ρS (C/m2), placed in xy-plane, at z = 0 as shown below. dS is an element of S, carrying a total charge of dq, where dq = ρS.dS. Note that dS = ρ.dφ.dρ. The element of charge, dq, carried by dS creates an element of electric

field dE at the point of observation P located at (0, 0, h).

From the expression RS u

RdS

E ∫∫= 204.

περr

, the electric field can be defined at point P.

( ) 2/122 hRR

uhuR z

+==

+−=

ρ

ρ ρr

r

hence,

( )

( )

( ) ( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

+

−=

+−+

=

==

∫∫∫∫

∫∫

∫∫∫∫

zS

zS

SR

S

uh

ddhuh

dd

uhuh

dd

RR

ddu

RdS

E

2/3222/322

2

0

2/3220

30

20

.....4

..4

4...

4.

ρ

ρφρ

ρ

ρφρπερ

ρρ

ρφρπερ

περφρρ

περ

ρ

ρ

rr

Remark:

For an infinite surface charge, due to the symmetry of the distribution, for each 1dE ,

there is 2dE , hence the component along ρ of the vector sum of 1dE and 2dE will

vanish. The resultant vector Er

has only z-component.

19

Hence, Changing the variable

(Consider m = ρ2),

( )

( )

( )( )[ ]

zS

zS

zS

zS

zS

u

uhmh

udmhm

h

uh

ddh

uh

ddhE

0

02/12

0

0 2/320

0 2/322

2

00

2/3220

2

2.

.211)2(

4.

.4

.

...4

ερε

ρ

ππερ

ρ

ρρφπερ

ρ

ρφρπερ

π

=

+−=

⎟⎠⎞

⎜⎝⎛

+=

+=

+=

∞−

∞

∞

∫

∫∫

∫∫r

dmdmmmd

dmm

d

m

m

21

2.

21

2

==

=

=

=

ρρ

ρ

ρ

ρ

(electric field due to infinite surface charge)

III. Electric flux density Given vector A

r, continuous in a region containing the smooth surface S, the surface integral

(or flux) of Ar

through S is given by : ∫∫∫∫∫∫ ===

SS nSdSAdSuAdSA .cos... θψ

rrr

We define the electric flux density (also called electric displacement) as:

EDrr

0ε= C/m2 and, ∫∫=S

dSD.r

ψ Coulombs

IV. GAUSS’S LAW G A U S S ’ S L A W The total electric flux ψ through any closed surface is equal to the total charge enclosed by that surface. enclosedQ=ψ

20

Remark: The definition of flux is given by: ∫∫=

SdSD.

rψ

Knowing that the surface is a closed surface bounding a volume, the expression can be written as: ∫∫=

SdSD.

rψ

The total charge enclosed in a closed surface (bounding a volume v) is generally defined by: ∫∫∫=

v venclosed dvQ .ρ

where ρv is the volume density of charge in (C/m3).

Hence, we can write ∫∫∫∫∫ ===v enclosedvS

QdvdSD .. ρψr

Applying the divergence theorem, where ( )∫∫∫∫∫ ∇=

vSdvDdSD

rrr.. , leads us

to write

( ) ∫∫∫∫∫∫ =∇v vv

dvdvD .. ρrr

hence; vD ρ=∇

rr.

MAXWELL

(the first of the Maxwell’s Equations to be derived) Remark: • Gauss’s law is an alternative statement of Coulomb’s law. • Gauss’s law can be used to determine E

r or D

r when the charge distribution is

symmetric. (If it is not symmetric, we must resort to Coulomb’s law to determine Er

or ). D

r

• Whether the charge distribution is symmetric or not, Gauss’s law always hold.

V. Applications of Gauss’s law If the charge distribution is symmetric (in order to be able to use Gauss’s law), construct a closed surface (Gaussian Surface) passing through the observation point where we want to find the electric field. Gaussian surface is chosen such that D

r is normal or tangential to the gaussian surface.

So that, if tangential/parallel to Dr

dS , dSDdSD .. =⇒r

if normal/orthogonal to Dr

dS , 0. =⇒ dSDr

21

V.1. A point charge Consider a point charge Q placed at the origin. Due to the symmetry, the flux density D

r

is along ru (the electric field is directed outward the point charge), where ruDD .=r

,

with D is the magnitude of Dr

, depending only on the distance between the observation point and the source (the point charge). In other words, the magnitude D depends only on the coordinate r of the spherical coordinate system.

rr uDD .=r

rn udSudSdS .. == Gauss’s law : enclosedQ=ψ

2

2

4

4

...

.

rQD

QrD

QdSD

QuudSD

QQdSD

r

r

Sr

S rrr

enclosedS

π

π

=

=

=

=

==

∫∫∫∫∫∫

r

hence, rur

QD 24π=

r

Remark: 2

0

2

0

2 4.sin rddrdSS

πθθφππ

== ∫∫∫∫

V.2. Infinite line charge Consider an infinite line charge of density ρl , placed at z-axis. Due to the symmetry, the

flux density is along Dr

ρu (the electric field is directed outward the line), where

ruDD .=r

, with D depends only on the distance between the observation point and the source (the line charge). In other words, the magnitude D depends only on the coordinate ρ of the cylindrical coordinate system.

ρρ uDD .=r

).(.. zbottomztopside udSudSudSdS −++= ρ

Remark : 0. =zuu ρ

z

P (r, φ, θ)

Dr

y r

Q

Gaussian Surface (a sphere of radius r)

x

P

Gaussian Surface (a cylinder of radius ρ)

l

ρl

ρ

x

z

r

22

D

y

Gauss’s law : enclosedQ=ψ with l

rπρψ ρρ 2. DdSDdSD sideS

=== ∫∫∫∫

and ll ll .. ρρ == ∫ dQenclosed

hence, ρρρ πρρ

uuDD2

lr

==

V.3. Infinite surface charge Consider an infinite plane surface charge of density ρS , placed in xy-plane. Due to the

symmetry, the flux density Dr

is along zu (the electric field is directed outward the

surface), where zuDD .=r

, with D depends only on the distance between the observation point and the source (the surface charge). In other words, the magnitude D depends only on the coordinate z.

⎪⎩

⎪⎨⎧

<−>=

0)z ( ).(0)z ( .

foruDforuDD

zz

zzr

).(.. zbottomztopsideside udSudSudSdS −++= Remark :

• bottomtop dSdS =

• 0. =zside uu • The top and bottom surfaces of the Gaussian surface are considered as 2 different

surfaces

Gauss’s law : enclosedQ=ψ

Gaussian Surface x

ρS

z

r

23

P area : A

D

y

r

with ( ) ADdSdSDdSD ztoptopzS

2. =+== ∫∫∫∫r

ψ

and AdSQ SSenclosed .. ρρ == ∫∫

hence, zS

zz uuDD2ρ

==r

V.4. Uniformly charged sphere Consider a uniformly charged sphere of radius R and with the density ρν , centered at the origin. Two cases will be treated : 1) a point inside the sphere, 2) a point outside the sphere.

charged sphere of density ρν

y

z

Gaussian Surface (a sphere of radius r1 for 1st case and r2 for 2nd case)

r1

x

R r2

1) a point inside the sphere

2111 4.

1

rDdSDdSD rrSπψ === ∫∫∫∫

r

3111 3

4.111

rdvdvQ vvvenclosed πρρρ === ∫∫∫∫∫∫

Where rvrr ur

uDD ρ31

1 ==

2) a point outside the sphere 2

222 4.2

rDdSDdSD rrSπψ === ∫∫∫∫

r

3

34. RdvdvQ vvvenclosed πρρρ === ∫∫∫∫∫∫

Where rvrr urRuDD ρ2

2

3

2 3==

24

Dr

Remark : Generalisation, for D

r everywhere :

⎪⎪⎩

⎪⎪⎨

⎧

≥

≤<=

Rrfor 3

Rr0for 3

2

3

rv

rv

ur

R

ur

Dρ

ρr

vR ρ3

r

R

VI. Electric potential – MAXWELL’S EQUATION Consider a point charge Q in an electric field E

r. The force on Q is EQF

rr= . The work

done in displacing the charge Q by ld in the region of electric field Er

is: l

rl

rdEQdFdW .. −=−=

Thus, the total work done in displacing the charge Q from a form A to B in the electric field Er

is:

∫−=B

AdEQW l

r.

We define the potential difference between points A and B, as:

∫−==B

AAB dEQWV l

r.

If Er

is due to a point charge q,

ABAB

AB

r

r rrAB

VVr

qr

qrr

qudrur

qV B

A

−=−=

⎥⎦

⎤⎢⎣

⎡−=−= ∫

14

14

114

..4

00

02

0

πεπε

πεπε

VA and VB are the potentials (or absolute potentials) at A and B respectively. VAB may be regarded as the potential at point B with reference to A. Thus, the potential at a point P due to a point charge q is given by

rqV

04πε=

where r is the distance between P and Q, and the infinity is taken as the reference.

ld

A

B

Er

25

Remark: • For an electric field due to n point charges (q1, q2, …. qn) located at points

with position vectors r1, r2, … rn, the potential V created at point P is:

∑= −

=n

k kp

k

rr

qPV

1041)(πε

(for point charges)

• For continuous charge distributions,

∫=L r

dV

ll .4

1

0

ρπε

(due to line charge)

∫∫=S

S

rdS

V.

41

0

ρπε

(due to surface charge)

∫∫∫=v

v

rdv

V.

41

0

ρπε

(due to volume charge)

vS ρρρ ,,l : density of charge (source) r : distance between the source ( ) and the observation

point P dvdSd ,,l

We have seen that ∫−=B

AAB dEV lr.

but BAAB

B

A

A

BVVdEdE =−==− ∫∫ l

rl

r..

hence,

0

..

..

=

=−=−=

−−=+

∫∫∫∫

AA

A

A

A

B

B

ABAAB

VdEdE

dEdEVV

lr

lr

lr

lr

Where, 0. =∫ lr

dE

Applying Stoke’s Theorem: ( )∫∫∫ ∧∇= dSEdErr

lr. ,

We find: 0

rrr=∧∇ E

26

0. =∫ lr

dE or 0rrr

=∧∇ E

2nd MAXWELL’S EQUATION to be derived Remark: The circulation of the vector electrostatic field along a closed path is zero

= the vector electrostatic field is conservative = no net work is done in moving a charge along a path in an electrostatic field = the line integral of E

r between two points doesn’t depend on the path taken

Since 0

rrr=∧∇ E , E

r is said to be derived from the gradient of a scalar field V, and from the

way we define potential, it follows that, VE ∇−=

rr

(note that the curl of the gradient of a scalar function is always zero)

VII. Electric dipole Electric dipole : and assemble of two point charges of equal magnitude but opposite sign

separated by a small distance. Potential at point P :

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=

−+

+=

+= −+

21

12

0

210

2010

4

114

44

rrrrq

rrq

rq

rq

VVV

πε

πε

πεπε

y

z

d

r2

r

r1

-q

+q

θ2

r2 - r1 ≈ d cos θ

θ

θ1

P (r, θ, φ)

x

27

Approximation: If r >> d,

2012

221 cos

4cos rqdV

drrrrr θ

πεθ=⇒

⎭⎬⎫

⎩⎨⎧

≈−≈

We define

qdppdqp

dud r

==

=

=

r

rr

r

.

cos. θ

pr : dipole moment,

hence, 20

20 44

.r

pr

upV r

πεπε==

r

The electric field due to the dipole (with centre at the origin),

( )θθθπε

uur

pVE r .sin.cos24

. 30

+=∇−=rr

VIII. Electric flux lines and equipotential surfaces Electric flux lines: the lines to which the electric field is tangential at every point. Properties: The line starts at positive charges (sources) and terminates at negative charges (sinks).

No two flux lines can intersect, except at singular or equilibrium points (points at which the resultant electric field is zero)

Cartesian:

Let zzyyxx uDuDuDD ++=r

The flux lines are given by: zyx D

dzDdy

Ddx

==

Cylindrical:

The flux lines are given by: zD

dzDd

Dd

==φρ

φρρ .

Spherical:

The flux lines are given by: φθ

φθθD

drDdr

Ddr

r

.sin.==

28

Equipotential surface: any surface on which the potential is the same. Remark: • No work is done in moving a charge from one point to another along an equipotential

surface:

0.

0

=⇒

=−⇒

∫ lr

dE

VV BA

• The flux lines (or the direction of E

r) are always normal to equipotential surfaces.

• The equipotential surfaces are determined by: V = constant, (if V is given)

Exe: point charge Exe: electric dipole

Flux line Equipotential surface

Equipotential surface

Flux line

IX. Electrostatic energy In introducing electrostatic potentials, we defined them in terms of the work done to move a positive charge from one point to another. We will quantify the potential energy present in a system of charges by determining the amount of work that must be done to assemble such a system of charge distribution. Consider a space completely empty and does not contain any charge. No work is done to bring a point charge Q1 from infinity to a point in the free region. The work required to bring a charge Q2 from infinity against the field of a charge Q1 in free space to a distance R12 is:

120

12222 4 R

QQVQW

πε==

29

The work required to bring a third charge Q3 from infinity to a point that is R13 from Q1 and R23 from a Q2 is:

⎟⎟⎠

⎞⎜⎜⎝

⎛+==

230

2

130

13333 44 R

QR

QQVQW

πεπε

The total work required to place the three charges is (which is also the potential energy stored in the system of charge distribution):

[ ]'''21

44444421

444

4440

332211

130

1

230

23

230

3

120

12

130

3

120

21

130

13

230

32

120

21

230

2

130

13

120

12321

VQVQVQ

RQ

RQ

QR

QR

QQ

RQ

RQ

Q

RQQ

RQQ

RQQ

RQ

RQ

QR

QQWWWWtotal

++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=++=

πεπεπεπεπεπε

πεπεπε

πεπεπε

In general, (for N point charges): ∑=

=N

kkktotal VQW

1

'21

where ∑≠=

=N

kjj jk

jk R

QV

,1041'πε

For a continuous charge distribution (line charge, surface charge, volume charge), the potential energy stored is given by:

∫∫∫

∫∫

∫

=

=

=

dvVW

dSVW

dVW

v

S

..21

..21

..21

ρ

ρ

ρ ll

ELECTROSTATIC ENERGY IN TERMS OF FIELD QUANTITIES:

We use Gauss’s Law : vD ρ=∇rr

.

and the following vector identity : ( ) ( ) ( )VDDVDV ∇+∇=∇rrrrrr

.. hence;

( )

( ) ( )[ ]∫∫∫

∫∫∫∫∫∫

∇−∇=

∇==

dvVDDV

dvVDdvVW v

rrrr

rr

..21

..21..

21 ρ

30

Using the divergence theorem, where ( )∫∫∫∫∫ ∇=vS

dvAdSArrr

.. , and the relation

VE ∇−=rr

, the expression of the energy is reduced to:

( ) ∫∫∫∫∫∫∫∫∫∫ +=−−= dvDEdSDVdvEDdSDVWSS

..21..

21.

21.

21 rrrrrr

Note that R

V1

∝ and 2

1R

D ∝r

, and the element dS varies as R2 , where the

term ∫∫S dSDV ..r

will vanish over a spherical surface at infinity.

Hence, the energy becomes:

∫∫∫∫∫∫ ==vv

dvEdvDEW 202

1..21 ε

rr

We define the electrostatic energy density ωE (J/m3) as:

∫∫∫∫∫∫∫∫∫ ===

====

v EvvE

EE

dvdvDdvEW

DEEDdv

dW

ωε

ε

εεω

0

22

0

2

0

20

21

21

21

21.

21 rr

31

32

Chapter 5 E L E C T R I C F I E L D S I N M A T E R I A L S P A C E

In this chapter, we develop the theory of electric phenomena in material spaces. Materials are classified in terms of their conductivity σ (mhos/m) or (siemens/m), as conductors and non-conductors (or dielectrics or insulators). Conductors (σ >> 1) Dielectrics (σ << 1) Microspically, the major difference between conductros and dielectrics is the amount of the (free) electrons available for the conduction of current : dielectrics have few electrons available for conduction of current, in contrast to conductors. I. Convection and Conduction Current

Electric current is generally caused by the motion of electric charges. Convection current occurs when electric current flows through a dielectric. By definition, the current through a given area is the electric charge passing through the area per unit time:

dtdqI =

1 Ampere of current: the transfer of charge at a rate of 1 coulomb per second

Consider a flow of charge of density ρv , accros an element of surface S∆ , with velocity vr . In time , the charge moves at distance vt∆ r

. t∆ .

The amount of charge passing through the surface S∆ is:

( )( ) SvtStvvolumeQ vvv ∆∆=∆∆×=×=∆ ... rr ρρρ

Hence, SvtQI v ∆=∆∆

=∆ ..rρ

We define the current density; vJ vrr

.ρ= Convection Current

(A/m2)

Hence, SJI ∆=∆ .r

Where ∫∫=S

dSJI .r

Convection Current (A)

S

dS

vr

r

nu

33

J

Density

When an electric field is applied to a conductor, conduction current occurs due to the drift motion of electrons. The electrons move and encounter forces called resistance. The average drift velocity of the electrons is directly proportional to the electric field intensity. This is explained by:

τvmEeF =−= where F : the force on the electron –e due to E

E : the electric field applied

v : the electron drift velocity m : mass of the electron τ : average time interval between collisions

Or, Emev τ−

=

For n electrons per unit volume, the density of charge is given by: nev −=ρ Hence, from the definition of convection current, the conduction current density can be defined:

EEm

nevJ v στρ ===2

.rr

EJrr

.σ= Conduction Current Density (A/m2) (also known as the point form of OHM’S LAW)

II. Conductors Conductor has a large amount of free electrons (mobile and free to move) in its body, described by its high value of conductivity. An electric field applied to a conductor introduces coulombs forces on the charges. Free to move electrons will then move against the direction of the positively applied electric field. These electrons will then leave behind the holes (protons) which are positively charged and attract other electrons. This process continues until a certain amount of electrons are accumulated at one side of the conductor, leaving positive charges at the other side. The amount of accumulated charges is proportional to the magnitude of the electric field applied. The time taken for the conductor and its charges to reach the equilibrium is called the relaxation time. The charge accumulation leads to the creation of internal electric field whose direction is to oppose the direction of the external electric applied. This makes the total electric field inside the conductor zero. The total charge density inside the conductor is also maintained at zero in dynamic case. Conductors are then said to be equipotential body since no electric field is formed inside the it even if an external electric field is applied (hence no change in potential is found along any line inside the conductor). The presence of free to move electrons in the conductor play the role in “neutralizing” the total charge and electric field in its body.

34

RESISTANCE OF CONDUCTOR

Consider a conductor of uniform cross-sectional area S, and of length h. A voltage V is applied to the conductor producing an electric field E

r.

The electric field applied is unif

The cross-sectional area is unifo

Ohm’s law:

Jr

Hence,

Sh

IVR

IRIS

hV

ρσ

σ

===

==

.

..

area) R : res

σ

ρ 1=c : res

For a non-uniform cross-sectoi

Er

I

orm: hEdEV .. == ∫ lr

rm: SJdSJIS

.. == ∫∫r

hV

SI

EJE

σ

σσ

=⇒

=⇒= ..r

Sh

c

(if the conductor has the uniform cross-sectional

istance

istivity

n conductor, ∫∫∫

==dSE

dE

IV

R.

.rl

r

σ

l V

35

III. Dielectrics III.1. Polarization

In ideal dielectrics, there is no electrons available for conduction of current. Charges in dielectrics are not able to move freely and they are bound by finite forces but we can expect a displacement when an externel force is applied. An atom (or molecule) of a dielectric is composed of a positively charged nucleus, surrounded by negatively charged electrons. The whole atom (or molecule) is electrically neutral (same amount of positive and negative charges). When an external electric field is applied, the positive charge is displaced in the direction of E

r and the negative charge is displaced in the opposite

direction. These displacements create electric dipoles and polarize a dielectric. Dipole Moment : dqp

rr .= (Cm) Non-polar dielectric : dielectric whose molecules do not posess dipoles until the

application of electric field (the dipole moment is in the direction of E

r).

(exe : oxygen, hydrogen, rare gases) Polar dielectric : dielectric which has built-in permanent dipoles which are

randomly oriented (exe : water) We define polarization P

r(C/m2) as the dipole moment per unit volume of the

dielectric ;

vp

v

dQP

N

kkk

v

rr=

∆=

∑=

→∆

1

0lim ( pr : average dipole moment)

The major effect of the E

r on a dielectric is the creation of dipole momentswhich align

themselves in the direction of Er

.

III.2. Equivalent charge distributions of polarized dielectrics We’ve seen in the previous section that for a dipole whose center is at the origin, the potential at point P is given by :

204

.R

upV R

πε

r

=

The potential dV at point P due to the dipole moment dvPdp .r

= is :

204..RudvP

dV R

πε

r

=

-q +q

dr

36

We have : 2

1Ru

RR=⎟

⎠⎞

⎜⎝⎛∇r

; hence, ⎟⎠⎞

⎜⎝⎛∇=

RP

RuP R 1..

2

rrr

Using the vector identity , ( ) ( ) ( )fAAfAf ∇+∇=∇

rrrrrr...

We find : ( )PR

PRR

PRuP R

rrrrrrr

.111..

2 ∇−⎟⎠⎞

⎜⎝⎛∇=⎟

⎠⎞

⎜⎝⎛∇=

The potential V is given by (and using the divergence theorem) :

( )

( )

( )

( )∫∫∫∫∫

∫∫∫∫∫

∫∫∫∫∫∫

∇−+=

∇−

+⎟⎠⎞

⎜⎝⎛=

∇−

+⎟⎠⎞

⎜⎝⎛∇=

⎥⎦

⎤⎢⎣

⎡∇−⎟

⎠⎞

⎜⎝⎛∇=

vS

vS

vv

dvPRR

dSP

dvPR

dSPR

dvPR

dvPR

V

PR

PR

dvdV

rrr

rrr

rrrr

rrrr

.4

14

.

.4

1.14

1

.14

114

1

.114

00

00

00

0

πεπε

πεπε

πεπε

πε

Remark :

∫∫S RdSP

04.πε

r

: Potential due to surface charge (closed surface bounding the

volume of the dielectric given), with density nPS uP.r

=ρ .

( PSρ : polarization/bound surface charge density)

( )∫∫∫ ∇−v

dvPR

rr.

41

0πε : Potential due to volume charge (the volume of the dielectric

given), with density PPv

rr.∇−=ρ

( Pvρ : polarization/bound volume charge density) In other words : A polarized dielectrics can be represented by an equivalent

polarization/bound surface charge density PSρ and an equivalent

polarization/bound volume charge density Pvρ for field calculations.

Bound charges are those which are not free to move within the dielectric material ; they are cused by the displacement which occurs during the polarization.

37

If vρ is the free charge volume density, the total volume charge density in the dielectric is :

ED PvPvvt

rrrr0.. ερρρρ ∇=+∇=+=

Hence

( )PED

PED

PED

EPD

rrr

rrrrr

rrrrrr

rrrrrr

+=

+∇=∇

∇+∇=∇

∇=∇−∇

0

0

0

0

.

...

...

ε

ε

ε

ε

PEDrrr

+= 0ε (When E

r is applied to a dielectric, D

r is increased by amount P

r inside the dielectric)

Note that in free space, P

r= 0, so that ED

rr0ε= .

To simplify the discussion, we except that P

r varies directly as the applied E

r, which is

usually the case for some dielectrics and we have :

EP e

rr0εχ=

χe : susceptibility (how sensitive a given dielectrics to electric fields)

III.3. Dielectric constant From the results found above, we may conclude that ;

EDrr

.ε= rεεε .0= 0

1εεχε =+= er

Where : ε : permittivity of the dielectric (F/m) ε0 : permittivity of free space (F/m)

εr : relative permittivity / dielectric constant (dimensionless) -depends on the frequency -equal to 1 for free space and non-dielectric materials

III.4. Linear, isotropic, homegeneous Linear : if D

r varies linearly with E

r

(otherwise, nonlinear) Homogeneous : if ε or σ is the same at all points in the region being considered (otherwise, inhomogeneous) Isotropic : if D

r and E

r are in the same direction

38

IV. Continuity equation The principle of charge conservation:

The time rate of decrease of charge within a given volume must be equal to the net outward current flow through the closed surface of the volume:

∫∫=Sout dSJI .r

: current coming out the closed surface

dtdqin− : the time rate of decrease of charge within a given volume

dt

dqdSJI in

Sout −==⇒ ∫∫ .r

but, ∫∫∫∫∫ ∇=

vSdvJdSJ ...

rrr (divergence theorem)

∫∫∫∫∫∫ ∂∂

−=−=−v

v

v vin dv

tdv

dtd

dtdq

..ρ

ρ

∫∫∫∫∫∫ ∂∂

−=∇⇒v

v

vdv

tdvJ

ρ..rr

t

J v

∂∂

−=∇ρrr

.

CONTINUITY OF CURRENT EQUATION

Remark: For steady current, 0=∂∂

tvρ , hence 0. =∇ J

rr, showing that the total

charge leaving a volume is the same as the total charge entering it (which leads to Kirchoff’s current law).

V. Boundary conditions

In this section, we are going to obtain mathematical relations that describe the transitional properties of the electrostatic field from one region to another. Consider the electric field E

r existing in two different regions characterised by ε1 = ε0. εr1

and ε2 = ε0. εr2 , separated by and interface. The electric field E

r can be decomposed as:

nt EEE 111 += in the first region

nt EEE 222 += in the second region

tE1 tangential component of 1E (to the interface)

nE1 normal component of 1E (to the interface)

39

Let us consider first the boundary condition of the tangential component of the electric field. Starting form the Maxwell’s equation 0. =∫abcda

dE lr

, where we construct a closed path abcda (a

rectangular between the two regions) with the length ab = cd =∆w and bc = da = ∆h.

( )( ) ( )

0

02222

0..

12

211122

=∆−∆=

=⎟⎠⎞

⎜⎝⎛ ∆

+∆

+∆−+⎟⎠⎞

⎜⎝⎛ ∆

−∆

−+∆=

=+++= ∫∫∫∫∫

wEwE

hEhEwEhEhEwE

dEdE

tt

nntnnt

dacdbcababcdal

rl

r

where tt EE 21 = or ( ) 012

r=−∧ EEun

Similarly, we consider the boundary condition of the normal component of the electric field. Starting form the Maxwell’s equation (Gauss’s law) encS

QdSD =∫∫ .r

, where we construct a

closed surface (a pillbox/cylinder between the two regions) with the length ∆h.

nu

tu

interface

ε1

ε2

E2t

E2n

E1t

E1n

2E

1E

∆h

∆w

d c

a b

∆S

nu

tu

interface

ε1

ε2

E2t

E2n

E1t

E1n

2E

1E

∆h

40

( ) ( SDSDhDhD

dSDdSD

nntt

bottomtopsideS

∆−∆+⎟⎠⎞

⎜⎝⎛ ∆∆

±⎟⎠⎞

⎜⎝⎛ ∆∆

±=

⎟⎠⎞⎜

⎝⎛ ++= ∫∫∫∫∫

1221 2.

2.

..

ll )

rr

Allowing ∆h → 0, gives: ( ) ( )SDSDdSD nnS

∆−∆=∫∫ 12.r

The total charge enclosed by the closed surface: SQ Senclosed ∆= .ρ

where Snn DD ρ=− 12 or ( ) Sn DDu ρ=− 12. (ρS : the free charge density at the interface separating the two regions) We shall consider the boundary conditions at an interface separating:

• 2 perfect dielectrics • perfect dielectric and perfect conductors

DIELECTRIC – DIELECTRIC

Continuity of the tangential component of electric field E

r at the interface:

tt EE 21 = or ( ) 012

r=−∧ EEun

•

• The free charge in any perfect dielectric is zero. Therefore, at the interface between two such dielectrics, ρS should be zero, where:

nn DD 21 = or ( ) 0. 12 =− DDun

•

1E

2E

E1n

E1t

E2n

From the results found above, it can be shown that:

2

1

2

1

tantan

r

r

εε

θθ

= Law of Refraction

E2t

θ2

ε2

interface

ε1θ1

41

DIELECTRIC – CONDUCTOR

For a perfect conductor, σ → ∞. But EJrr

.σ= and ∞≠Jr

, hence 0rr

=E . This causes ρv = 0 (a perfect conductor cannot contain an electrostatic field within it). If the region 1 is taken as a dielectric and the perfect conductor is the region 2, we have:

•

01 == tt EE E

r can only be external and normal to the

conductor

• Conductors are characterised by the presence of free electrons, so it is expected that free charges would exist at the interface separating a dielectric and a conductor. Because

0222

r== ED ε inside the conductor, we have then:

Snn DD ρ==1

42

Chapter 6

POISSON & LAPLACE EQUATION - CAPACITANCE

In this chapter, we shall consider electrostatic problems where only electrostatic conditions (charge and potentials) at some boundaries are known and it is desired to find E

r and V throughout the

region. Poisson’s and Laplace’s equations are usually to solved this problems. I t willl also cover the concepts of resistance and capacitance. I. POISSON’s and LAPLACE’s equations

Starting from the Gauss’s law (Maxwell’s equation) : vED ρε =∇=∇

rrrr..

and the relation field-potential : VE ∇−=rr

, we obtain : ( ) vV ρε =∇−∇

rr.

For the general case (for inhomogeneous medium), the equation above can be written as : ( ) vV ρε −=∇∇

rr. ,

and the ε is constant in term of the position for homogeneous material, where :

ερ vV

−=∇ 2

r POISSON’S EQUATION

which is the Poisson’s Equation. A special case of the Poisson’s Equation occurs when 0=vρ , and the equation is reduced to :

ερ vV

−=∇ 2

r LAPLACE’S EQUATION

which is the Laplace’s Equation. The symbol 2∇

r is called the Laplace operator (see chapter 3).

GENERAL PROCEDURE OF SOLVING POISSON’S OR LAPLACE’S EQUATION

• • •

•

Identify the variable coordinate (on which the potential V depends). Chose the suitable coordinate system. Solve the Poisson’s or Laplace’s equation (to get the general expression of the potential V in terms of the constants). Determine the constants from the boundary conditions of the problem.

43

II. Resistance - Capacitance II.1. Resistance

(See chapter 5 (II)).

II.2. Capacitance

Capacitance is a measure of the amount of charge of a particular configuratoin of two (or more) conductors is able to retain per unit voltage applied between them (the capacitance describes the ability of the configuration to store electrostatic energy). Consider two conductors surrounded by a homogeneous dielectric of ε = ε0 + εr . The total charge on each conductor is Q, with conductor 1 carries a total charge of +Q and conductor 2 carries –Q. V is the potential difference between the two conductors. The capacitance is defined as the ratio of the amount of positive charge to the resulting potential betweev the conductors :

VQC = Capacitance ( F )

1 Farad = 1 C/V There are generally two procedures to find the capacitance :

Assuming Q and calculate the electric field Er

• (exe : using Gauss’s law). V can be

derived from ∫−= lr

dEV or . VE ∇−=rr

and apply VQC = .

Assuming V and calculate Q (exe : using Laplace’s equation). Apply VQC . =•

+Q

r

44

ε

E-Q

V

Chapter 7

M A G N E T O S T A T I C F I E L D S Electric field characterized by E

r and D

r while magnetic field is characterized by H

r and B

r, where

Hr

is the magnetic field intensity and Br

is the magnetic flux density. For linear material space, HBrr

.µ= where µ is the permeability of the material, that will be discussed later in this chapter. We have seen in the previous chapters that static charge gives electrostatic field. If the static charge is moving with a constant velocity, a static magnetic field (or magnetostatic field) is produced. We will consider a magnetostatic field in free-space. There are similarities and dissimilarities between electrostattic and magnetostatic fields. Below is the analogy between electrostatic and magnetostatic fields : Term Electric Magnetic

Basic laws RuRqq

F 221

4πε=

r

24.

Rud

IdB R

πµ

∧=

l

enclosedSQdSD =∫∫ .

r enclosedL

IdH =∫ lr

.

Force law EqFrr

= BvqFrrr

∧=

Source element dq lr dIvq .=

Field intesity l

r VE = (V/m) l

r IH = (A/m)

Flux density S

D ψ=

r (C/m2)

SB φ=

r (Wb/m2)

Relationship between fields EDrr

.ε= HBrr

.µ=

Potentials VE ∇−=rr

)0( if =∇−= JVH m

rrr

∫−= lr

dEV . ABrrr

∧∇=

Flux ∫∫=S

dSD.r

ψ ∫∫=S

dSB.r

φ

dtdVCI = (C : capacitance)

dtdILV = (L : inductance)

Energy density EDwE

rr.

21

= HBwM

rr.

21

=

Poisson’s equation ερ vV −=∇ 2

r JA

rrr.2 µ−=∇

45

I. BIOT-SAVART’s Law B I O T - S A V A R T ’ S L A W The magnetic field intensity dH produced at a point P, by the differential current element I.dl, i. is proportional to the product I. dl and the sine of the angle between the element and the line joining

P the element. ii. and is inversely proportional to the square of the distance between P and the element.

2

2

2

4sin..

sin...

sin..

RdI

RdIk

RdIdH

πα

α

α

l

l

l

=

=

∝

k: constant of proportionality, k=1/4π

where,

32

32

4.

4.

4.

4.

RRdI

RudI

dB

RRdI

RudI

dH

R

R

πµ

πµ

ππr

ll

rll

∧=

∧=

∧=

∧=

remark: unit of H is A/m uint of B is Wb/m2 or Tesla Remark :

The direction of dH can be determined by the Right Hand Rule with the rigth hand thumb

pointing the direction of the current, the fingers encircling the wire in the direction of dH :

dH

I

I

ld

α

Rr

P

I

dH

46

If we define Kr

: the surface current density (amperes/meter) J

r : the volume current density (amperes/meter2)

we can write, dvJdSKdI ...rr

l == Thus, in terms of the distributed current sources,

∫∫∫

∫∫

∫

∧=

∧=

∧=

v

R

SR

LR

RudvJ

H

RudSK

H

RudI

H

2

2

2

4.4.

4.

π

π

π

rr

rr

lr

[for line, surface and volume current]

Exe : Line current

Consider a conductor with finite length AB, carrying a current I in the z direction. Using Biot-Savart’s law, we define the magnetic field intensity at a point of observation P, placed at z=0.

Biot-Savart’s law : 32 4.

4.

RRdI

RudI

dH R

ππ

rll ∧

=∧

=

with z

z

uzuR

udzd

−=

=

ρρrl .

where φρ udzRd ..=∧r

l

and we have : α

ρ3

33

sin=R

Rr

P

α

αA

αB

A

B

z

ρ

47

We write dz in term of dα,

ααρα

ρ

ρα

ddz

z

z

.sin

tan

tan

2

−=

=

=

Thus,

∫

∫∫−

=

−==

B

A

udI

duIudzI

HB

A

B

A

α

α φ

φφ

ααπρ

ααρ

ρα

πρ

ρα

πρ

..sin4

.sin

sin4

..sin4

...23

3

3

3r

( ) φααπρ

uIH AB coscos4

−=r

For a conuctor of infinite length,

1cos0

1cos→⇒→

−→⇒→

BB

AA

αααπα

⇒ φπρuIH

2=

r

II. AMPERE’s CIRCUIT Law

A M P E R E ’ S L A W

The line integral of the tangential component of Hr

around a closed path is the same as the net current Ienc enclosed by the path (the circulation of H

r equals Ienc ) :

∫ =L encIdH lr

.

(applied to determine Hr

when the current distribution is symmetrical. Just like Gauss’s law, Ampere’s law is a special case of Biot-Savart’s law) From the Stokes theorem, ( )∫∫∫ ∧∇=

SLdSHdH ..

rrl

r

and we have ∫∫=Senc dSJI .r

Where: JH

rrr=∧∇

MAXWELL

(the third of the Maxwell’s Equations to be derived) Remark : 0

rrr≠∧∇ H (magnetostatic filed is not conservative)

48

II.1. Infinite line current Consider an infinite line current carrying a current I, placed at the z-axis. Due to the

symmetry, the magnetic field Hr

is along φu (since the magnetic field encircles the

current source following the right-hand rule), with H is the magnitude of Hr

, depending only on the distance between the observation point and the line current. In other words, the magnitude H depends only on the coordinate ρ of the cylindrical coordinate system.

φ

φφ

φρ udd

uHH

..

.

=

=

l

r

Ampere’s law : ∫ =L encIdH lr

.

πρ

πρ

φρ

φρ

φ

φ

φ

φφφ

2

2

....

IH

IH

IdH

IIuduH

L

encL

=

=

=

==

∫∫

φπρuIH

2=

r

II.2. Infinite surface current

Consider an infinite surface current carrying a current Kr

. The ma

along xu if the point of observation is above the surface and it is alopoint of observation is below the surface (from the application of Biis the magnitude of H

r, depending only on the distance between th

and the surface current. In other words, the magnitude H depends onlz of the cartesian coordinate system.

⎪⎩

⎪⎨⎧

<−>=

0)(z .0)(z .

x

x

uHuHH

r

the surface current is along yu ,

yy uKK .=r

where the total current enclosed is ; bKI yenc .=

x

ρ O

amperian path, through P

infinite line current

z

I

y

P

z inc

x b

4

2

1

3

a

Kr

49

gnetic field Hr

is

ng xu− if the the ot-Savart), where H e observation point y on the coordinate

ld

finite surface urrent

y

Ampere’s law : ∫ =L encIdH lr

.

y

y

y

y

KH

bKbH

bKbHabHa

bKdH

21

...2

.))(()(0))(()(0

..1

4

4

3

3

2

2

1

=

=

=++−−+−

=⎟⎠⎞⎜

⎝⎛ +++ ∫∫∫∫ l

r

⎪⎩

⎪⎨

⎧

<−

>=

0)(z .21

0)(z .21

xy

xy

uK

uKHr

III. Magnetic flux density

We define magnetic flux as : ∫∫=S

dSB.r

φ unit (Weber) or (Wb)

and the magnetic flux density as : HB

rr.0µ= unit (Wb/m2) or (Tesla)

The magnetic flux line is the path to which B

r is tangential at every point in a magnetic field.

The magnetic flux lines always close upon themselves, which means that it is impossible to have isolated poles (or magnetic charges) of magnetic field – isolated magnetic charge does not exist. The total flux through a closed surface in a magnetic field must be zero, that is,

0. =∫∫S dSBr

Law of conservation of magnetic flux (Gauss’s law for magnetostatic field) Remark : • Magnetostatic flux is conservative, but magnetostatic field is not. • Electrostatic field is conservative, but electrostatic flux is not. Divergence theorem, ( )∫∫∫∫∫ ∇=

vSdvBdSB

rrr.. where:

0. =∇ B

rr

MAXWELL

(the forth of the Maxwell’s Equations to be derived) Magnetostatic fields have no source or sink (magnetostatic field lines are always continuous).

50

Summary : MAXWELL’S EQUATIONS FOR STATIC EM FIELD Differential (or point) form Integral form Remarks

vD ρ=∇rr

. ∫∫∫∫∫ ==v venclosedS

dvQdSD .. ρr

Gauss’s law for electrostatic field

0. =∇ Brr

0. =∫∫S dSBr

Gauss’s law for magnetostatic field. Nonexistance of magnetic monopole.

0rrr

=∧∇ E 0. =∫L dE lr

Faraday’s law. Conservativeness of electrostatic field.

JHrrr

=∧∇ ∫∫∫ ==SenclosedL

dSJIdH ..r

lr

Ampere’s law.

IV. Magnetic vector potential We define the magnetic scalar potential, Vm as: mVH ∇−=

rr (if 0

rr=J )

We know that the divergence of magnetic flux is zero, 0. =∇ B

rr, and since the divergence of

the rotational (or curl) of any vector Br

is zero, ( ) 0. =∧∇∇ Brrr

, hence we define a vector

Ar

, the magnetic vector potential as: AB

rrr∧∇= A

r: magnetic vector potential

We’ve seen that (for line, surface and volume current), that:

∫∫∫

∫∫

∫

∧=

∧=

∧=

v

R

SR

LR

RudvJ

H

RudSK

H

RudI

H

2

2

2

4.4.

4.

π

π

π

rr

rr

lr

or

AR

udvJB

AR

udSKB

AR

udIB

v

R

S

R

L

R

rrr

r

rrr

r

rrlr

∧∇=∧

=

∧∇=∧

=

∧∇=∧

=

∫∫∫

∫∫

∫

20

20

20

4.4

.4

.

πµ

πµ

πµ

We can define:

∫∫∫

∫∫

∫

=

=

=

v

S

L

RdvJA

RdSKA

RdIA

πµ

πµ

πµ

4.

4.

4.

0

0

0

rr

rr

lr

(magnetic vector potential for line, surface and volume

current)

We have ∫∫=S

dSB.r

φ , and we can show that : ∫= LdA l

r.φ (using Stoke’s

theorem).

51

52

Chapter 8

MAGNETIC FORCES AND MATERIALS - INDUCTANCE -

In this chapter, we study the force a magnetic field exerts on charged particles, current elements and loops. We will also consider magnetic fields in material media, and discuss on the definition of inductance. I. Magnetic Forces

I.1. Force on a charged particle

A magnetic field can only exert force on a moving charge. We define, mF

r, the magnetic force experienced by a charge q moving with a velocity vr

in a magnetic field , as: Br

BvqFm

rrr∧=

(Remark: vFm

rr⊥ and BFm

rr⊥ )

L O R E N T Z F O R C E

Let the electric force, eFr

ee EqF =rr

The total force on a moving charge q in the presence of both electric and magnetic field, me FFF

rrr+=

hence, )( BvEqF

rrrr∧+=

LORENTZ FORCE EQUATION

I.2. Force on a current element As we have defined in the chapter 5, the convection current density is given by vJ v

rrρ= .

Consider a current element ldIr

. of a current-carrying conductor, and since dvJdSKldI ...

rrr== , we can write,

vdqdvvldI v

r

rr

....

== ρ

where, vdqldI rr.. =

53

L A P L A C E F O R C E

We determine the force on a current element ldIr

. of a current carrying conductor due to the magnetic field , B

r

BvdqFdrrr

∧= . hence, BldIFd

rrr∧= .

LAPLACE FORCE

Remark: If the current I is through a closed path L or circuit, ∫ ∧= BldIF

rrr.

(the current element ldIr

. doesn’t exert force on the element itself. It’s the field Br

(which is external) who exerts force o dn I lr

. ). For the case of surface and volume current:

BdvJFd

BdSKFdrrr

rrr

∧=

∧=

.

.

I.3. Force between two current elements

)( 1Fddr

: the force on element 11. ldIr

due to 2Bdr

produced by 22 . ldIrr

, A magnetic field can only exert force on a moving charge.

Laplace: 2111 .)( BdldIFddrrr

∧= , with Biot-Savart’s: 221

2202 4

21

R

uldIBd R

π

µ rrr ∧

=

∫∫∧∧

=

∧∧=

22

21

21

1

2101

221

221101

)(4

4

)()(

21

21

L

R

L

R

R

uldldIIF

R

uldIldIFdd

rrrr

rrrr

πµ

π

µ

1Fr

= total force on current loop 1 due to current loop 2.

Note that, 21 FFrr

−=

11. ldIr

21Rr

22 . ldIr

)( 1Fddr

I2I1

54

II. Magnetization in materials These electronic motions produce internal magnetic fields iB

r that are similar to the

magnetic field produced by a current loop. We define the magnetic moment for a current loop: nb uSIm rr .=

S: the area of the loop Ib: the bound current The sum of magnetic moments is zero in a material if no magnetic field is applied on it. If an external magnetic field is applied, the magnetic moments more align themselves with the external magnetic field so that the net magnetic moment is not zero. Considering that there are N atoms in a given volume <v, we define the magnetization as

v

mM

N

kk

v ∆=

∑=

→∆

1

0lim

rr

A medium is said to be magnetized if 0

rr≠M .

We can show that: MJ b

rr∧∇= and nb uMK rr

∧= , where;

bJ : bound volume current density / magnetization volume current density.

bK : bound surface current density.

and ∫∫ +='

0

'

0 '4

'4 S

b

v

b

RdSK

Electron orbiting around the nucleus

Electron spin

IbCircular current loop

RdvJ

Aπµ

πµr

: magnetic potential vector of a magnetic body

due to bJ throughout thebody

due to bK onthe surface of the body

55

Using the same reasoning as in the chapter 5, we find that:

)(0 MHBrrr

+= µ where ;

0=Mr

: in free space 0≠M

r : in a material medium

Note: (for linear and isotropic materials)

We define, χm : magnetic susceptibility (dimensionless) as

HM m

rrχ=

where,

H

H

HB

r

m

r

r

rr

µ

µµ

χµ

=

=

+=

0

0 )1(

with:

mr

r

χµµµ

µµµ

+==

=

10

0

µ : permeability of the material (H/m) µr : relative permeability of the material

III. Classification of magnetic materials

Nonmagnetic materials : χm = 0 or µr = 1 Magnetic materials : χm ≠ 0 or µr ≠ 1 Remark: Free space is regarded as nonmagnetic.

Paramagnetics χm > 0 µr ≥ 1

Ferromagnetics χm >> 0 µr >> 1

Linear Nonlinear

Diamagnetics χm < 0 µr ≤ 1

MAGNETIC MATERIALS

56

Diamagnetics & Paramagnetics: For most practical purposes, we may assume that µr ≈ 1 for paramagnetic and diamagnetic material. Thus, diamagnetic and paramagnetic materials may be regarded as linear and nonmagnetic. Example - Diamagnetic: bismuth, lead, copper, silicon, diamond Paramagnetic: air, platinum, tungsten, potassium

Ferromagnetics: Materials whose atoms have relatively large permanent magnetic moment. Properties: 1. Capable of being magnetized very strongly by a magnetic field. 2. Retain a considerable amount of magnetization when removed from the field. 3. Lose ferromagnetic properties to become paramagnetic when the temperature

raised above the curie temperature (depends on the material). 4. Nonlinear. The relation HB r

rrµµ 0= doesn’t hold for ferromagnetic material

because µr depends on Br

and cannot be represented by a single value. B-H curve (magnetization curve)

The closed curve: hysteresis loop (dependent on the materials) The area of the hysteresis loop: hysteresis loss (energy loss per unit volume during one

cycle of the periodic magnetization of the ferromagnetic material)

B

Hysteresis loop

-Br

-Hc

Hc

Br

Coercive field intensity

Permanent flux density

Initial magnetization curve

H

IV. Magnetic boundary conditions The conditions that (or B

rHr

) satisfy at the boundary between two different media.

We use: Gauss’s law for magnetic fields, ∫∫ = 0.dSBr

and

Ampere’s circuit law, ∫ = encIdlH .r

57

We show that: Boundary between 2 magnetic media 1 and 2

B1n = B2nor µ1H1n = µ2H2n

Continuity of the normal component of B

r at the boundary

Discontinuity of the normal component of H

r at the boundary

And,

µ2B1t = µ1B2tor H1t = H2t

Discontinuity of the tangential component of at the boundary B

r Continuity of the tangential component

of Hr

at the boundary

V. Inductors and inductance Consider a closed loop, Co, carrying current I. It produces B

r, which causes a flux,

∫= dSB.r

φ (Wb) If the circuit Co has N identical turns, we call the flux linkage, λ, as: φλ N= We call the INDUCTANCE, L:

IN

IL φλ

== (Weber/Ampere) or (Henry)

The circuit is called INDUCTOR. Magnetic energy stored in an inductor, Wm:

2

21 LIWm = (Joules)

Consider 2 neighbouring closed loops, C1 and C2, bounding surface S1 and S2 respectively. I1 flows in C1, creating 1B .

I2 flows in C2, creating 2B .

S

C0

I

S2

C2

I2S1

C1

I1

58

φ1 = φ11 + φ12 : total flux passing through C1

φ2 = φ22 + φ21 : total flux passing through C2

∫∫=

1 1111 .S

dSBφ : the flux passing through C1 due to C1.

∫∫=1 1212 .

SdSBφ : the flux passing through C1 due to C2.

Self-Inductance

We define the SELF-INDUCTANCE of loop C1 as the magnetic flux linkage per unit current in the loop itself:

1

1111 I

Lλ

= SELF-INDUCTANCE of loop C1

If C1 has N1 turns, λ11 = N1φ11 , where λ11 : flux linkage of C1

Mutual-Inductance

The MUTUAL INDUCTANCE between 2 circuits is the magnetic flux linkage with one circuit per unit current in the other:

2

1212 I

Mλ

= MUTUAL INDUCTANCE between 1 and 2

If C1 has N1 turns, λ12 = N1φ12 , λ12 : flux linkage between C1 and C2, due to C2. We show that; M12 = M21

The total magnetic energy in the magnetic field:

2112

222

211

21211221

..21

21 IIMaILIL

WWWWWWWm

++=

++=++=

with:

+1 if I1 and I2 flow such that the magnetic fields of the two circuits strengthen each other

a = -1 if their magnetic fields oppose each other

59

VI. Magnetic energy Consider a volume covered with conducting sheets at the top and bottom surfaces.

We have I

zxHISB

IL

∆∆∆

=∆∆

=∆∆

=∆.... µφ

Since 2

21 LIWm = , hence:

vH

zyxH

IzxH

ILWm

∆=

∆∆∆=

∆∆∆=

∆∆=∆

..21

....21

....21

.21

2

2

2

µ

µ

µ

The magnetostatic energy density, Wm (J/m3): v

Wmvm ∆

∆= →∆ 0limω

µ

µω22

121 2

2 BBHHm ===

The energy in a magnetostatic field in a linear medium ∫∫∫= dvW mm .ω

Remark (electrostatic): ∫∫∫∫∫∫ == dvEdvHBWe ..21..

21 2ε

rr

z

Br

Conducting sheet

Conducting sheet

∆I ∆x

∆z

∆y

∆I

x

y

60

VII. Magnetic circuits

ANALOGY BETWEEN ELECTRIC AND MAGNETIC CIRCUITS

ELECTRIC MAGNETIC

Conductivity σ Permeability µ

Field intensity Er

Field intensity Hr

Current ∫∫= dSJI .r

Magnetic flux ∫∫= dSB.r

φ

Current density ESIJ .σ== Flux density H

SB .µφ

==

Electromotive force (emf) V Magnetomotive force (mmf)

∫== dlHNIF .r

Resistance R

Reluctance Slµ

=ℜ

l: the mean length of the magnetic core S: the cross-sectional area

Conductance G = 1/R Permeance ℘ = 1/ℜ

Ohm’s law RIEVSI

VR

...

==

==

l

l

σ Ohm’s law

SF

.µφl

==ℜ

Kirchoff’s law:

∑ ∑∑

=−

=

0.

0

IRV

IKirchoff’s law:

∑∑

∑=ℜ−

=

0

0

φ

φ

F

I φ φ

F = NI

N turns

I

R V

ℜ

61

62

Chapter 9

M A X W E L L ’ S E Q U A T I O N S We examine situations where electric and magnetic fields are dynamic (time-varying). In static: Electric and magnetic fields are independent of each other. Electric fields are due to static electric charges. Magnetic fields are due to: Motion of electric charges (uniform velocity) Magnetic poles (north and south) In dynamic: Electric and magnetic fields are interdependent. Electromagnetic (EM) fields are due to accelerated charges/time-varying currents. EM fields are represented by :

),,,(

),,,(

tzyxH

tzyxEr

r

Stationary charges electrostatic fields Steady current magnetostatic fields Time-varying currents EM fields (or waves) I. Faraday’s Law

F A R A D A Y ’ S L A W A time-varying field produces an induced voltage “electromotive force” (emf) in a closed circuit, which causes a flow of current. The induced emf, Vemf, in any closed circuit is equal to the time rate of change of the magnetic flux linkage by the circuit,

dtdN

dtdVemf

φλ−=−= (Volt)

λ: the magnetic flux linkage N: the number of turns in the circuit φ: the flux through each turn

L E N Z ’ S L A W

The negative sign in dtdNemfφ

−=V is an assertion that the induced emf will caused a current to flow in

the closed loop in such a direction as to oppose the change in the linking magnetic flux.

63

For a circuit with a single turn (N=1), dtdVemfφ

−=

The variation of flux with time may be caused by: 1) having a stationary loop in a time-varying B

r.

2) having a time-varying loop area in a static Br

. 3) having a time-varying loop in a time-varying B

r.

I.1. Stationary loop in time-varying B

r (transformer emf)

We have ∫∫∫

=

=

S

Lemf

dSB

dlEV

.

.r

r

φ

Hence, ∫ ∫∫ ∂∂

−==L Semf dS

tBdlEV ..

r

(transformer emf)

Invoking Stoke’s theorem: ( )dSEdlEL S∫ ∫∫ ∧∇=

rrr.

⇒tBE∂∂

−=∧∇rr

MAXWELL’S EQUATION (for time-varying EM field)

Remark: the time-varying E

r is not conservative, 0≠∧∇ E

rr.

I.2. Moving loop in static Br

(motional emf) When a conducting loop is moving in a static B

r, an emf is induced in the loop.

We have: BvqFm

rr∧= .

(the force on a charge moving with uniform velocity vr in a magnetic field Br

)

We define the motional electric field as: Bvq

FE m

m

rr∧==

64

We define the motional emf (or flux-cutting emf) induced in a conducting loop, moving with uniform velocity v in static

r Br

field,

( )∫ ∫ ∧−==L Lmemf dlBvdlEV

rr.

Stoke’s: ( )

( ) ( )∫∫∫∫ ∫∫

∧∧∇=∧

∧∇=

SL

L S mm

dSBvdlBv

dSEdlE

..

..rrrrr

r

⇒ ( )BvEm

rrrr∧∧∇=∧∇

I.3. Moving loop in time-varying B

r

Combining the transformer emf and the motional emf, gives the total emf:

( )∫∫∫∫ ∧+∂∂

==LSLemf dlBvdS

tBdlEV ...

rrr

Using Stoke’s:

⇒ ( )BvtBE

rrrrr∧∧∇+

∂∂

−=∧∇

II. Displacement current For static EM fields, we recall that,

JHrrr

=∧∇ …………………………………..(1) But ( ) JH

rrrrr.0. ∇==∧∇∇ ………………………..(2)

(the divergence of the curl of any vector field is zero) And the continuity of current equation imposes that,

0. ≠∂∂

−=∇t

J vρrr……………………………(3)

(not equal to zero for time-varying) This shows the incompatibility of the equation (2) and (3) for TIME-VARYING EM field. We must modify equation (1) to agree with (3)

65

Consider dJJH +=∧∇rrr

( dJ : to be determined)

( )

( )

tD

Dt

t

JJ

JJH

v

d

d

∂∂

∇=

∇∂∂

=

∂∂

=

∇−=∇

∇+∇==∧∇∇

rr

rr

rrr

rrrrrr

.

.

..

..0.

ρ ⇒

tDJ d ∂∂

=r

dJ : displacement current density

Jr

: conduction current density

Where, tDJH∂∂

+=∧∇r

rrr

MAXWELL’S EQUATION (for time-varying field Hr

). We can now define the displacement current;

∫∫∫∫ ∂∂

==SSd dS

tDdSJI ..r

r

Displacement current

III. Maxwell’s equation

DIFFERENTIAL FORM INTEGRAL FORM

vD ρ=∇rr

. ∫∫∫∫∫ =v vs

dvdSD .. ρr

0. =∇ Brr

0. =∫∫s dSBr

tBE∂∂

−=∧∇r

rr ∫ ∫∫∂

∂−=

L SdSB

tdE ..

rl

r

tDJH∂∂

+=∧∇r

rrr ∫ ∫∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+=L S

dStDJdH ..r

rl

r

66

IV. Time-varying potentials

For static EM fields: Electric scalar potentials: ∫∫∫=v

v

Rdv

Vπερ4

.

Magnetic vector potentials: ∫∫∫=v R

dvJAπ

µ4

..r

r

We start from: AB

rrr∧∇= (for both static and time-varying)

Faraday’s law:

( )

VtAE

tAE

tA

At

tBE

∇−=∂∂

+

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∧∇

⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

∧∇−=

∧∇∂∂

−=

∂∂

−=∧∇

rr

r

rr

rr

rr

rr

rrr

0

Where, tAVE∂∂

−∇−=r

rr

Remark: we can define B

r and E

r for time-varying if A

r and V given.

We have vD ρ=∇rr

. (for both static and time-varying) and εDEr

r= ,

hence,

( )

( )ερερ

v

v

At

V

At

VE

−=∇∂∂

+∇

∇∂∂

−∇−==∇

rrr

rrrrr

.

..

2

2

Using

tAVE

tDJH

AB

∂∂

−∇−=

∂∂

+=∧∇

∧∇=

rrr

rrrr

rrr

for time-varying, with ED

HBrr

rr

ε

µ

=

=

67

We can write:

2

2

tA

tVJ

tAV

tJ

tEJ

tDJ

HBA

∂∂

−⎟⎠⎞

⎜⎝⎛∂∂

∇−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

−∇−∂∂

+=

∂∂

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+=

∧∇=∧∇=∧∇∧∇

rrr

rrr

rr

rr

rrrrrrr

µεµεµ

µεµ

εµµ

µ

µ

applying the vector identity, where: ( ) AAA

rrrrrrrr2. ∇−∇∇=∧∇∧∇

hence, ( ) 2

22 .

tA

tVJAA

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

∇+−=∇∇−∇r

rrrrrrrµεµεµ

We let:

tVA∂∂

−=∇ µεrr

.

LORENTZ CONDITION FOR POTENTIALS (or LAURENTZ GAUGE)

so that we have:

JtAA

tVV v

rr

rr

r

µµε

ερ

µε

−=∂∂

−∇

−=∂∂

−∇

2

22

2

22

NONHOMEGENEOUS WAVE EQUATION

Remark: for static 0. =∇ A

rr

It can be shown that the solutions of the last 2 equations are:

[ ]

∫∫∫=v

v

Rdv

Vπερ4

. : retarded electric potential

[ ]

∫∫∫=v R

dvJAπ

µ4

.r

r : retarded magnetic vector potential

68

Remark: Notation of [ ]vρ and [ ]J

r means that, the time t in their expressions

( ) ( ) tzyxJtzyxv ,,,,,,,r

ρ is replaced by:

vRtt −=' : retarded time

with µε1

=v : the velocity of the wave propagation

and 'rrR −=r

: distance between the source point r’ and the observation point r

V. Time-harmonic fields Time-harmonic quantities: quantities that vary periodically or sinusoidally with time. Sinusoids are easily expressed in phasors.

Recall (phasors) -given phasor z is a complex number,

( )

⎟⎠⎞

⎜⎝⎛=

+==

=ℑ=ℜ

−=

+==

∠=+=

−

xy

yxzr

yzxz

j

jrerrjyxz

j

1

22

tan

)()(

1

sincos.

φ

φφ

φφ

square root: ⎟⎠⎞

⎜⎝⎛∠=

2φrz

complex conjugate: φ−∠=−= rjyxz*

Given 22222

11111

φφ∠=+=∠=+=

rjyxzrjyxz

Addition: ( ) ( )212121 yyjxxzz +++=+ Subtraction: ( ) ( )212121 yyjxxzz −+−=− Multiplication: ( )212121 .. φφ +∠= rrzz

Division: ( )212

1

2

1 φφ −∠=rr

zz

69

We let: θωφ += t , hence,

( ) ( ) ( )( ) ( )θω

θωφ

θωφ

+=ℑ=ℑ

+=ℜ=ℜ=ℜ

trermzmtreereereze

j

jtjj

sin.)(cos...)(

Exe:

( ) ( )( ) ( )tjj

tjj

eeImtItI

eeIetItIωθ

ωθ

θω

θω

00

00

sin)('

cos)(

ℑ=+=

ℜ=+=

We define the phasor current (form), , hence: θθ ∠== 00 IeII j

S

( )

( )θω

ω

+=ℜ=

tIeIetI tj

S

cos)(

0

(Instantaneous form)

Given ),,,( tzyxA

r: time-harmonic field,

and ),,( zyxAS : the phasor form of Ar

,

Hence, ( )tjS eAeA ω.ℜ=

r

Notice that ( ) ( tjS

tjS eAjeeAe

ttA ωω ω ... ℜ=ℜ )

∂∂

=∂∂r

, showing that,

ω

ω

jA

dtA

AjtA

S

S

→

→∂∂

∫ .

.

r

r

TIME HARMONIC MAXWELL’S EQUATION ASSUMING TIME FACTOR ejωt :

DIFFERENTIAL FORM INTEGRAL FORM

vSSD ρ=∇.r

∫∫∫∫∫ =v vSs S dvdSD .. ρ

0. =∇ SBr

0. =∫∫s S dSB

SS BjE ω−=∧∇r

∫ ∫∫−=L S SS dSBjdE .. ωl

SSS DjJH ω+=∧∇r

( )∫ ∫∫ +=L S SSS dSDjJdH .. ωl

70

Chapter 10

ELECTROMAGNETIC WAVE PROPAGATION Maxwell’s fundamental concepts of electromagnetic theory have been established in the previous chapters, allowing us to derive the electric and magnetic wave equations in this chapter. The principles of electromagnetic plane waves are based on the relationships between electricity and magnetism. A changing magnetic field will induce an electric field, and a changing electric field will induce magnetic field. The front of wave is sometimes referred to as an equiphase surface. A plane wave has a plane wavefront, a cylindrical wave has a cylindrical wavefront, and a spherical wave has a spherical. A major goal of this chapter is to solve Maxwell’s equations and EM motion using some approximations in the following media:

1. Free space 2. Lossless dielectrics 3. Lossy dielectrics (very low-loss) 4. Good conductors

I. Waves equation in source-free region

In radiation problems, the solution of the electric and magnetic fields (Maxwell’s Equations) starts by specifying the current or the charge distributions on the source and the radiated fields are then calculated. We start with the point form of Maxwell’s equations as listed below:

0.

.

=∇

=∇∂∂

=∧∇

∂∂

−=∧∇

B

DtDH

tBE

vrr

rr

rrr

rrr

ρ

with the constitutive relationships as given by:

r

r

EJHB

ED

µµµεεε

σµ

ε

...

0

0

====

=rr

rr

For sinusoidal time function, Maxwell’s can be written as:

( )

0.

.

=∇

=∇

+=+=∧∇

−=−=∧∇

B

D

EjDjJH

HjBjE

vrr

rr

rrrr

rrrr

ρ

ωεσω

ωµω

Taking the curl of the curl of Er

, we have: ( )HjE ∧∇−=∧∇∧∇ ωµrrr

In the other hand, we have a vector identity: ( ) EEErrrrrrrr

2. ∇−∇∇=∧∇∧∇ By member identification of the above, we may write:

( ) ( ) EEHjrrrrr

2. ∇−∇∇=∧∇− ωµ

71

In propagation problems, electromagnetic fields are studied under the assumption the space of propagation is source free. Another way of looking at it is to assume that the sources are sufficiently far away from the propagation region of interest. Assume now that the region containing the wave is source-free, where there is no charge distribution to give birth to any field, or 0=vρ . From the application of Gauss’s law, this situation leads to the fact that the

divergence of Er

is also zero: ED v

rrrr.0. ∇===∇ ερ (for a linear, homogeneous and

isotropic region) The previous equation hence yields:

( ) EHjrr

2∇−=∧∇− ωµ

⇒ ( ) Ejjrr

2∇−=+− ωεσωµ

⇒ ( )EjjErrr

ωεσωµ +=∇2 Defining γ : constant of propagation (complex number)

α : attenuation constant (Nepers/m) β : phase constant (rad/m)

where ( )ωεσωµγ

βαγ

jjj

+=

+=2

Thus we have: EErrr

22 γ=∇ ELECTRIC WAVE EQUATION

and similarly, HHrrr

22 γ=∇ MAGNETIC WAVE EQUATION Remark:

( )( )

2

2

2

22

tE

tE

tEj

tE

EjEj

EjjEE

∂∂

+∂∂

=

∂∂

+∂∂

=

+=

+==∇

rr

rr

rr

rrrr

µεµσ

ωµεµσ

µεωωµσ

ωεσωµγ

The time-domain wave equations can be written as :

2

22

2

22

tH

tHH

tE

tEE

∂∂

+∂∂

=∇

∂∂

+∂∂

=∇rr

rr

rrrr

µεµσ

µεµσ

Note that, in Cartesian; 2

2

2

2

2

22 .

zyx ∂∂

+∂∂

+∂∂

=∇∇=∇rrr

72

II. Electromagnetic Power and Poynting Theorem

The dimensions of Er

and Hr

are V/m and A/m respectively. Hence the vector HErr

∧ has VA/m2 as dimension, which is power density W/m2. The cross product HE

rr∧ results

in a vector along the direction of propagation that is the direction of the energy flow. The vector HE

rr∧ is known as the power density vector or Poynting Vector,

HEtzrrr

∧=℘ ),( POYNTING VECTOR (Power density vector)

which is directed along the direction of the electromagnetic energy flow.

Poynting Theorem Poynting theorem is an expression of the electromagnetic power balance that includes the relationship between the generated, transmitted, stored and dissipated electromagnetic powers. We start with a vector identity that involves the divergence of the Poynting Vector;

( ) ( ) ( )

JEtDE

tBH

tDJE

tBH

HEEHHE

rrr

rr

r

rrr

rr

rrrrrrrrr

..

..

...

−∂∂

−∂∂

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=

∧∇−∧∇=∧∇

where J is the conduction current.

r

We know that, Bt

HtBHBH

t

rrr

rrr

∂∂

+∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

21

21

2.

HBrr

µ= and the medium is assumed independent of time,

tBH

tHH

Ht

Ht

HHBHt

∂∂

=∂∂

=

∂∂

+∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

rr

rr

rrr

rrr

µ

µµ21

21

2.

similarly, we obtain :

tDEDE

t ∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

rr

rr

2.

73

Substituting equations ;

( ) JEDEt

BHt

HErr

rrrrrrr

.2.

2.. −⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−=∧∇

Integrating the equation over a closed volume in which we would like to examine the power balance equation ;

( ) ∫∫∫∫∫∫∫∫∫∫∫∫ +⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∧∇−VVVV

dvJEdvDEt

dvBHt

dvHE ..2.

2..

rrrrrr

rrr

Using divergence theorem ;

( ) ∫∫∫∫∫∫∫∫∫∫∫ +⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∧−VVVS

dvJEdvDEt

dvBHt

dSHE ..2.

2. rr

rrrrrr

( )∫∫ ∧−S

dSHErr

: the power density HEtzrrr

∧=℘ ),( entering the surface S

enclosing the volume of interest V.

∫∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛V

dvDE2.rr

and ∫∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛V

dvBH2.rr

: electric and magnetic energies

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

∫∫∫∫∫∫ VVdvDEdvBH

t 2.

2.

rrrr

: the rate of increase of the electric and magnetic

energies stored within the volume V.

∫∫∫V dvJE ..rr

: power dissipated (ohmic losses) within the volume V.

Time-average Poynting Vector The Poynting Vector HEtz

rrr∧=℘ ),( describes the intantaneous power density associated

with the propagation of electromagnetic waves. Frompractical viewpoint it is more appropriate to discuss time-average rather than the instantaneous Poynting Vector. For periodic functions, such as sinusoidal excitation, it is necessary to integrate over a time interval equal to one period, hence ;

∫ ℘=℘T

ave dttzT

z0

).,(1)(r

74

III. Solution of Uniform, Time-Harmonic Plane Wave Motion

A uniform plane wave is a wave whose magnitude and phase are both constant over a set of planes. By approximation, a spherical wave in free space is a uniform plane wave as observed at a far distance. Note that, a wave whose amplitude (not phase) varies within a plane normal to the direction of propagation is said to be plane but non-uniform. Let consider a uniform plane wave moving in z-direction. The wave front will then be in the plane orthogonal to z-direction, which is xy-plane. Since the wave is uniform, the electric and magnetic field have no variation in the plane orthogonal to z-direction or in other words, the vectors of electric and magnetic fields do not depend on x and y. Hence:

0=∂∂

=∂∂

=∂∂

=∂∂

yH

xH

yE

xE

Under these conditions (where the wave is plane, uniform and directed along z), other consequences resulted:

1. EZ = HZ = 0 There are no electric and magnetic field components along z-direction (or along the direction of the wave movement).

2. ( )npropagatio ofdirection ,, HE direct

The directions of E , H and direction of wave follow the right hand rule.

Without losing the generality, the case is simplified further by fixing the direction of E to

be along x, and, or automatically, H will be along y.

In this case, the laplacian of vector E in Cartesian is written as:

( )

( )

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

00.

.

zE

zE

yE

xE

Ezyx

EEEzyx

E

x

xxx

x

zyx

∂∂

=

∂∂

+∂∂

+∂∂

=

++⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=

++⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=∇r

The wave equations are then reduced from

⎪⎪⎩

⎪⎪⎨

⎧

∂∂

+∂∂

=∇

∂∂

+∂∂

=∇

2

22

2

22

tH

tHH

tE

tEE

µεµσ

µεµσ

to:

2

2

2

2

2

2

2

2

tH

tH

zH

tE

tE

zE

yyy

xxx

∂

∂+

∂

∂=

∂

∂∂∂

+∂∂

=∂∂

µεµσ

µεµσ

75

III.1. Uniform plane wave in free space Free space is considered lossless, with σ = 0, ε = ε0 and µ = µ0.

The wave equations are then reduced to:

⎪⎪⎩

⎪⎪⎨

⎧

∂

∂=

∂

∂∂∂

=∂∂

2

2

002

2

2

2

002

2

tH

zH

tE

zE

yy

xx

εµ

εµ

With no loss in generality, the first differential equation gives a solution in the form of:

( )

(

( )ztj

ztj

ztjx

eE

eE

eEtzE

0

00

00

0

0

0),(

βω

εµωω

εµω

−

−

−

=

=

=) where 000 εµωβ =

Remark to be made is that, given above is the complex notation of the electric field magnitude. The real time notation is:

),( tzEx

( )ztEtzEx 00 cos),( βω −=

where E0 is the amplitude of the field and ω is the angular frequency. The plot of electric field magnitude against coordinate z at several fixed times shows that the wave is moving towards positive z, or we say that the wave is propagating along z.

From the first Maxwell’s equation, HjE 0ωµ−=∧∇ and we know that the only component of the electric field in this case is along x which depends only on coordinate z. The equation is then reduced to:

( )y

ztj

yx

HjeEj

Hjz

EHjE

000

0

0

0 ωµβ

ωµ

ωµ

βω −=−

−=∂∂

−=∧∇

−

76

The magnitude of the magnetic field can then be derived:

( )

( )

( )ztj

ztj

ztjy

eE

eE

eEjjH

0

0

0

00

0

00

00

00

0

βω

βω

βω

µε

ωµµεω

ωµβ

−

−

−

=

=

=

Thus: xy EH0

0

µε

=

Basic Properties – uniform plane wave in free space:

The phase constant in free space: 000 εµωβ = radians per meter The attenuation constant in free space: 00 =α Nepers per meter

To find the phase velocity of the wave, we analyse the movement of a point of

the curve of the field magnitude defined by a constant value of the cosine:

( )

00

0

0cos

ββω

βωβω

Ktz

KztKzt

−=

=−=−

The phase velocity is the variation in time of the position of the point represented by z, thus:

0βω

==dtdzv p

Remark: With 000 µεωβ = , the phase velocity of a wave in free space is:

8

00

1031×=== cv p εµ

meters per second

c is called the velocity of light in vacuum/air. The phase constant in free space can also be written in term of c, where:

cωβ =0

77

The intrinsic wave impedance in free space:

3771200

00 === π

εµ

η ohms

Remark:

HHE 00

0 ηεµ

==

As the wave propagates along z, the phase changes by the amount of zje 0β−

z0β . The distance z that the wave must travel so that the phase changes by 2π radians is called the wavelength:

πλβ 20 =

fc

fvp ====

000

22εµω

πβπλ meters

III.2. Uniform plane wave in lossless dielectric For lossless materials, the conductivity, permittivity and permeability are given by: σ = 0, ε = ε0.εr and µ = µ0. µr leading us to the following results:

0=α : attenuation constant

rrrr cεµωεµεµωµεωβ === 00 : phase constant

r

r

r

r

εµη

εεµµ

εµη 0

0

0 === : intrinsic wave impedance

rr

pcvεµµεβ

ω===

1 : phase velocity

f

vp=λ : wavelength

78

III.3. Uniform plane wave in lossy media The conductivity of lossy materials is not zero σ ≠ 0. The presence of a loss in the medium introduces wave dispersion by conductivity. Dispersion makes a general solution in the time domain impossible except by Fourier expansion methods. Thus, only solutions for the frequency domain (or steady state) will be given. The wave equations: ( ) EEjjE

rrrr22 γωεσωµ =+=∇

( ) HHjjHrrrr

22 γωεσωµ =+=∇

For uniform plane wave in z-direction: ( ) xxx EEjj

zE 2

2

2

γωεσωµ =+=∂∂

( ) yyy HHjj

zH 2

2

2

γωεσωµ =+=∂

∂

The complex solutions:

( )

( )ztjz

tjzjz

tjzx

eeE

eeE

eeEtzE

βωα

ωβα

ωγ

−−

−−

−

=

=

=

0

0

0,

( ) ( )ztjzy eeHtzH βωα −−= 0,

The real time solutions: ( ) ( )zjeEtzE z

x βωα −= − cos, 0

( ) ( )zjeHtzH zy βωα −= − cos, 0

Low Loss Dielectric

The Loss Tangent of the material is defined as: ωεσθ =tan

For very small conductivity, σ << ωε (or when θtan is smaller than 0.1), an approximation can be done, resulting simplified expressions of the following quantities:

εµσθ

λπα

2tan ==

µεωβ =

⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ += θ

εµ

ωεσ

εµη tan

211

21 jj

µε1

=pv

f

v p=λ

79

Good Conductors

Good conductors have very high conductivity, σ >> ωε , allowing the following approximation to be made:

( )

( ) µσπ

ωµσ

ωµσωεσµεω

ωεσµεω

ωεσωµγ

fj

jj

jj

jj

jj

jj

+≈

⎟⎟⎠

⎞⎜⎜⎝

⎛−≈

−≈

−≈

−=

+=

12

12

1

1

Hence,

µσπβα f== The Skin Depth is the length a given wave travels in a given conductor so that the magnitude of the field is reduced by the factor of (around 0.368), which means at z = 1/α = δ : the skin depth.

1−e

Thus, µσπβα

δf111

===

And by approximation,

( )

( )

( ) SRj

j

j

jj

j

+=

+=

+=

°∠≈≈+

=

1

11

21

45

σδ

σωµ

σωµ

σωµ

ωεσωµη

where RS is the Skin Effect (or the Surface Resistance of the conductor),

σωµ

σδ 21

==SR

80

TU

TO

RI

AL

S

Elect

romag

netic

Theor

y

KJE

44

2

ELECTROMAGNETIC THEORY KJE442

TUTORIAL 1 I) Given 2 vectors jiA

rrr53 += and jiB

rrr24 += , find the values of BA

rr+ , BA

rr− , AB

rr−

and sketch the results. ( is a unit vector along x-axis and ir

jr

is a unit vector along y-axis in the positive direction)

II) Show that kjiA

rrrr−+= 24 and kjiB

rrrr422 +−= are perpendicular to each other.

III) Give the cylindrical and spherical coordinates of the point whose cartesian coordinates are x =

3, y = 4, z = 5 and show it on a sketch.

IV) Given yx uuA 42 +=r

and zy uuB 46 +=r

. Find the smaller angle between Ar

and Br

.

V) Write points M, N and vectors A , B and C in Cartesian, cylindrical and spherical coordinate systems.

VI) Demonstrate the relation ( ) ( ) ( )CBABCACBA

rrrrrrrrr.. −=∧∧ , and deduce the relation

( ) ( ) ( )ABCBACCBArrrrrrrrr

.. −=∧∧ .

VII) Given a vector Rr

(see figure), determine the unit vector Ru of Rr

.

Rr

P (ρ,φ,0)

h

Tutorial 1

ELECTROMAGNETIC THEORY KJE442 VIII) Use the spherical system to find the area of the strip of βθα ≤≤ on the spherical shell of

radius a (see figure). What results when 0=α and πβ = ?

β α

IX) Develop the equation for the volume of a sphere of radius a from the differential volume. X) Use the cylindrical coordinate system to find the area of the curved surface of a right circular

cylinder where radius = 2 m, height = 5 m and (see figure). 00 12030 ≤≤θ

30° y

2m

5m

z

φ = 2π/3 x

Tutorial 1

ELECTROMAGNETIC THEORY KJE442

TUTORIAL 2 I) A total charge of 15 x 10-6 coulombs is uniformly distributed on a line of 5 cm length. Find the

line charge density. II) Find the total charge contained in a cylinder (of length 30 cm and of radius 10 cm) if the

volume charge density is 21

100−−= ρρν

ze c/m3. III) Given a surface of density c/m510−=Sρ

2, uniformly distributed on sphere of radius 10 cm. Find the total charge.

IV) State Coulomb’s Law of force between any two point charges.

Point charges of 3 x 103 µC are situated at each of the three corners of a square whose side is 0.2 m. Find the magnitude and direction of the electric field at the vacant corner-point of the square.

V) Determine the resultant force on charge q3 shown below.

Given: q1 = - 4 x 10-6 C q2 = 3 x 10-6 C q3 = - 2 x 10-6 C d13 = 0.08 m d23 = 0.12 m

x

y

45° d13

d23

q1

q2q3

VI) The force on a point charge situated 10 cm away from another point charge of the same magnitude is 1 Newton. Determine the magnitude of the charge.

ρL

a

P

h

VII) Charge is distributed uniformly along a circular line with density ρL. Develop the expression of E

r at point P,

h meters from the origin. VIII) Find the electric field E

r and the force on a point charge

of 50 µC at (0,0,5) due to a charge of 500π µC that is uniformly distributed over the circular disk ρ ≤ 5 m, at z = 0.

(0,0,5)

ρS 5m

Tutorial 2

ELECTROMAGNETIC THEORY KJE442

P

a

H

O

P’

H’

b

IX) Given a uniform distribution of electric

surface charge ρS on the surface of a disk, at a<ρ<b as shown in figure.

Determine E

r at point P.

Deduce Er

’ at point P’. What is E

r0 at origin?

X) Given surface S as shown in figure, find the

surface area if the radius of the sphere is R, in terms of θ1, θ2, φ1 and φ2.

A surface charge density of θρ sin1RS =

is found to be distributed on surface S. Given that , )2cos(1sin2 2 AA −=determine the total charge Q carried by surface S. What would be the total charge QT if surface S was the total surface of the sphere in question?

z

φ2

φ1

θ1

θ2

Surface S

φ1

Surface S

ρ2

ρ1

z XI) Shaded surface S in shown in figure. Find the total surface area of surface S. Electric surface charge of density

φρρ cos.=S is found distributed over surface S. Determine the total charge Q carried by surface S. What would be the total charge QT if surface S was the total surface of the cylinder in question?

φ2

h

Tutorial 2

ELECTROMAGNETIC THEORY KJE442 XII) Two point charges q1 and q2 are placed at points A and B respectively, in a free space, as

shown in Figure 1α. Given the distance between A and B is d, answer the following questions:

a) Determine the Coulomb’s force exerted on charge q1 at point A, as a function of q1, q2, d

and xu . b) Deduce the force exerted on charge q2.

Another charge Q is now placed at the origin of the system as shown in Figure 1β.

c) If q1 = q2 = q, determine the expression of Q as a function of q, so that the system is in equilibrium (so that the total force at every charge is zero).

d) Determine the total absolute electric potential at point C due to all the charges at equilibrium.

XIII) Figure 1 shows a line charge of constant density lρ and of length 2a. The line charge is put

parallel to x-axis at y = a.

a) If ld is a vector differential length of the

line charge, write ld in Cartesian coordinate system.

b) Write vector R in Cartesian coordinate system.

c) Show that the electric field intensity at point P due to the line charge is as below:

( )yz uauhhak

aE ..

22

.222

0

−+

=πε

ρl

where k2 = a2 + h2

Given: ( )∫

+=

+ 2222/322 mumu

mudu

( )∫

+

−=

+ 222/322

1.mumu

duu

Figure 1

Tutorial 2

ELECTROMAGNETIC THEORY KJE442

Referring to Figure 2, a square loop of charge of sides 2a is centered at the origin with the same line density lρ .

d) From the results found in c), deduce

the electric field intensity at point P due to the square loop.

e) What would be the electric field at the origin?

Figure 2

Tutorial 2

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 3 I) Consider an infinite line source of strength ρL Coulombs per meter. Find the electric field

intensity at a distance ρ from the line. Find the potential at that point. II) Consider a sphere of radius R carrying a constant surface charge density on its surface. Find

the expression of the electric flux density inside and out side the sphere. III) Suppose that a charged sphere of charge Q and of radius a is placed concentric with and

insulated from a larger conducting sphere of radius b, which is uncharged. Find the Dr

field for : a) r < a b) r > b c) a < r < b

IV) Suppose that we have 2 concentric spheres of radii a and b, where b > a, and that the inner

sphere has a positive charge of Q.

a) Sketch the flux pattern and calculate Dr

for r < a, a < r < b, r > b for a charge of –Q on the outer sphere.

b) Repeat for a charge of +Q on the outer sphere. V) A positive charge of 100 x 10-6 µC is uniformly distributed throughout a spherical surface 30

cm in diameter. Calculate and plot the variation of the electric field intensity Er

and the absolute potential V as a function of the radius r from the center of the sphere to a distance of 1 meter.

VI) The electric field from a charged sphere of 10 cm radius is 20 kV m-1 at a distance of 20 cm

from the center of the sphere. Assuming uniform charge distribution on the surface of the sphere, find: a) the D

r field.

b) The total charge on the sphere. VII) Find the potential difference between 2 parallel line charges of ρL and - ρL separated a

distance d. [Assume a radius a from each line, and integrate only from a to d-a]

VIII) The voltage difference between two parallel wires is 1000 V. The radius of each wire is 1 mm,

and the separation is 1 m. a) Find ρL (Cm-1) on each wire. b) Find the E

r field between wires along a straight line joining the two.

IX) A volume charge distribution of density

3rk

v =ρ C/m3 is found in the region between

two concentric spheres of radius a and b as shown in figure, where k is a constant. a) Determine the total charge in the region. b) Using gauss’s law, determine the electric

flux density for r < a, a < r < b and r >b.

Tutorial 3

ELECTROMAGNETIC THEORY - KJE442

X) A surface charge distribution ρS is carried by a plane disk of radius R centred at the origin, in

xy-plane, as shown in Figure. The non-uniform surface charge density is given by:

22 ρ

ρ−

=R

KS C/m2, where K is a constant.

A closed surface S encloses the total charge

(i) State and explain Gauss’s law. (ii) What is the unit of the constant K?

z

(iii) Given:

( )

( )21

22

21

22

xM

xdx

xMd

−−=

⎥⎦

⎤⎢⎣

⎡−

where M is a constant. Find the total

charge carried by the disk. (iv) What is the total electric flux passing through the closed surface S?

XI) A spherical distribution of charge ⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

2

0 1br

v ρρ C/m3 exists in the region br ≤≤0 .

Using Gauss’s law, determine the electric field intensity E in the region;

i) br ≤≤0ii) br >

R O ρS

Electric Flux Lines

Surface S

Tutorial 3

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 4 I) The difference of potential at 20 m from a point charge and the potential at 30 m from the

same charge is 50 V. Find the value of the point charge, and at what r is the potential equal to 25 V?

II) Find E

r, D , and ρr

v for the following potential difference: a) V = 10 x2 b) V = 2 ρ sin φ c) V = (5/r ) cos θ

III) The ratio b/a for a particular coaxial cable is 3. Find the potential between conductors if the

radial electric field is 3000 Vm-1 at the surface of the inner conductor of 1 cm radius. IV) The electric field between two coaxial cylinders is 500 V/m at the inside surface of the outer

conductor. Find the potential difference between conductors if the radii are 2 cm and 5 cm.

V2

V1

V) Find the potential function for the region between the parallel circular disks (see figure). Neglect fringing.

VI) Find the potential function and the electric field

intensity for the region between two concentric right circular cylinders, where V = 0 at ρ = 1 mm and V = 150 Volts at ρ = 20 mm. Neglect fringing.

V = 150 Volts

V = 0

VII) Solve Laplace’s equation for region between coaxial cones, as shown in figure. A potential V1 is assumed at θ1, and V = 0 at θ2.

V1

V = 0

θ2

θ1

Tutorial 4

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 5 I) Find the relative permittivity of the dielectric material used in a parallel-plate capacitor if :

a) C = 40 nF, d = 0.1 mm, and S = 0.15 m2 b) d = 0.2 mm, E = 500 kV/m, and ρS = 10 µC/m2

II) Find the capacitance of :

a) 20 cm of coaxial cables having an inner conductor 1 mm in diameter, an outer conductor having an inside diameter of 2.5 mm, and a polyethylene dielectric.

b) A conducting sphere 1 cm in diameter, covered with a layer of polyethylene 1 cm thick, in free space.

c) A conducting sphere 1 cm in diameter, covered with a layer of polyethylene 1 cm thick, and surrounded by a concentric conducting sphere in 3 cm radius.

III) A parallel-plate capacitor contains three dielectric layers.Let εR1 = 1, d1 = 0.2 mm, εR1 = 2, d2 =

0.3 mm, εR3 = 3, d3 = 0.4 mm, and S = 20 cm2. Find C.

z IV) Figure shows two conductors, conductor 1 and 2, carrying a total charge of +Q and –Q respectively. Use Gauss’s law to find the expression of electric field intensity anywhere between the conductors. Deduce the potential difference between the conductors. Find the capacitance of the combination of the conductors.

ρ = b

ρ = a φ0

φ = 0 Conductor 1 +Q

h

z

V) Two conductors are maintained at V1 and V2 as shown in Figure. Using

Laplace’s equation, determine the potential function anywhere in between the conductors. Deduce the electric flux density anywhere between the conductors. Find the capacitance of the combination of the conductors.

ρ = b

ρ = a φ0

φ = 0 Conductor 1 V = V1

Conductor 2 -Q

h

Conductor 2 V = V2

Tutorial 5

ELECTROMAGNETIC THEORY - KJE442

VI) Figure a shows two (2) identical parallel conducting plates 1 and 2, separated by a dielectric

of permittivity ε, and maintained at the potential level V1 and V2 respectively. S is the surface area of the plates, and d is the distance between the plates. (i) Using Laplace’s equation, find the potential function in the dielectric, as a function of

V1, V2, and d. (Neglect the fringing effect). (ii) What is the electric flux density D

v in the dielectric?

(iii) Deduce the total charge found on one of the surface of the plate. (iv) Find the expression of the capacitance between the plates.

A parallel plate capacitor is shown in Figure b, where two (2) dielectric layers of permittivity ε1 and ε2 separate the parallel plates.

Figure a

z

d

Plate 2

Plate 1

ε

S

V2

V1

dielectric

(i) Show that the total capacitance CT of a two parallel-connected capacitors of capacitance C1 and C2 is CT = C1 + C2.

(ii) Determine the capacitance of the capacitor in Figure b if a = 1 cm, b = 2 cm, c = 1 cm, d = 1 cm, εr1 = 2 and εr2 = 4.

ε1 ε2

Conducting plate

Conducting plate

dc

a b

Figure b

Tutorial 5

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 6 I) Find H

r at point P, due to a finite straight-line

constant current I (see figure). Discuss for the case of an infinitely long straight-line current.

αA

αB

P

A

B

I

II) Find Hr

due to an infinitely long straight-line current using Ampere’s law.

h

a

P III) Find Hr

on the axis of a circular loop of radius a. Specialize the result to the center of the loop.

IV) A circular loop constant current I of radius a is shown in Figure. Using Biot-Savart’s law, determine the magnetic field intensity at point P as a function of the angle α. What must be the value of α so that |H| can be maximum? What would be |H| if α tends to infinity?

α

P

a

z

IV) A thin cylindrical hollow conductor of radius a, infinite in length carries a current I. Find H

r at

all points. a

+ ∞

- ∞

Tutorial 6

ELECTROMAGNETIC THEORY - KJE442 VI)

rent

ot uniform an

A solid cylindrical cable of radius a carries a constantcurrent I as shown in Figure 3Alpha. The curdistribution over the cross sectional surface is n d the surface current density

is ρueKJ ..= , where K andρn n are constants,

and ρ is the radial coordinate.

ic field

intensity inside and outside the conductor.

Given:

Using Ampere’s law, determine the magnet

∫ ⎟⎠⎞

⎜⎝⎛ −=

αα

αα 1.. xedxex

xx

VII) Determine H

r for a solid cylindrical conductor

of radius a, where the current I is uniformly distributed over the cross section.

Find H

r at all points, for an infinite

cylindrical coaxial cable.

VIII) and carrying a current I. The toroid has a n radius b, and the radius of

a

I

a

I

I I

a

b c

Determine the magnetic flux density inside a closely wound toroidal coil with an air core having N turns meaeach turn is a.

I

a

b

z

I

Tutorial 6

ELECTROMAGNETIC THEORY - KJE442 IX)

,

be current

passing through coil B?

Two narrow circular coils A and B have a common axis and are placed 10 cm apart as shown in Figure. Coil A has 2 turns of radius 2 cm, with a current of 0.5 A passing through it. Coil B is of radius 5 cmand has 10 turn. If the magnetic field at the center of the coil A is tozero, what should be the

10 cm

2 cm

5 cm

z

10 turns

2 turns Coil A

Coil B

Tutorial 6

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 7 I) Evaluate the inductance of an infinitely long solenoid with air-core having n closely wound

turns per unit length and carrying a current I. (Using approximations, show that the magnetic field can be considered zero outside the solenoid to simplify the calculations).

II) Based on question 1, evaluate the inductance of a solenoid of 2500 turns wound uniformly

over a length of 0.5 m on a cylindrical tube 4 cm in diameter. The medium is air inside the solenoid is air.

III) Assume that N turns of wire are tightly wound on a toroidal frame of rectangular cross-section

with dimensions as shown in figure. The medium is air. Find its self inductance.

Develop expressions for flux density and inductance of a toroid of circular cross section. (Assume the mean radius R of the toroid is much bigger than the radius r of each turn of wire. Hence, by approximation, the magnetic flux density can be assumed constant along radial coordinate inside the toroid)

I h

z

b a

IV) A solenoid (air-core) has 2000 turns of copper wire wound on a former of 1 meter length and 4 cm diameter. It is placed coaxially within another solenoid with the same length and number of turns but with a diameter of 7 cm. Determine the mutual inductance between the two solenoids.

V) Two (2) coaxial loops 1 and 2 lie in parallel planes separated by a distance h, having radii a

and b, with a<<b and a<<h (see Figure b). (i) Using the approximations a<<b and a<<h, find the flux through the loop 1 due to a

current I in the loop 2. (ii) Determine the mutual inductance of the loops.

h

z

Loop 2 (radius b) Loop 1

(radius a)

I

Figure b

Tutorial 7

ELECTROMAGNETIC THEORY - KJE442 VI) Figure shows a coaxial cable of infinite length with inner conductor having a radius a and outer

conductor of internal radius b. The relative permeability of the dielectric in between the conductors is assume equal to 1. Each conductor carries a current I in opposite direction. (a) Using Ampere’s law show that the magnetic field intensity H is zero at any point

outside the coaxial cable. (b) Using Ampere’s law, determine the expression of the magnetic field intensity H at a

point in between the inner and outer conductor of the coaxial cable.

(c) Find the magnetic flux between the conductors in a length d (the flux crossing the radial surface S).

(d) Hence, or otherwise, derive the self-inductance per unit length of the coaxial cable

(not including the internal inductance of the inner conductor).

Inner conductor Radius = a Dielectric

µr = 1

Outer conductor

d

S

b

Tutorial 7

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 8

I) Given a magnetic induction, xuB 210 −=r

Weber/m2, find the force on an electron whose velocity a) in the x-direction b) in the y-direction c) in the z-direction d) in the xy-plane at 450 to the axis.

II) The force experienced by a test charge q for three different velocities at a point in a region

characterized by electric and magnetic fields are given by:

( )[ ]( )[ ][ ] zyx

yyx

xyx

uvvuEuEqF

uvvuEuBvEqF

uvvuBvEuEqF

03003

0200002

0100001

for

for

for

=+=

=++=

=−+=

where, v0, E0, B0, are constants. Find E

r and B

r at the point.

III) A test charge q, moving with a velocity ( )yx uuv +=r

, experiences no force in a region of

electric and magnetic fields. If the magnetic flux density is ( )zx uuB 2−=r

Weber/m2, find Er

.

Br

R x

y IV) The circuit shown in figure is situated in a

magnetic field ( ) zutxB αω += sin.r

(T). Determine the voltage induced in the circuit.

h

w

V) Two metal bars slide over a pair of conducting rails in a uniform magnetic field

zzxx uBuBB +=r

with constant velocities xx uvv =1 and xx uvv −=2 . Determine the induced voltage in the circuit.

Br

1v 2v x

y

O

Tutorial 8

ELECTROMAGNETIC THEORY - KJE442 VI) A time-varying magnetic field is given by

( ) yutxB π2cos10=r

. Find the induced emf around a rectangular loop in the x-z plane as shown in Figure.

Br

b

a

z

y

x

VII) A rectangular loop of wire three (3) sides fixed and the fourth sid movable is situated in a

plane perpendicular to a uniform magnetic field

e

( )zy uuB += 5.1r

, refer to Figure 4b. The movable side consists of a conducting bar moving with a velocity 10 m/s in the y-direction. Find the emf induced in the loop.

Movable side

vr

Br

z

(0,0,2)

(0,0,0) y

x

VIII) A time varying magnetic field intensity is given by:

zutH ).5sin(..10 2ρ= i) Find the induced emf around a circular loop within radius a on x-y plane. Refer to

Figure Q4a. ii) If radius a is 10 mm and the loop has resistance R = 10 Ω, determine the current

induced in the loop and its direction.

H

R a

z

Tutorial 8

ELECTROMAGNETIC THEORY - KJE442 IX) Consider the metal bar and slider arrangement shown in Figure. The slide starts at the top and

fall under the influence of gravity. The length b of the bars is equal to 1 meter and the spacing a between bars is equal to 0.2 meter. A uniform magnetic field with a magnetic flux density

vector B equal to 20 Wb/m2 is passing between the bars.

i) The acceleration due to the gravitational force is zutg .8.9)( −= m/s2.

Determine the velocity )(tv of the slide. ii) Calculate the voltage V(t) generated as the slide falls from the top until it leaves

the bars and breaks contact.

iii) Assuming the initial position of the slide is at z0 = 0 and time t0 = 0, determine the time when the slide breaks contact with the bars.

iv) Plot the voltage V versus time t.

V(t)

b

a

B

Slider

z0 = 0

z y

x

bar

z0 = -b

Tutorial 8

ELECTROMAGNETIC THEORY - KJE442

TUTORIAL 9 I) Given a magnetic field in free space where there is neither charge nor current density (ρv = J =

0),

( ) ( ) yx unxtynaunxtB −+−= ωω cos...sin.r

with a, n, and ω are constants.

a) Use a Maxwell equation to derive the time-dependent electric field Er

. b) Determine the Poynting vector.

II) An electromagnetic wave propagates through a lossless insulator with a velocity 1.8 x 1010

cm/s. Calculate the electric and magnetic properties of the insulator if its intrinsic impedance is 260 Ω.

III) The wavelength of a 600 MHz wave propagating through a non-magnetic dielectrics is 20 cm.

What is the dielectric constant of the material? IV) A 3 GHz uniform plane wave propagates through rexolite in the positive z direction. The E

r

field at z = 0 is 100∠0 V/m. a) Calculate the rms value and phase of E

r at z = 4 cm.

b) Determine the total wave attenuation (in dB) over a distance of 6 wavelengths. (Given, for rexolite, εr = 2.54 and tan θ = 0.005)

V) A sinusoidal electric intensity of amplitude 250 V/m and frequency 1 GHz exists in a lossy

dielectric medium that has a relative permittivity of 2.5 and a loss tangent of 0.001. Find the average power dissipated in the medium per cubic meter.

VI) Determine the skin depth at 1 GHz of a conductor if the conductivity is 5 x 105 mho/m.

Assume it’s relative permittivity equal to 1. VII) The differential form of Maxwell’s Equations for harmonic electromagnetic field are given by:

vD

BjE

ρ

ω

=∇

−=∧∇

.

0. =∇

+=∧∇

B

DjEH ωσ

a) Starting from the Maxwell’s Equations for harmonic fields and using a vector identity:

( )EEE .2

∇∇+∇−=∧∇∧∇ , show that the electric wave equation for the source-free case (charge density, ρv = 0) can be written as:

( )EjjE ωεσωµ +=∇2

Deduce the electric wave equation for the case where the propagating medium is free space.

Tutorial 9

ELECTROMAGNETIC THEORY - KJE442

b) Assuming wave propagation along +x, and fixing the electric field along +y, the solution of the wave equation is given by

ytjxjx ueeeEE ωβα −−= .0

A uniform plane wave is propagating in +x-direction in free space as shown in Figure Q5b. At interface A (at x = 0), the wave enters a lossy dielectric medium of εr = 81 and µr = 1. The wave phase constant in the free space is found to be 20π rad/m. The magnitude of the wave at Interface A is measured to be 10 mV/m.

i) Determine the frequency of the plane wave. ii) Determine the phase velocity and wavelength of the plane wave in the lossy

dielectric. iii) If the magnitude of the electric field at 10 m from the Interface A is 9.5

mV/m, determine the attenuation constant in the lossy dielectric and the conductivity of the lossy dielectric.

Interface A

x

x = 0 |E|(x=0) = 10 mV/m

x = 10 m |E|(x=10) = 9.5 mV/m

Plane wave

yE

Free space Lossy dielectric εr = 81 µr = 1

VIII) Given an electromagnetic wave of magnetic field intensity:

( ) ( ) yutjzHH .102exp.exp. 60 πγ−=

(a) Determine the expression of the electric field intensity of the wave. (b) Determine the Poynting Vector. (c) If the wave is propagating in a lossless material with ε = 2ε0 and µ = µ0,

(i) determine the attenuation constant α. (ii) find the phase constant β. (iii) find the phase velocity vp. (iv) find the wavelength λ. (v) determine the intrinsic wave impedance of the material, η.

(d) If the wave is propagating in a lossy dielectric with conductivity σ = 10-15 S/m,

permittivity ε = 2.2ε0 and permeability µ = µ0,

(i) determine the loss tangent of the material. (ii) determine the attenuation constant α of the material. (iii) find the distance the wave travels in the material such that the magnitude of

the electric field intensity is reduced by half.

Tutorial 9

A P P E N D I X 1 ( 2 ) M a x w e l l ’ s E q u a t i o n s :

D I F F E R E N T I A L F O R M I N T E G R A L F O R M

vD ρ=∇.r

∫∫∫∫∫ =v vs

dvdSD .. ρ

0. =∇ Br

0. =∫∫s dSB

tBE∂∂

−=∧∇r

∫ ∫∫ ∂∂

−=L S

dStBdE .. l

tDJH∂∂

+=∧∇r

∫ ∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+=L S

dStDJdH .. l

C o n s t a n t s : Permittivity of free-space: ε0 = 8.854 x 10-12 F/m = (36π x 109)-1 F/m Permeability of free-space: µ0 = 4π x 10-7 H/m C a r t e s i a n : zyx uuu ,, are unit vectors along x, y and z directions respectively.

zyx uzVu

yVu

xVV ⎟

⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇r

zA

yA

xA

A zyx

∂∂

+∂

∂+

∂∂

=∇rr

zxy

yzx

xyz u

yA

xA

ux

Az

Au

zA

yA

A ⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥⎦

⎤⎢⎣

⎡∂∂

−∂∂

+⎥⎦

⎤⎢⎣

⎡∂

∂−

∂∂

=∧∇rr

2

2

2

2

2

22

zV

yV

xVV

∂∂

+∂∂

+∂∂

=∇r

zyx udzudyudxd ... ++=l

z

y

x

udydx

udzdx

udzdydS

..

..

..=

dzdydxdv ..=

A P P E N D I X 2 ( 2 )

C y l i n d r i c a l :

zuuu ,, φρ are unit vectors along ρ, φ and z directions respectively.

zuzVuVuVV ⎟⎠⎞

⎜⎝⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=∇ φρ φρρ1r

( )z

AAAA z

∂∂

+∂

∂+

∂∂

=∇φρ

ρρρ

φρ

1.1rr

( )z

zz uAA

uA

zA

uz

AAA ⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂

∂+⎥

⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂∂

=∧∇φρρ

ρρρφρ

ρφφ

ρρ

φ 111rr

2

2

2

2

22 11

zVVVV

∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

=∇φρρ

ρρρ

r

zudzududd .... ++= φρ φρρl

zudd

udzd

udzddS

...

..

...

ρφρ

ρ

φρ

φ

ρ=

dzdddv ... φρρ=

S p h e r i c a l :

φθ uuur ,, are unit vectors along r, θ and φ directions respectively.

φθ φθθuV

ruV

ru

rVV r ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

+⎟⎠⎞

⎜⎝⎛∂∂

=∇sin11r

( ) ( )φθ

θθθ

φθ ∂

∂+

∂∂

+∂∂

=∇A

rA

rAr

rrA r sin

1sinsin1.1 2

2

rr

( ) ( ) ( )φ

θθ

φθφ

θφθφθθ

θu

Ar

rAr

ur

rAAr

uAA

rA rr

r ⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂+⎥

⎦

⎤⎢⎣

⎡∂

∂−

∂∂

+⎥⎦

⎤⎢⎣

⎡∂∂

−∂

∂=∧∇

1sin

11sinsin1rr

2

2

2222

22

sin1sin

sin11

φθθθ

θθ ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∇V

rV

rrVr

rrV

r

φθ φθθ udrudrudrd r ..sin... ++=l

φ

θ

θ

φθ

φθθ

uddrr

uddrr

uddrdS r

...

...sin

...sin2=

φθθ dddrrdv ...sin.2=