Sem5 em skee 2523 electromagnetic field theory lect 01-24

Copyright of M.H. Ibrahim, 2008

WELCOME TO THE WORLD OF

ELECTROMAGNETICS

Mohd Haniff Ibrahim

Room Number: P03-510

Email: [email protected]

H/P: 013-7146367

Notes are downloadable at mail.fke.utm.my/~hanif/teaching/SEE2523.htm


• Course Outline

• Evaluation Method and Important Dates

• Contents

Electromagnetics Field Theory: SEE 2523

• Reference Books


LECTURE #1

Topics to be covered: Revision on vectors

Motivation:

To enable the students to refresh their knowledge on vectors and its operation

Reference: Hayt & Buck, page 1-14


Why Vectors?

AB

A

B


Vector Operations

• Vector Components in Rectangular Coordinates

• The Dot Product

• The Cross Product


Vector Components in Rectangular Coordinates

x

y

z

• A (2,4,5)

• B (4,5,6)

=AB

=BA

=AB

=BA

=r

=r


Example: D1.1 (Hayt pg.8)

Given points M(-1,2,1), N(3,-3,0) and P(-2,-3,-4). Find:

a)

b)

c)

d)

e)

MNMPMN +

MPrM

NM 32 −


The Dot Product

• The first type of vector multiplication

• The yield is a scalar component

•Definition:

ABcosBABA θ=•

• In rectangular coordinates:

•In general:

z,y,x

zzyyxx BABABABA ++=•


Example: D1.3 (Hayt pg. 12)

The three vertices of a triangle are located at A (6,-1,2), B(-2,3,-4) and C(-3,1,5). Find:

a)

b)

c) The angle at vertex A

AB

AC

BACθ


The Cross Product

• The second type of vector multiplication

• The yield is a vector component

• By definition:

nsinBABA ABθ=×

• The yield vector is normal to both A and B

• In rectangular coordinates:

• In general: z,y,x

zyx

zyx

BBB

AAAzyx

BA =×


Example: D1.4 (Hayt pg.14)

The three vertices of a triangle are located at A(6,-1,2), B(-2,3,-4) and C(-3,1,5). Find:

a)

c) A unit vector perpendicular to the plane in which the triangle is located.

ACAB×


Next Lecture

Please have a preliminary observation on the following topics:

a) Coordinate systems:

(i) Rectangular coordinate

(ii) Cylindrical coordinate

(iii) Spherical coordinate

Please refer to Hayt and Buck, page 14-22

LECTURE #2

Topics to be covered: Coordinate Systems

Motivation:

To familiarize the students with common coordinate systems used in

solving electromagnetic problems

Reference: Hayt & Buck, page 14-22

Why Coordinate Systems???

• Electromagnetic problems can be in any forms.

e.g.

Find electric/magnetic field in the wire and its surrounding.

Find electric/magnetic field induced by the radio antenna

Find electric/magnetic field induced by the telco tower

Need proper coordinate systems to analyze the electromagnetic problems

Important parameters of coordinate systems

• Differential line length, (scalar and vector)

Application example: Obtain the electric potential (Chapter 4)

dl

• Differential area, (scalar and vector)

Application example: Obtain the electric field using Gauss’s Law (Chapter 3)

ds

• Differential volume, (scalar)

Application example: Obtain the electric field using Coulomb’s Law (Chapter 2)

dv

∫= dl.EV

enQdsD =∫ .

∫= 24 rdvEo

v

περ

Rectangular Coordinate System (x, y, z)

x

y

z

A

B

From A to B: =

Surface C: ds =

Surface D: ds =

Surface E: ds =

dx

dz

dy

surface C

surface D

dl

surface E

x

y

z

Cylindrical Coordinate System (r, , z)φ

y

dr

rdφ

dz

z

x

z

r

φ

H (r, , z)φ

r

φ

z

Spherical Coordinate System (r, θ, )φ

y

z

x

φH (r, θ, )φ

φ

θr

r

θ

rdθ

drφrsinθd

Next Lecture


(i) Type of electrostatic charges

(ii) Electric field intensity

(iii) Coulomb’s Law


Lecture #3

Topics to be covered

(i) Type of electrostatic charges

Reference: Hayt and Buck, page 26-48

Motivation:

To familiarize the students with the type of sources for

electrostatic field

Electric charges

• Smallest magnitude of charge,

q = 1.602 ×10-19 (Coulomb)

• Charge can be positive or negative

• In electromagnetic problems: electric charges can be of type:

(i) Point charge

(ii) Line charge

(iii) Surface charge

(iv) Volume charge

Point charge

•Simplest electric charge

•Normally labeled as Q. Unit of Coulomb (C)

Q

Line charge

• Accumulated point charges along a thin line

• Accumulated point charges line charge density, ρl (C/m)

ρl (C/m)dldQ lρ=

A

B

Total charge along AB ???

dl

Surface charge

• Accumulated point charges on a surface

•Accumulated point charges surface charge density, ρs (C/m2)

ρs(C/m2)

dsdQ sρ=

ds

Total charge on the surface ???

Area A

Volume charge

• Accumulated point charges in a volume

•Accumulated point charges volume charge density, ρv (C/m2)

ρv(C/m2)

dvdQ vρ=

dv

Total charge in the volume ???

Volume V

Example: D2.4 (b) (Hayt pg.37)

Calculate the total charge for the following volume:

φρπφ 6.0sin;42,0,1.00 22zrzr v =≤≤≤≤≤≤

Example: Problem 4.5 (Sadiku pg. 155)

Determine the total charge for the followings:

(i) On line 0<x<5m if ρl =12x2 mC/m

(ii) On the cylinder r=3, 0<z<4m if ρs=rz2 nC/m2

(iii) Within the sphere r=4m if ρv=10/rsinθ C/m3

Example: Problem 2.1a (Skitek pg. 51)

Find the charge within a sphere of radius 0.03m when the charge density is given by:

( )3223

/sin

cos102 mCrv θ

φρ−×

=

Next Lecture


(i) Electric field intensity

(ii) Coulomb’s Law


Lecture #4

Topics to be covered:

(i)Electric Field Intensity

(ii) Coulomb’s Law


Motivation:

To understand the concept of electric field intensity and

its method of calculation

Electric (Electrostatic) Field Intensity

• Field generated by the static charges mentioned in the last lecture.

• Symbol of . Unit of volt/meter, (V/m).

• The first law to calculate the is Coulomb’s Law.

E

Electro+ static

Electric field Static sources

E

Coulomb’s Law

• The law was developed by French physicist, Charles Augustin de Coulomb in 1780s.

“ The magnitude of electrostatic force between two point electric charges is directly proportional to the product of the

magnitudes of each charge and inversely proportional to the square of the distance between the charges”

221

rQQF ∝

Coulomb’s Law

Q1 Q2

2F1F12r21r

12221

2 rrQQkF =e.g.

Coulomb’s experiments: Constant, o

kπε41

=

εo : Permittivity in free space/Dielectric constant of vacuum.

≈ 8.854 × 10-12 Farad/meter (F/m)

(Newton)

Coulomb’s Law

With regard to Lorentz Force Law (Chapter 9):

EQF =

Hence,

1221

1 ˆ4

rr

QEoπε

=

2122

2 ˆ4

rr

QEoπε

=

Q1

12r

1EA

Q2

21r

2EB

Field at A;

Field at B;

V/m

V/m

r

Coulomb’s Law for Point Charge

Electric field intensity incur by any point charge, Q

rr

QEo

ˆ4 2πε

=

Q

r

EB

r

Unit of voltage/meter (V/m)

Example: Problem 4.1 (Sadiku)

Point charges, Q1=5µC and Q2=-4µC are placed at (3,2,1) and (-4,0,6) respectively. Determine the force on Q1.

Example: Problem D2.2 (Hayt)

A charge of -0.3µC is located at A(25,-30,15) (in cm) and a second charge of 0.5µC is at B(-10,8,12) cm. Find electric field intensity at (a) origin, (b) P(15,20,50) cm.


Point charges Q1 and Q2 are respectively located at (4,0,-3) and (2,0,1). If Q2 = 4 nC, find Q1 such that:

a) The electric field at (5,0,6) has no z component.

b) The force on a test charge at (5,0,6) has no x component.


Five identical 15 µC point charges are located at the center and corners of a square defined by -1<x,y<1, z=0.

(a) Find the force on 10µC point charge at (0,0,2)

(b)Calculate the electric field intensity at (0,0,2)

Next Lecture


(i) Electric field intensity of continuous charge distribution.


Lecture #5


Electric Field Intensity of Continuous Charge

Distribution


Motivation:

To calculate the electric field intensity based on continuous

charge distribution

Recall (Coulomb’s Law)

Previously, Coulomb’s Law for point charge:

At A;

Rao

ar

QE ˆ4 2πε

=

Q A

Ra

ra

What if Q is a line charge, surface charge or volume charge ???

(V/m)

Electric Field Intensity-Line Charge Distribution

What is the field at A as incurred by the line charge?

A

( )mCl /ρ

l

Step 1: Identify small line length, dl with element of charge, dQ

dldQ lρ=dl

Ro

l aR

dlEd ˆ4 2περ

=

Step 2: Identify the electric field intensity, dE as contributed by dQ (based on Coulomb’s Law).

∫=z

o

total EdE

Step 3: Integrate dE, over the length of line charge, z, to obtain total electric field.


For line charge, normally cylindrical coordinate system was utilized.From previous,

A

'dzdQ lρ=dz’

( )zr ,,φ

lρ

r

x

y

z

φ

z

E field at A ???

Step 1: ???

Step 2: ???

Step 3: ???

z’

a

b


Step 1: 'dzdQ lρ=

Step 2:

( )( )( )( )( )212'2

'

2'2

' ˆˆ

4 zzr

zzzrr

zzrdzEd

o

l

−+

−−

−+=

περ

Step 3: Using ;

( ) ( ) mVrz

rrE

o

l /coscosˆ

sinsinˆ

4 1212⎭⎬⎫

⎩⎨⎧ −++= αααα

περ

( ) ( )21

22223

22 xcc

x

xc

dx

+=

+∫

( ) ( )21

2223

22

1

xcxc

xdx

+

−=

+∫and


For a ∞ and b ∞ (line charge of infinite length)

( )mVrr

Eo

l /ˆ2περ

=

Problem 2.19 - Hayt

A uniform line charge of 2µ C/m is located on the z axis. Find electric field in Cartesian coordinates at P(1,2,3) if the charge extends from (a) z = -∞ to z= ∞; (b) z=-4 to z=4

Example: Problem 2.5.1 (Skitek)

Find the electric field intensity on the axis of a circular ring of uniform charge, ρl and radius a. Let the axis of the ring be along the z axis.


Find the electric field at P(0,0,1.5) due to two infinite and parallel line charges of uniform ρl. Let one ρl=10-6

C/m line be located at y=2 m and other ρl =-10-6 C/m line be located at y=-2m, with both lines parallel to the z axis and in the z=0 plane.

Problem 2.26: Hayt

A uniform line charge density of 5 nC/m is at y=0, z =2 in free space, while -5 nC/m is located at y=0, z=-2m. Find the electric field at the origin.


Two identical uniform line charges, with ρl =75 nC/mare located in free space at x=0, y=±0.4m. What force perunit length does each line exert on the other.

Example: D2.5 (Hayt page 43)

Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes in free space. Find electric field intensity at (a) (0,0,4) ; (b) (0,3,4)

Electric Field Intensity-Surface Charge Distribution

ρs (C/m2)

BWhat is the electric field at B ???

As in the case of line charge distribution;

dS

Step 1: Identify small surface, dS having charge element of dQ

dSdQ sρ=

Step 2: Identify electric field intensity, dE as contributed by dQ

Ro

S aR

dSEd ˆ4 2περ

=

Step 3: Integrate dE over the total surface, to obtain total electric field

∫=surface

EdE


)/(ˆ4 2 mVa

RdSE Ro

s

περ

=

For a finite surface area, electric field incurred at any desired location:

What if the surface is having infinite area ???

Consider a circular surface charge having density of ρS (C/m2) with radius a, centered at origin and lying on the z=0 plane.

To solve


x

y

z

a

C (0,0,z)Find electric field at C ???

dS

( )φρρ rdrddSdQ sS ==

( )( ) ⎥

⎥

⎦

⎤

⎢⎢

⎣

⎡

+

+−=

23

224 zr

zzrrrdrdEdo

s

πεφρ

Using cylindrical coordinate;

∞→aConsider;

???E

Electric Field Intensity- Infinite Surface Charge Distribution

For surface charge distribution of infinite area:

)m/V(nEo

s

ερ2

=

Example:D2.6 (Hayt, page 45)

Three infinite uniform sheets of charge are located in free space as follows: 3 nC/m2 at z =-4, 6 nC/m2 at z=-1 and -8 nC/m2 at z=-4. Find electric field intensity at the point (a) (2,5,-5); (b) (4,2,-3); (c) (-1,-5,2) and (d) (-2,4,5)

Example: (Exercise from Skitek)

Find the electric field along the z axis, as incurred by the annular ring with surface charge density of ρs, having inner radius of a and outer radius of b.

Example:Problem 2.6-1 (Skitek)

A semi-infinite sheet of charge density, ρS is described by -∞< x <0, -∞< y <∞, in the z=0 plane. Calculate the component of electric field normal to the sheet at a distance a, directly above the edge at x=0.

Electric Field Intensity-Volume Charge Distribution

As previously noted:

)/(ˆ4

)/(ˆ4 22 mVa

RdVmVa

RdQE R

o

vR

o περ

πε==

However, we are not going to cover on detail…..

However, electric field due to volume charge distribution will be elaborated in Chapter 3.

End of Chapter 2…..

Next Lecture


(i) Electric Flux

(ii) Electric Flux Density

(iii) Gauss’s Law


Lecture #6


(i) Electric Flux/ Flux Density

(ii) Gauss’s Law


Motivation:

To understand the concept of electric flux/ flux density and to apply the Gauss’s Law in electric

field calculation

Electric FluxSketched field lines (imaginary lines) pointed outwards from the electric charge. Symbol of ΨE (psi). A scalar component.

Flux is in the same direction as the electric field

QElectric flux about a point charge

Relation: ΨE=Q Unit of electric flux: Coulomb

E

Electric Flux

-∞

∞

Electric flux about an infinite line charge

∞

∞

∞

Electric flux about an infinite surface charge

∞

Electric Flux Density

Symbol of : vector quantity

Unit of Coulomb/m2

D

Total electric flux, ΨSurface with area A

( )2m/CA

D Ψ=∴


Consider a sphere with radius r, and we locate a point charge, Q at the centre of the sphere:

Q

Flux density at radius r:

( )222 44

m/Crr

Qrr

D E

ππψ

==

Electric field intensity at radius r:

( )m/Vrr

QEo

24πε=

Hence: ED oε=


Consider a small portion of sphere (radius of r ) surface, dS and flux (from point charge, Q, at the centre) pass through the surface.

Small element of flux that pass through dS :

∫=dS,surface

E dS.Ddψ

Total flux that pass through the sphere:

∫ ==Ssurface

EE Qd,

ψψ

∫ ==Ssurface

enclosedQdSD,

.Gauss’s Law

dS

Flux density,

D

Q

Gauss’s Law

The law was developed by Carl Friedrich Gauss, a German mathematician in the early 19th century.

The law states that:

“ Total electric flux passing through any closed imaginary surface, enclosing the charge Q(Coulomb), is equal to Q (Coulomb)”

How to apply Gauss’s Law in calculating electric field???

Mathematically;

∫ =surface

enQdSD .

Gauss’s Law-Point Charge

Q

Step 1: Draw a close imaginary surface with radius of r, enclosing the point charge Q. Spherical surface can be used.

Imaginary surface

ensurface

QdSD =∫ .

The aim is to find an electric field at a distance r from the point charge.

r

Step 2: Apply the Gauss’s Law, and obtain the flux density, DStep 3: Use the relation of and obtain . The obtained electric field is the electric field at a distance r from the point charge.

ED oε= E

Gauss’s Law- Line ChargeStrictly for special case of line charge – Infinite length of line charge.

The aim is to find an electric field at a distance r from the line charge.

-∞

∞

r

Step 1: Draw a close imaginary surface with radius of r, enclosing the line charge ρl. Cylindrical surface can be used.

Step 2: Apply the Gauss’s Law, and obtain the flux density, D

Step 3: Use the relation of and obtain . The obtained electric field is the electric field at a distance r from the line charge.

ED oε=E

L

( )mCl /ρ

Gauss’s Law- Surface ChargeStrictly for special case of surface charge – Infinite area of surface charge.

The aim is to find an electric field above and below the surfacecharge.

aStep 1: Draw a close imaginary surface enclosing the surface charge ρS.Cylindrical surface with radius r can be used.

ρS(C/m2)

Step 2: Apply the Gauss’s Law, and obtain the flux density, DStep 3: Use the relation of and obtain . The obtained electric field is the electric field above the surface charge.

ED oε=E

r

Exercise #1: Skitek

Through the use of Gauss’s Law, find the expression for , inside and outside of a sphere of ro radius and of uniform ρV(C/m3) distribution.

D E

Exercise #2: Skitek

Through the use of Gauss’s Law, find the expression for , inside and outside an infinite cylinder of ro radius and of uniform ρV (C/m3) distribution.

D E

Example 1

Let (nC/m2) in free space.

(a) Find at r = 0.2 m.

(b)Find the total charge within the sphere, r = 0.2 m.

(c) Find the total electric flux leaving the sphere, r = 0.3 m.

rrD ˆ3

=

E

Example 2

The cylindrical surfaces, r = 1, 2 and 3 cm carry uniform surface charge densities of 20, -8 and 5 nC/m2, respectively.

(a) How much electric flux passes through the closed surface, r=5 cm, 0<z<1m

(b)Find at P(1cm, 2 cm, 3 cm)D

Example 3

The spherical surfaces r = 1, 2 and 3 cm carry surface charge density of 20, -9 and 2 nC/m2, respectively.

(a) How much electric flux leaves the spherical surface r =5 cm.

(b)Find at P (1cm, -1 cm, 2 cm). D

Example 4: (Hayt)

Spherical surfaces at r = 2, 4 and 6 m carry uniform surface charge densities of 20 nC/m2, -4 nC/m2 and ρso, respectively.

(a) Find at r = 1, 3 and 5 m.

(b)Determine ρso such that = 0 at r = 7m.

DD

Example 5: D 3.5 (Hayt)

A point charge of 0.25 µC is located at r =0, and uniform surface charge densities are located as follows: 2 mC/m2 at r =1 cm and -0.6 mC/m2 at r = 1.8 cm. Calculate at:D

(a) r = 0.5 cm

(b) r = 1.5 cm

(c) r = 2.5 cm

(d)What uniform surface charge density should be established at r = 3 cm to cause = 0 at r = 3.5 cm. D

Example 6: (Hayt)

A uniform volume charge density of 80 µC/m3 is present throughout the region 8 mm < r < 10 mm. Let ρV = 0 for

0 < r < 8 mm.

(a) Find the total charge inside the spherical surface, r = 10mm.

(b)Find Dr at r = 10 mm.

(c) If there is no charge for r > 10 mm, find Dr at r = 20 mm.

Next Lecture


(i) Divergence

(ii) Divergence Theorem


Lecture #7


(i) Divergence

(ii) Divergence Theorem


Motivation:

To understand the divergence concept and the application of divergence theorem

Divergence

Consider the Gauss’s Law;

ensurface

QdSD =•∫

Graphically;

∫ •surface

dSD

Qen

What if the closed surface volume shrink to zero ?

Let the charge be located at (xo, yo, zo)

(xo, yo, zo)

Divergence

Consider the closed surface volume shrink to zero. Take ∆v as the closed surface volume.

Divided both sides of Gauss’s Law with ∆v and take the limit as ∆v 0.

⎥⎦⎤

⎢⎣⎡∆

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∆

•

→∆→∆

∫v

Qv

dSDen

v

surface

v 00limlim

Mathematically;

Charge density, ρV at point (xo, yo, zo)Divergence of orD D•∇

Divergence

Using a divergence concept, the Gauss’s Law can be stated as:

VD ρ=•∇

Physically;

1st Maxwell’s equation for static

field

“ It relates the rate of change of the component to the charge density, ρV at a point”

D

Positive divergence

“flux source”

Negative divergence

“sink of flux”

+ρV

D

-ρV

D

VD ρ=•∇ VD ρ−=•∇

Divergence

Divergence operator, is a vector quantity:( )•∇Heavily depending upon coordinate system.

Rectangular:

Cylindrical:

Spherical:

zD

yD

xDD zyx

∂∂

+∂

∂+

∂∂

=•∇

( )z

DDr

rDrr

D zr ∂

∂+

∂∂

+∂∂

=•∇φφ11

( ) ( )φθ

θθθ

φθ ∂

∂+

∂∂

+∂∂

=•∇D

rD

rDr

rrD r sin

1sinsin11 2

2

Example 1

Find the volume charge density that is associated with each of the following fields:

( )222 /ˆˆˆ mCzzyyxxxyD ++=(a)

(b) ( )222222 /ˆsinˆcossinˆsin mCzzrrzrrzD φφφφφ ++=

Example 2

If , find:( )23 /ˆ2ˆ2ˆ4 mCzyyzxxD −−=(a) div D

(b) ρV at P(x, y, z)

(c) the total charge lying within the region -1<x,y,z<1

(d) the total charge lying within that region without finding ρV first.

Example 3

Evaluate above an infinite sheet of uniform charge distribution, ρS (C/m2) located at the z = 0 plane.

D•∇

Example 4

Evaluate at a distance r from an infinite length of line charge of uniform charge distribution, ρl (C/m). The line charge is located at z-axis.

D•∇

x

y

z

b

Example 5

a

ρV (C/m3)

(a) Find the expression of at r<a, a<r<b and r>b

(b)Evaluate at r<a, a<r<b and r>bD

D•∇

Divergence Theorem

A theorem used to relate the surface integral to a volume integral involving the same vector.

ρV

∫ •surface

dSD

∫=volume

Ven dVQ ρ

Consider a volume charge and applying the Gauss Law.

D•∇

Divergence Theorem∫∫ •∇=•volumesurface

dVDdSD⇒

∫∫ ==•⇒volume

Vensurface

dVQdSD ρ

Gauss’s Law:

Summary of Chapter 3

Students are expected to be well-versed in the followings:

(d) Extension of Gauss’s Law that lead to the concept of divergence (1st Maxwell’s equation for static field)

(a) Concept of electric flux and flux density

(b) Relation between flux density and electric field

(c) Application of Gauss’s Law in calculating electric field induced by special cases of charge distribution

The end of chapter 3

Next Lecture


(i) Potential Difference

(ii) Absolute Potential


Lecture #8


(i) Potential difference

(ii) Absolute Potential


Motivation:

To enable the students to understand the concept of potential difference and

absolute potential induced by point charge and continuous charge system

Work in Electric Field

E

Q

Point B

Point A

( )JouledlFdW •−=

Small work done in moving Q along the length dl:

dl

Work done in moving Q from point B to point A:∴

( )JouledlEQWA

B∫ •−=


Vector of length; (Discussed in our first and second lecture)dl

Depending on the coordinate used:

( )( )

( )Sphericaldrrdrdrdl

lCylindricazdzrdrdrdl

Cartesianzdzydyxdxdl

φφθθθ

φφˆsinˆˆ

ˆˆˆ

ˆˆˆ

++=

++=

++=

Example

(2,0,0)

x

y

z

(0,4,0)

(0,0,4)

Path 1

Path 2

Obtain the work done in moving 2 C charge in existence of

(i) From (2,0,0) to (0,4,0) along path 1

(ii) From (2,0,0) to (0,4,0) along path 2

(iii) Round trip movement at (2,0,0) along path 1 and path 2.

zzyyxxE ˆˆˆ ++=


From previous example:

(i) = (ii)The work done does not depend on the path taken.

From (iii);

0=•−= ∫ dlEQW 2nd Maxwell’s equation for

static field

Potential Difference/Potential

Work done in moving a unit of charge from point B to point A:

=QW ∫ •−

A

B

dlE (volt)

Absolute potential (potential): (if point B is in infinity)

From above:

∫∞

∞ •−=−=A

A dlEVVQW

Potential difference between point A and B

BA

A

BAB VVdlEV −=•−= ∫

Potential Difference (Point Charge)

Q

rA

rB

AB

Let us move a unit of positive charge, 1C from point B to point A:

The potential difference between point B and point A

∫ •−=A

BAB dlEV

( ) rdrdl;m/Vrr

QEo

== 24πε

From:

???VVV BAAB =−=∴

Q

rA

∞

A∞

( ) ∫∞∞ •−=A

A dlEV

Absolute Potential/Potential (Point Charge)

What if B=∞

???VVA ==∴

Potential (Charge System)

Q4

What is total absolute potential/potential at A ???

A

Q1

Q2 Q3

R1

R2R3

R41

1

4 RQ

oπε

2

2

4 RQ

oπε

3

3

4 RQ

oπε4

4

4 RQ

oπε

∑=

=+++=∴4

14

4

3

3

2

2

1

1

44444 i io

i

ooooA R

QR

QR

QR

QR

QVπεπεπεπεπε


What if we have a system of charges; i.e. line charge, surface charge, volume charge ???

ASurface charge; ρS (C/m2)

From previous slide:

∑=

=4

1 4i io

iA R

QVπε ∫=∴

surface oA R

dQVπε4

What is the potential at A due to the surface charge ??

Continuous charge system


In general; potential at a point (let point A) due to continuous system of charges:

Line Charge:

Surface Charge:

Volume Charge:

∫=surface o

SA R

dSVπερ

4

∫=line o

lA R

dlVπερ

4

∫=volume o

vA R

dvVπερ

4

Example 1: Skitek

Obtain the expression for the absolute potential along the z-axis of a ring of uniform ρl (C/m) and whose radius is ro. The ring is located at z = 0.

( )z

zr

zrE

oo

ol

23

222 +=

ε

ρ

Given; electric field at any point on z axis

Example 2: Skitek

Obtain the absolute potential inside and outside a thin shell, centered at the origin, of uniform ρS (C/m2) and of radius ro. Obtain the potential at:

( )

( )( ) o

o

o

rriiirrii

rri

2

2

==

=

Next Lecture


(i) Potential Gradient

(ii) Energy Density in the Electrostatic Field


Lecture #9


(i) Potential Gradient

(ii) Energy Density in Electrostatic Field


Motivation:

To enable the students to calculate the electric field using the potential value

and to understand the concept of energy density in electrostatic field

Recall …

Charge Distribution

( )vSlQ ρρρ ,,,

Coulomb’s Law

Electric Field Intensity

E

Gauss’s Law

Electric Potential

V

∫= RdQV

oπε4

Potential Gradient

Potential Gradient

Consider…

RQV

oA πε4=

Q

Point A

R

Alternatively, electric field at point A can be obtained using gradient of potential at point A

AA VE ∇−=

gradient or grad

Gradient operator:

zzVy

yVx

xVV ˆˆˆ

∂∂

+∂∂

+∂∂

=∇

zzVV

rr

rVV ˆˆ1ˆ

∂∂

+∂∂

+∂∂

=∇ φφ

φφθ

θθ

ˆsin1ˆ1ˆ

∂∂

+∂∂

+∂∂

=∇V

rV

rr

rVV

(Cartesian)

(Cylindrical)

(Spherical)

Example 1

Please refer to Example 1 in Lecture #8. Given the potential at any point on z-axis is:

( )21222 zr

rV

oo

ol

+=

ε

ρ

Obtain the electric field term at any point on z-axis and prove that the answer is similar to the given term in Example 1 (Lecture #8).

Example 2

Using Example 2 in Lecture #8, obtain the electric field terms at the following location (radius) based on the previously obtained potential terms:

( )

( )( ) o

o

o

rriiirrii

rri

2

2

==

=

Verify your results using the common law.

Energy Density in Electrostatic Field

ρl (C/m)

ρS (C/m2)

ρv (C/m3)

What is the total energy conserved by the charge distributions?


For simplicity, consider…

We have a system of four point charges.

The work done by an external source in positioning (locating) the charges

What is the energy (potential energy) present in such system of charges ???

Q4

Q3

Q2

Q1


How to calculate the work done by external source in positioning the charges ???

Step 1: Assume that the system is empty (no charge exist)

Step 2: Bring point charge, Q1 from infinity into the system. No work required !!!!

Q101 ==WW


Step 3: Bring point charge, Q2 from infinity into the system. Requires work !!!!

Q1

Q2R21

21

12

212

21

40

0

RQQ

VQWWW

oπε+=

+=+=


Q1

Q2R21

Q3

R31

R32

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

+++=++=

32

23

31

13

21

12

323313212

321

4440

0

RQQ

RQQ

RQQ

VQVQVQWWWW

ooo πεπεπε



Q1

Q2R21

Q3

R31

R32Q4 R41

R43

R42


( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛++

+⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

++++++=+++=

43

34

42

24

41

14

32

23

31

13

21

12

434424414323313212

4321

444

4440

0

RQQ

RQQ

RQQ

RQQ

RQQ

RQQ

VQVQVQVQVQVQWWWWW

ooo

ooo

πεπεπε

πεπεπε

(1)


Notice that…

12112

21

21

12212 44

VQR

QQR

QQVQoo

===πεπε

By interchange the subscripts in (1), the following energy term can be obtained:

343242141232131121 VQVQVQVQVQVQW +++++=(2)

Hence, the average energy: ( ) ( )2

21 +

( ) ( )( ) ( ) ⎥

⎦

⎤⎢⎣

⎡+++++

++++++=

44342413343231

22423211141312

21

QVVVQVVVQVVVQVVV

W


It can be written that the total energy required:

( )

∑=

=

+++=

4

1

44332211

21

21

kkkVQ

VQVQVQVQW

In terms of continuous charge distribution:

ldVWline

l∫= ρ21 SdVW

surfaceS∫= ρ

21 vdVW

volumev∫= ρ

21

V is the potential at the charge distribution !!!


Alternatively; the previously obtained equations of energy can be written as: (Using vector identity)

( )dvEDWvolume

•= ∫ 21

The equation state that energy in the system of charges can be obtained if the electric field distribution (which is incurred by the charge itself) is known

Example

A spherical charge distribution of radius ra and uniform ρv is centered at the origin. If the total charge is Q:

Find the energy needed to build up the charge system.

Summary of Chapter 4

The students should be able to understand the followings:

(i) The concept of work and energy in electrostatic field

(ii) The concept of potential difference and absolute potential due to point charges and continuous charge system.

(iii) Able to apply the potential gradient concept in calculating the electric field at a specific point

(iv) Ability to calculate the stored energy in a electrical charge system

Next Lecture


(i) The concept of electric current

(ii) Conductors and conductivity


Lecture #10


(i) Electric Current

(ii) Conductors and conductivity


Motivation:

To refresh the students understanding regarding the concept of electric

current, conductors and conductivity

Electric Current

Current: Moving electric charges.

: Source of magnetostatic field. (Chapter 8)

Definition: Rate of movement of charge passing a given reference point/plane per second.

dtdQI = (Ampere)

Let us introduce the current density, . Unit of (A/m2) J

QQ

Q dS dI( )2/ mA

dSdIJJ ==

∫=∴surface

dsJI .

x

y

z

Electric Current

ρv

dvdQ vρ=

yUU y ˆ=

From:dtdv

dtdQdI vρ==

dSdy

dtdydSvρ=

yU

dSU yvρ=yJ

( )2/mAUJ vρ=

In general:

Charge in motion constitutes a current

E

Continuity of Current

Let us consider:

ρV ∫=volume

Ven dVQ ρ

Close surface

Rate of charge flow outwardly from the close surface current

dtdQI en−=

Q

Reduction of charge

Due to close surface, current flow outwardly:

∫ •=surface

dSJI

Continuity of Current

It can be written:

dvdt

ddvdtd

dtdQdSJ

volume

v

volumev

en

surface∫∫∫ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=−=•

ρρ

( )∫ ∫−=•∇volume volume

v dvdt

ddvJ ρDivergence

Theorem

dtdJ vρ−=•∇⇒ Continuity current

equation

Current per unit volume, emanating from a point equals the time rate decrease of volume charge density at the same

point

J

Will be used in the

study of conductor

Conductors & Conductivity

Consider the atomic structure of conductor:

E

EqF e−=

The electron will moved with constant velocity, called drift velocity, dUThe electron drift velocity and the applied electric field are linearly related by the electron mobility, µe

EU ed µ−=

Conductors & Conductivity

From relation between current density and charge velocity:

UJ vρ=

In conductor:EUJ eede µρρ −==

Conductivity;

Unit: Siemens per meter

σEJ σ=

Relation between induced current flow and applied electric field in conductors

Conductor Properties in Electrostatic Field

E

conductor

What will happen to the free charge, ρv in

conductor due to applied electrostatic field ?

vρ

Consider the continuity current equation:

dtdJ vρ−=•∇

Due to applied electric field, current density will exist; EJ σ=

dtdE vρσ −=•∇∴ (1)


From: vD ρ=•∇ (Divergence)

01=+

dtd v

o

v ρσε

ρ

(1) Can be further written as:

0=+o

v

o

v

dtd

ερ

εσρ

PDE !!!

Solution will give:

( )t

vovoet⎟⎟⎠

⎞⎜⎜⎝

⎛−

= εσ

ρρ

Initial charge density at t= 0

⎟⎟⎠

⎞⎜⎜⎝

⎛=

o

rt

εσ1

Relaxation time constant


For bad conductor ( ↓ tr ↑ charge (ρv) decay slowly)

For good conductor ( ↑ tr ↓ charge (ρv) decay rapidly)

E

conductor

vρE

Sρ−Sρ+

0=vρ

0

0

=∴

=•∇

E

D

(inside)

Next Lecture


(i) Conductor – Free Space (Boundary condition)

(ii) Resistance in Conductor


Lecture #11

Topics to be covered:(i) Conductor – Free Space

(Boundary condition)

(ii) Resistance in Conductor


Motivation:

To enable the students to understand the concept of boundary condition and to calculate the resistance in conductor

Boundary Conditions

Consider heat flow (thermodynamics…)

conductor

heater

air at 27oC

To calculate the temperature at conductor-air boundary, need to use the boundary conditions.

Boundary conditions: Essential in finding physical quantities (temperature, electric field, magnetic field, etc..) if two different materials are in contact.

Boundary Conditions (Conductor – Free Space)

Let us consider conductor in free space:

conductor

Free space

E

What is the resultant electric field at the conductor-free space boundaries if we apply an external electric field ???

Consider the upper boundary:

#2 (conductor)

#1 (free space)

2E

1E

tE 2

tE1

nE 2

nE1


(1) Consider the normal component:

#1

#2

nE1

nE 2

Relation between & ? nE1 nE2

Use Gauss’s Law

ensurface

QdSD =•∫

ρS

o

SnE

ερ

=1

(1) Consider the tangential component:

tE1

tE 2

#1

#2

Relation between & ? tE1 tE2

Use conservation of electric field

∫ =•loop

dlE 0

a b

cd 01 =tE


From previous slide:

( )mVn

EEE

o

S

tn

/ˆ

111

ερ

=

+=

In general:

conductor

free space

E

Example 1: Skitek

At a point on a conductor-free space boundary,

Find: (a) ρS at the point on the boundary (b) at the boundary.

( )mVyxE /ˆ3ˆ2 +=

D

Example 2: D5.5 (Hayt, pg 128)

Given the potential field in free space, V = 100 sinh 5x sin 5y V and a point P(0.1,0.2,0.3). Find at P: (a) V (b) (c) ρS if it is known that P lies on a conductor surface.

E

Resistance in Conductor

R

How to calculate the resistance for the conductor ???

aa b

I

( )Ω=I

VR ab

l

J

E

y

( )ASE

dSE

dSJI

y

surface

surface

σ

σ

=

•=

•=

∫

∫ lEdlEV y

a

bab ∫ =•−=

SlRσ

=∴Conductor with uniform cross section

Resistance in ConductorFor conductor with non-uniform cross section:

∫

∫

∫

∫•

•−=

•

•−==

surface

a

b

surface

a

bab

dSE

dlE

dSJ

dlE

IVR

σ

I

ab

J

E

Example 1: Skitek

Find the resistance between the =0 and = π/2 surfaces of the truncated wedge section shown below.

Given, = 4x10-7 S/m and V/m

φ φ

( )φ1−=

rE

x

y

z0.9 m0.1 m

1 m

Example 2: Skitek

For the truncated wedge of example 1, find the resistance between faces at z = 0 and z = c , when the inner radius is ra and the outer radius is rb.

Example 3: Skitek

For the truncated wedge of example 1, find the resistance between the curved faces at radius of ra and that at a radius of rb, where rb > ra and the length in the z direction is c meter.

Next Lecture


(i) Dielectrics

(ii) Polarization in Dielectrics


Lecture #12

Topics to be covered:(i) Dielectrics(ii) Polarization in Dielectrics


Objectives

1) To understand the concept of dielectric material

2) To investigate the effect of applied electric field to the dielectric material Polarization

3) To understand the concept of bound charge densities in dielectric due to polarization

4) To investigate the effect of polarization on the propagating electric fields in dielectric material

Dielectric

Previously, we learn about conductors free charges to produce conduction current, J=E

Dielectrics or insulators differ from conductors no free charges and charges are confined/bounded to molecular structure. Hence very small value of . For example, glass, quartz, polymer, etc…

+ve

Bound charges

Atomic structure of dielectric

Polarization in Dielectric Material

Dielectric

Macrosopic view Microscopic view0=aE

+ve

No effect on atomic structure

Polarization in Dielectric Material

DielectricMacrosopic view

0≠aE⊕

-⊕

-

⊕

-⊕

-

⊕

-⊕

-⊕

-

+ve

⇔⊕-

Microsopic view

Q

( )CmdQp =dipole moment 0≠aE

0≠aE-Q

d

pnP =

Electric polarization vector

Bound Charge Densities

0≠aE

Consider the dielectric being polarized by the application of external field

Dipole moment will appear and thus the polarization vector

⊕-

⊕-

⊕-

⊕-⊕-⊕-

⊕-⊕-⊕-

⊕-⊕-⊕-

⊕-⊕-⊕-

⊕⊕⊕⊕⊕

⊕⊕⊕ ⊕

Two type of bound charges appear:

P

Bound surface charge density, ρsb

sbρ+

sbρ−

( )2m/CnPsb •=ρ

vbρBound volume charge density, ρvb

( )3m/CPvb •−∇=ρ

Effect of Polarization on Fields

From previous, we have identify the existence of bound surface charge density, ρsb and bound volume charge density, ρvb due to polarization.

Let us apply the Maxwell’s equation in polarized dielectric to find the electric field.

From: vo ED ρε =•∇=•∇

Due to existence of ρvb: ( )vbvo E ρρε +=•∇

Free charge density

From definition of ρvb: ( )PE vo •∇−=•∇ ρε

Further manipulation:vo PE ρε =•∇+•∇

Hence: PED o += ε

Effect of Polarization on Fields

Previously, in free space:

PED o += ε

ED oε=Presently, in polarized dielectric:

Alternatively, it can be written as:

EED ro εεε ==Polarization vector is eliminated by introducing εr

Relative permittivity of dielectric

Depending on type of dielectric. For example:

εr (air)=1.0006

εr (glass) =6.0

Example: Skitek

An ungrounded spherical configuration of concentric spherical dielectric shells, enclosed by a conductor shell, is shown below. Regions 1 and 3 are free space, region 2 is a dielectric whose relative permittivity is εr1and region 4 is a conductor. If a charge Q is placed at the center, find (a) D in all region (b) E in all region (c) P in all region (d) ρS on the conductor surface (e) ρsb on the dielectric surface (f) ρvb within the dielectric

a

b

c

d

#4

#3

#2

#1

Next Lecture


(i) Boundary Conditions (Dielectric-Dielectric)

(ii) Capacitance


Lecture #13


(i) Boundary Conditions (dielectric-dielectric)(ii) Capacitance


Objectives

1) To understand the concept of boundary conditions between dielectrics

2) To investigate the fundamental aspects of capacitance and to calculate the capacitance for a given structure

Boundary Conditions (Dielectric-Dielectric)

Let us consider two bounded dielectrics :

What is the resultant electric field at the dielectrics boundary if we apply an external electric field ???

Consider the upper boundary: Assume surface charge, ρS at the boundary

2E

1E

tE 2

tE1

nE 2

nE1

#2 (Dielectric 2), ε2

#1 (Dielectric 1), ε1

ρS

Dielectric 2E

Dielectric 1

Boundary Conditions (Dielectric-Dielectric)


#1

#2

nE1

nE 2

Relation between & ? nE1 nE2

Use Gauss’s LawρSen

surface

QdSD =•∫

Snn EE ρεε =− 2211


tE1

tE 2

#1

#2

tE1 tE2Relation between & ?

Use conservation of electric field

a b

cd∫ =•

loop

dlE 0 tt EE 21 =

Example 1: Skitek

In the region y>0, we find εr1=4, and in the region y<0, we find εr2=3. If

at the boundary, find:

(a)

(b)

(c)

(d)

(e)

(f) ρsb

Assume that ρS =0 at the boundary

( )m/VzE 101 =

1D

2E2D1P

2P

Example 2: Skitek

A boundary between two dielectrics is found at x=0 plane. If material 1 exists for x>0 with εr1=4 and material 2 exists for x<0 with εr2=7, and when (at boundary), find:

(a)

(b)

(c)

(d)

(e)

(f) ρsb

Assume that ρS =0 at the boundary

( )m/VzyxE 6321 −+=

1D1P2E

2D

2P

Example 3: Skitek

Region 4x+3y >10 consists of a dielectric material with permittivity ε1 and region 4x+3y<10 is consists pf dielectric material with permittivity ε2. At boundary, the following electric field is recorded:

( )mVzyxE /ˆ5ˆ7ˆ21 +−=

Obtain and ρsb at the boundary. Assume ρS =0 at the boundary.

2211 ,,, PDPD

Capacitance

++++

++++

+++

+ ++

----

----

--- -- -

-

Conductor a

Conductor b

Dielectric, ε

E

Capacitance of a two conductor capacitor is defined as the ratio of magnitude of charge on one of the conductors to the magnitude of potential difference between the conductors.

( )FaradVQ

Cab

=

Vab∫=+ dSQ Sρ ∫=− dSQ Sρ

Sρ+ Sρ−

Capacitance

From:abV

QC =

By definition:

∫∫

•−=

a

b

S

dlE

dSC

ρ

Or:

∫∫

•−

•=

a

bdlE

dSEC

ε

(1)

(2)

(1) or (2) can be used to calculate the capacitance for any given capacitor structure.

Example 1: Skitek

Calculate the capacitance of two concentric metallic spheres separated by a dielectric with permittivity ε. Let rbbe the inner radius of the outer sphere and ra be the outer radius of the inner sphere.

Example 2: Skitek

Find the capacitance of an L length coaxial capacitor shown below.

Conductor a

Conductor b

ε

rb

ra

Example 3: Skitek

Find the expression for the capacitance permeter length of a cylindrical capacitor whose cross section is given below.

a

b

c

d

#4

#3

#2

#1

#1: conductor

#2: dielectric, ε2=6εo

#3: free space

#4: conductor

End of Chapter 5

Thank you

Please recap the objectives stated in each lecture to ensure that you fully understood the topics.

Next Lecture


(i) Uniqueness Theorem

(ii) Laplace’s and Poisson Equations


Lecture #14


(i) Uniqueness Theorem(ii) Laplace’s and Poisson’s Equations


Objectives

1. To understand the concept of uniqueness theorem

2. To calculate the electric potential/electric field at a given point using the solution of Poisson’s and Laplace’s equations

Methods to find Electric Field

Electric FieldCharge Distribution

Coulomb’s Law

Gauss’s Law

Electric Potential

Potential Gradient∫= r

dQVπε4

Uniqueness Theorem

Boundary Condition

Area Properties

Uniqueness Theorem

Solution for electrostatic problem (electric potential) can be unique for a specified region if uniqueness theorem is obeyed.

Potential, V

Boundary Conditions Area properties

1. Poisson’s Equation

2. Laplace’sEquation

Poisson’s & Laplace’s Equation

From Maxwell’s equation:

v

v

v

E

E

D

ρε

ρε

ρ

=•∇

=•∇

=•∇

Consider homogeneous material

( ) vV ρε =∇−•∇

Laplacianoperator

(coordinate dependence)

Poisson’s

When ρv=0: 02 =∇ V Laplace’s

ερ vV −=∇ 2

Solution of Poisson’s and Laplace’s Equation

Electric potential, V can be obtained by solving the Laplace’s and Poisson’s equations.

For example, for Laplace’s equation:

0

0

2

2

2

2

2

2

2

=∂∂

+∂∂

+∂∂

=∇

zV

yV

xVV

3-dimensions

But, we only interested in solving a 1-dimensional problem

easier and less complicated.

000 2

2

2

2

2

2

=∂∂

=∂∂

=∂∂

zV@

yV@

xV

Solution of Poisson’s and Laplace’s Equation

For instance; V only varied in x direction:

So, we choose only:02

2

=∂∂

xV

AxV=

∂∂

⇒1st integration

2nd integration BAxV +=⇒

Variables (A and B) can be solved by using the given boundary conditions

Example: Skitek

A parallel plate capacitor has a separation of 0.5mm between plates and a potential difference between plates of 50V. If ρv =0 and ε=4εo between plates, find:

(a) V(y) for 0≤y≤0.5mm if the potential at y=0 is 25 V and at at y=0.5 mm is 75 V.

(b)Electric field between plates

(c) Capacitance per square meter

Example: Skitek

Two infinite length, concentric and conducting cylinders of radius ra=0.02 m and rb=0.05 m are located with axes on the z-axis. If ε=4εo, ρv =0 between the cylinders, V=50 V at ra, V=100 V at rb, find:

(a) V in the range 0.02≤r≤0.05

(b)Electric field intensity

(c) Electric flux density

(d)ρs at rb

(e) Capacitance per meter length

End of Chapter 6

&

End of Electrostatics Topics

Next Lecture


(i) Magnetostatics Field and Sources

(ii) Biot-Savart’s Law


Lecture #15


(i) Magnetostatic Field Intensity and Field Sources

(ii) Biot-Savart’s Law


Objectives

i) To understand the concept of magnetostatic field intensity

ii) To learn about the source of magnetostatic field intensity

iii) To calculate the magnetostatic field intensity using the Biot Savart’s law

Magnetic (Magnetostatic) Field Intensity

Magneto + static

Magnetic field intensity source

Symbol of . Unit of Ampere/meter (A/m).

Field sources: Steady current.

Type of field sources: (i) Filament current

(ii) Surface current

(iii) Volume current

H

Filament Current

I

dl Filamentary current element: dlI

Surface Current

sJ

dSSurface current element: dSJS

Surface current density : unit (A/m)

(A.m)

(A.m)

Volume Current

JVolume current element:

dvJ

dv

Those elements (filament, surface and volume) will create magnetic field.

But, how ???

(A.m)

volume current density : unit (A/m2)

Recall

Previously; in electric field calculation

Charge density; Electric fieldCoulomb’s Law

vSlQ ρρρ ,,,

Current density; JJI S ,,

Similarly, in magnetic field calculation

Magnetic fieldBiot-Savart’s

Law

Biot-Savart’s Law

The law was co-originated by Felix Savart, a professor at College de France and Jean Baptiste Biot, a French physicist in 1820.

I

R

Observations:

rradiuszI

H

ˆˆ

ˆ

→→→ φ

2

1R

H

IH

∝

∝

Biot-Savart’s Law

Hence, Biot-Savart’s Law can be stated as:

)/(4

ˆ

,2 mA

RadlIH

llength

R∫×

=π

(filament current)

)/(4

ˆ

,2 mA

RdSaJH

Ssurface

RS∫×

=π

)/(4

ˆ

,2 mA

RdVaJH

Vvolume

R∫×

=π

(surface current)

(volume current)

Biot-Savart’s Law

I

B

At any point, let B:

dl

R

Ra

)/(4

ˆ

,2 mA

RadlIH

llength

R∫×

=π

Consider a filamentary current;

Biot-Savart’s Law-Filamentary Current

Consider a finite length of filamentary current:

)',,( zr φ

y

x

z

φ

z’

r

Step 3:

Integrate over filament current length to get total H

Rdl

Step 1:Select current element

(0,0,z)

a

b

Step 2:

Identify dH

Hd

I


Step 1: Current element dlI

Step 2: Identify dH

24ˆ

RadlIHd R

π×

=zdzdl ˆ=

( )( ) rrzzzR ˆˆ' +−−=Hence,

( )( )2322 '4

ˆ

zzr

IrdzHd−+

=π

φ

Using:

( ) ( )∫+

=+ 2

12222

322 xcc

x

xc

dx


Step 3:

( )( ) ( )( )φ

πˆ

'

'

'

'4 2

1222

122 ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

−−

−+

−=

zar

za

zbr

zbr

IH

Alternatively:

( ) )/(ˆsinsin4 21 mA

rIH φααπ

+=1α

2αFor infinite length; ( )mAr

IH /ˆ2

φπ

=

r

Example 1

Consider AB in figure below as part of an electric circuit. Find H at the origin due to AB.

x

y

1

1 A

B

6 A

0

Example 2A square conducting loop of side 2a lies in the z=0 plane and carries a current I in the counterclockwise direction. Find a magnetic field intensity at center of the loop.

Example 3

An infinitely long conductor is bent into an L shape as shown in figure below. If a direct current of 5 A flows in the conductor, find the magnetic field intensity at (a) (2,2,0), (b) (0,0,2)

x

y

5 A0

5 A

Example 4The y and z axes, respectively, carry filamentary currents 10 A along and 20 A along . Find H at (-3,4,5).

y z−

Example 5

Find the expression for the H field along the axis of the circular current loop (radius of a), carrying a current I.

Next Lecture


(i) Ampere’s Circuital Law


Lecture #16


(i) Ampere’s Circuital Law


Motivation:

To understand the concept of Ampere’s Circuital Law and its

application in calculating magnetic field

Ampere’s Circuital Law (ACL)

Discovered by Andre-Marie Ampere, a French physicist in 1800s.

Analogous to Gauss’s Law.

Gauss’s Law Electric fieldSpecial case of charge distribution

Magnetic fieldSpecial case of current distribution Ampere’s

Circuital Law

Ampere’s Circuital Law (ACL)

Amperian loopGaussian surface

Infinite length of filament current

Infinite surface current

Infinite length of line charge

Infinite surface charge

Ampere’s Circuital LawGauss’s Law

∫ =

surfaceGaussian

enQdsD . ∫ =

loopAmperian

enIdlH .

Ampere’s Circuital Law- Filament current

I

z Step 1: Create an amperian loop (closed loop that surround the filament current)

Step 2: Identify the loop element, dl

Step 3: Use the ACL to find the magnetic field

∫ =

loopAmperian

enIdlH .

dl

∞+

∞−

Ampere’s Circuital Law- Filament current

From previous slide;

φφ ˆHH = φφ ˆrddl = IIen =

Hence:

)/(ˆ2

2

ˆ.ˆ

.

mAr

IH

rIH

IrdH

IdlH en

φπ

π

φφφ

φ

φ

=

=

=

=

∫∫

As proved from B-S Law

Ampere’s Circuital Law- Surface Current

yx

z

yJJ SS ˆ=

Step 1: Create an amperian loop

Step 2: Identify the loop element

Step 3: Use the ACL

( )xdx ˆ−

( )xdx ˆ ( )zdz ˆ

( )zdz ˆ−

1

2

3

4

Ampere’s Circuital Law- Surface Current

From previous slide;

( )

( ) ( ) ( )

)/(ˆ21

2

2

ˆ.ˆˆ.ˆ

ˆ.ˆˆ.ˆ

.

1

4

4

3

3

2

2

1

mAnJHJH

JH

LJzdzxHxdxxH

zdzxHxdxxH

IdlH

Ss

x

sx

sxx

xx

en

×=∴=⇒

=⇒

=−+−−

+−+⇒

=⇒

∫∫

∫∫

∫

Example 1

Plane x=10 carries current 100mA/m along while line x=1, y=-2 carries filamentary current 20π mA along . Determine at (4,3,2).

zz

H

Example 2

An infinitely long solid conductor of radius a is placed along the z-axis. If the conductor carries current I in the direction, show that:

φπ

ˆa

IrH 22=

z

Example 3

Through the use of Ampere’s Circuital Law, find the field in all regions of an infinite length coaxial cable carrying a uniform and equal current I in opposite directions in the inner and outer conductors. Assume the inner conductor to have a radius of a (m) and the outer conductor to have an inner radius of b (m) and an outer radius of c (m). Assume that the cable’s axis is along the z axis.

H

Example 4

Through the use of Ampere’s Circuital Law, find the field inside and outside an infinite-length hollow conducting tube whose radius is 0.002 m that carries a current I = 10-7 A, directed in the direction. Consider the thickness of the tube very small.

H

z+

Example 5

Consider two wire transmission line whose cross section is illustrated in figure below. Each wire is of radius 2 cm and the wires are separated by 10 cm. The wire centered at (0,0) carries current 5 A in the direction of , while the other centered at (10cm,0) carries the return current. Find at: (5cm, 0) and (10cm, 5 cm)

z+

H

x

y

10 cm

Next Lecture


(i) Stokes’ Theorem

(ii) Magnetic Flux Density

(iii) Maxwell’s equation for static field

(iv) Vector Magnetic Potential


Lecture #17

Topics to be covered:(i) Stokes’ Theorem

(ii) Magnetic Flux Density

(iii) Maxwell’s equation for static field

(iv) Vector Magnetic Potential


Motivation:

To understand the concept of magnetic flux density and magnetic potential

Recall…In electrostatic field analysis:

Divergence Theorem ∫ ∫∇=surface volume

dvD.dS.DGauss’s Law

∫ =surface

enQdS.D

Stokes’ Theorem

Ampere’s Circuital Law

enloop

Idl.H =∫ ( )∫ ∫ ×∇=loop surface

dS.Hdl.H

In magnetostatic field analysis:Another form of GL

Another form of ACL

Details on the curl operator: Hayt and Buck, page 225-232

Magnetic Flux and Flux Density

Electric Flux Density,

Electric Flux, ψE

Unit: Coulomb

Electrostatic field

D

∫=surface

E dS.Dψ

ensurface

E QdS.D == ∫ψ

Magnetic Flux Density,

Magnetic Flux, ψM

Unit: Weber

Magnetostatic field

B

∫=surface

M dS.Bψ

0== ∫surface

M dS.Bψ

ED oε= HB oµ=

µo : magnetic permeability (free space). 4π × 10-7 (H/m)

Example

Determine the magnetic flux through a rectangular loop (a × b) due to an infinitely long conductor carrying current I as shown below. The loop and the straight conductors are separated by a distance d.

z

d a

bI

Example: D8.7 (Hayt)

A solid conductor of circular cross section is made of a homogeneous nonmagnetic material. If the radius a = 1mm, the conductor axis lies on the z axis, and the total current in the direction is 20A, find:

(a) H at r=0.5 mm

(b) B at r=0.8 mm

(c) The total magnetic flux perunit length inside the conductor

(d) The total flux for r < 0.5 mm

(e) The total magnetic flux outside the conductor.

z

Maxwell’s Equations for Static Field

Electrostatic Field Magnetostatic Field

∫ ∫∫ ===∇surface volume

venvolume

dvQdS.DdvD. ρ

vD. ρ=∇ Point form

( ) 0==∇ ∫∫loopsurface

dl.EdS.E.

0=×∇ E Point form

( ) ∫ ∫∫ ===×∇loop surface

ensurface

dS.JIdl.HdS.H

JH =×∇ Point form

( ) 0==∇ ∫∫surfacevolume

dS.BdvB.

0=∇ B. Point form

The equations describe the relationship between electric field, magnetic field, static charges and steady current. For time-varying field, the equations distinct in certain aspects.

Example: D 8.5 (Hayt)

Calculate the value of current density:

(a) In rectangular coordinates at (2,3,4) if

(b) In cylindrical coordinates at (1.5, 90o, 0.5) if

(c) In spherical coordinates at (2, 30o, 20o) if

zxyyzxH 22 −=

( )r.cosr

H φ202=

θθ

ˆsin

H 1=

In electrostatic field analysis:

Vector Magnetic Potential

VE −∇=Charge Distribution

Electric Potential, V

Similarly, in magnetostatic field analysis:

Magnetic Potential, A AB ×∇=

From: (Maxwell’s) and (identity) 0. =∇ B ( ) 0. =×∇∇ A

Current Distribution

How ???

Vector Magnetic PotentialCaution: Mathematical derivation starts here

From Maxwell’s; JH =×∇

( ) JA

JB

o

o

µ

µ

=×∇×∇⇒

=×∇⇒

From the vector identity: ( ) ( )AAA ×∇×∇−∇∇=∇ .2

By simplifying: 0. =∇ A

JA oµ−=∇⇒ 2(Vector Poisson’s Equation) Recall: (PE)

∫=volume

v

o RdvV ρ

πε41

o

vVερ

−=∇ 2

Hence:

∫=volume

o

RdvJA

πµ4

(Wb/m)

Vector Magnetic Potential

In general;

Volume current

Surface current

Filament current

∫=volume

o

RdvJA

πµ4

(Wb/m)

∫=surface

So

RdSJA

πµ4

(Wb/m)

∫=line

o

RdlIA

πµ4

(Wb/m)

Example 1: Skitek

A conductor of radius a carries a uniform current with . Show that the magnetic vector potential for r < a is:

zJJ o ˆ=

Example 2: Skitek

( )mWbzrJA oo /ˆ41 2µ−=

Find for an infinite line of current and derive the field from that current.

A H

Example 3: Skitek

For the filamentary circular current loop carrying current I, with radius a at z=0, find the (a) the vector magnetic potential, (b) through the use of ; along the axis of the current loop.

A B A

Example 4: D 8.9 (Hayt)

The magnetic vector potential of a current distribution in free space is given by:

Find H at (3, π/4, -10). Calculate the flux through r =5,

0 ≤Φ ≤ π/2, 0 ≤ z ≤10.

( )mWbzeA r /ˆsin15 φ−=

Next Lecture


(i) Magnetic Forces

(ii) Magnetic Material


Lecture #18

Topics to be covered:(i) Magnetic Forces

(ii) Lorentz Force Equation


Motivation:

To demonstrate the force exerted by the magnetic field and to understand the nature

and application of Lorentz force equation

Recall…

In electrostatic field;

EQ

Exerted force;

EQF = (N)

Now, in magnetostatic field;

B I

Exerted force; dl

BdlIFd ×=

Magnetic Forces-Current Elements

Let two infinite and parallel filament current-carrying conductors:

x

y

z

d

I2

I1

Find force per unit length between the two conductors ???dl2

Force experienced by dl2:

1222 BdlIFd ×=

( ) ( )xdIzdzIFd o −×−=πµ2

122

( )m/NydII

dzFd o

πµ2

21=

Magnetic Forces-Current Elements

Similarly; from the previous equation:

BdSJFd s ×=

Force on a surface currentBdVJFd ×=Force on a volume current

BdlIFd ×=Force on a filament current

Example 1A rectangular loop carrying current I2 is placed parallel to an infinitely long filamentary wire carrying current, I1 as shown. Find the force experienced by the loop.

I1

I2

ro a

b

x

z

Example 2

A current element of length 2 cm is located at the origin in free space and carries current 12 mA along . A filamentary current of 15 is located along x =3, y =4. Find the force on the current element.

xz

Example 3

A three phase transmission line consists of three conductors that are supported at points A, B and C to form an equilateral triangle as shown below. At one instant, conductors A and B both carry a current of 75 A while conductor C carries a return current of 150 A. Find the force per meter on conductor C at that instant.

A

BC

x

y

2 m

Force on Moving Point Charge

Consider a volume current distribution, JCurrent – Moving point charges

Relation between volume current distribution and charge velocity, (chapter 5) U

UJ Vρ= (1)

Previously, magnetic force on volume current distribution:

BdvJFd ×= (2)

From (1) and (2):

( )BUdQ

BUdvFd v

×=

×= ρ

(3)

Magnetic force on moving charges

Extension from volume current to moving point charge

Force on Moving Point Charge

Equation (3) is a fundamental equation for working principle of electric motors and electric generators.

B

U

( )BUqF e ×=

B

B

conductor

B

a

baF U

BU ×

( )BUqe ×

Lorentz Force Equation

When the charge is immersed in combined and fields, the combined force becomes

E B

BUdQEdQFd ×+= (N)

Example: D 9.1 (Hayt)

The point charge, Q=18 nC has a velocity 0f 5×106 m/s in the direction . Calculate the magnitude of the force exerted on the charge by the field:

z.y.x. 300750600 ++

mTzyxB 643 ++−=(a)

(b) m/kVzyxE 643 ++−=(c) and acting together.B E

Next Lecture


(i) Magnetic Material

(ii) Magnetic Boundary Conditions


Lecture #19

Topics to be covered:(i) Magnetic Material

(ii) Magnetic Boundary Condition


Objectives

1. To understand the concept of magnetic polarization; i.e: the effect of atomic orientation due to applied magnetic field

2. To investigate on the bound magnetization current in polarized material

3. To investigate the effect of magnetic polarization to the resultant magnetic field in the material

4. To understand the concept of boundary condition between two magnetic materials

Magnetic Polarization

Magnetic material

0=B

Macroscopic view

0==∑mM

Magnetization vector (A/m)

(i) Dipole moment randomly oriented

(ii) Total dipole moment

From macroscopic view; :0=B

Microscopic view

Magnetic dipole

moment

Bound current due to electron movement

mI


Characteristics of dipole moment will determine type of magnetic materials:

(i) Diamagnetic

(ii) Paramagnetic

(iii)Ferromagnetic

(iv)Antiferromagnetic

(v) Ferrimagnetic

(vi)Superparamagnetic

*Interested readers: please refer to page 274-276


Magnetic material

0≠B

Macroscopic view

(i) Orientation of dipole moment according to the magnetic field direction (polarized)

(ii) Total dipole moment

From macroscopic view; :

0≠=∑mM

0≠B

0≠=∑mM

Bound Magnetization Current

0≠BM

Im

Bound magnetization surface current density (on surface)

( )mAnMJ sm /ˆ×=

Bound magnetization current density (within the material)

( )2/ mAMJ m ×∇=

Example 1: Skitek

A long cylinder of magnetic material of radius a (m) is found along the z axis. When the magnetization, within the cylinder, find (a) (b) at r = a on the magnetic material surface.

)/(ˆ10 mAzM =

mJ smJ

Example 2: Skitek

A long bar of magnetic material is found parallel to the y axis with its cross section defined by 0 ≤ x ≤ 0.1 and 0 ≤ z ≤0.2. If , find (a) within the material (b) on the four surfaces (c) total bound current that flows through a cross section of the bar at z = 0.1 plane for a length of 1 meter in the y direction.

)/(ˆ3 mAyxM = mJ smJ

Effect of Magnetization on Magnetic Fields

(Recall) From free space condition:

JB

o

=×∇µ

mJIn magnetic material, need to include (due to magnetization)

( )m

o

JJB+=×∇∴

µ

From ;MJm ×∇=

( )MJB

o

×∇+=×∇µ

Effect of Magnetization on Magnetic Fields

Further manipulation:

JMB

o

=⎟⎟⎠

⎞⎜⎜⎝

⎛−×∇

µ

MBHo

−=⇒µ

( )( )2m/WbHH

MHB

ro

o

µµµ

µ

==

+=⇒

rµ : magnetic permeability of the material

Example

In certain material for which µ=6.5µo,

( )m/AzyxH 402510 −+=Find:

The magnetic flux density and magnetization

Boundary Conditions (Two magnetic materials)

Let us consider two magnetic materials:

H

Magnetic material 1, µ1

Magnetic material 2, µ2

What is the resultant magnetic field at the boundary if we apply an external magnetic field ???

2H

1H

tH 2

nH 2

tH 1nH 1

Consider the boundary: Assume surface current in inwards direction (current due to free charges)

sJ

Magnetic material 2

Magnetic material 1

sJ⊗⊗⊗⊗⊗ ⊗

Boundary Conditions (Two Magnetic Materials)


#1, µ1

#2, µ2

nH 1

nH 2

Relation between & ? nH1 nH2

Use Gauss’s Law for magnetic field

0=•∫surface

dSBnn BB 21 =


tH 1

tH 2

#1

#2

Relation between & ? tH1 tH2

Use Ampere’s Circuital Lawa b

cd ∫ =•loop

enIdlH

stt JHH =− 21

⊗⊗ ⊗⊗⊗⊗⊗⊗ ⊗

sJ

l∆

Boundary Conditions (Two Magnetic Materials)

HB µ=From:

The boundary conditions can be further written as:

nn HH 2211 µµ = Normal component

stt JBB=−

2

2

1

1

µµTangential component

Example

The interface 2x+y=8 between two media carries no current. If medium 1 (2x+y≥8) is nonmagnetic with:

( )m/AzyxH −+−= 341

Find:

(a) The magnetic energy density in medium #1

(b) and in medium 2, 2x+y≤8, with µ2=10µo

(c) The angles and make with the normal to the interface

2M 2B1H 2H

Next Lecture


(i) Time Varying Fields

(ii) Faraday’s Law

(iii) Displacement Current


Lecture #20

Topics to be covered:(i) Time Varying Field

(ii) Faraday’s Law


Objectives

1. To identify the difference between the static field (electric and magnetic field) with the time varying field.

2. To exhibit the phenomena whereby the current is produced by the time varying magnetic field, i.e: Faraday’s Law.

3. To investigate on the changes of Maxwell’s equation for time varying field (due to Faraday’s Law).

Time Varying Field

Previously;

( ) ( )11111111 ,,,,, zyxHzyxE

Now;

( ) ( )tzyxHtzyxE ,,,,,,, 11111111

* Field at these points will also change with time, t

Field Source

Field Source

Time Varying Field

Two important issues in time varying fields:

Faraday’s Law Displacement Current

Modification of Maxwell’s Equation

Faraday’s Law

Basic discussion on time varying field will be based on Faraday’s Law.

Faraday’s Law: Generation of electric current due to time varying magnetic field. Discovered by Michael Faraday in 1831 after 10 years of scientific research.

+-

Flux linkageLoop

Galvanometer Battery

Faraday’s Law

From the experiment, three observations have been deduced by Faraday.

1) When the battery is turned on, there is a deflection in the galvanometer.

2) When the battery is turned off, there is also a deflection in a galvanometer but in an opposite direction from (1).

3) Later demonstration showed that a moving coil would also cause a deflection in the galvanometer.

Conclusion from Faraday’s experiment, the electromotive force (emf):

( )dt

dvoltV memf

Ψ−= Faraday’s Law

Faraday’s Law

( ) ( )memf dtdvoltV Ψ−=

magnetic flux

Lenz’s Law

Knowing that:∫ •=Ψ SdBm

3 conditional phenomena may produce emf:

1) Time varying field - Static circuit (Transformer emf)

2) Static field – Moving circuit (Motional emf)

3) Time varying field – Moving circuit

(Transformer emf + Motional emf)

Time Varying Field- Stationary Circuit

Transformer emf

I

2

1R

)(tB increases

B induced

)(tB

SddtBdVemf •−= ∫

Transformer emf

Vemf: Potential between point 1 and 2

(1)

Transformer emf

From electric field: ∫ •=l

emf dlEV (2)

(1)=(2)

∫∫ •−=• SddtBdldE

l

Using Stokes Theorem for LHS:

( ) ∫∫ •−=•×∇ SddtBdSdE

s

dtBdE −=×∇ Maxwell equation for

time varying field

Example 1

Given: A structure as followsz

i1(t)10Ω

10Ω

i2(t) 10 cm

10 cm

Obtain i2(t) if i1(t)=2.5cos (2π×104 t) A

5 cm

Moving Circuit- Static Field Motional emf

Moving bar

U

B

Direction of current: Fleming’s Right Hand Rule

z

xSd

Magnetic force on charge Q moving at velocity of :

BdvJFd ×=

BdvUFd v ×=⇒ ρ

U

Motional emf

BUdQFd ×=⇒

From ;EQF =

BUEdQ

Fdm ×==

Motional electric field intensity

( )∫ ∫ •×=•=l l

memf ldBUldEVHence:

Motional emf

Nonzero value only along the path in motion

Example 2

The loop in example1 is now moving with velocity

as shown below.

( )s/myU 5=

z

I110Ω

10Ω

10 cmyo

10 cm

1

2

4

3

I2

U

Find I2 if I1=10 A

Time Varying Field - Moving Circuit

Transformer emf + Motional emf

In the case of time varying field and moving circuit, the generated emf :

( ) ldBUSddtBdVemf •×+•−= ∫∫

Example 3

A conducting bar moving on the rail is shown below. Find an induced voltage on the bar if a bar moving with a velocity of and

0

6 cm

x

y

Q

P

Bu

( )s/my20 ( ) 26104 m/mWbzytcosB −=

Next Lecture

Please have a preliminary reading on the following topic:

1) Displacement Current

2) Maxwell’s Equations in Time Varying Field

3) Lossy Dielectrics

Reference: Hayt & Buck (page 313-321)

Lecture #21

Topics to be covered:(i) Displacement Current

(ii) Maxwell’s Equations for Time Varying Field

(iii) Lossy Dielectrics


Objectives

1. To investigate the phenomena of displacement current due to time-varying electric field.

2. To investigate on the changes of Maxwell’s equation for time varying field (due to the concept of displacement current).

3. To learn about the concept of lossy dielectric and how it used to classify the material.

Displacement Current

Recall: Capacitor in Chapter 5;

+ + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - -

Dielectric: How can the current go through the dielectric material (insulator) ??

The phenomena is to be discussed using the concept of displacement current


From continuity current equation:

vdtdJ ρ−=•∇

( )

( ) ( )DdtdH

dtdH v

•∇−=×∇•∇⇒

−=×∇•∇⇒ ρ

D is time varying field

The left handside is zero (by vector identity)

( ) ( ) ( )DdtdD

dtdH •∇−•∇=×∇•∇∴

( ) JdtDdH •∇+⎟⎟⎠

⎞⎜⎜⎝

⎛•∇=×∇•∇⇒

+ + + + + + + + + + + + + +

- - - - - - - - - - - - - - - -


dtDdJH +=×∇∴

Conduction current density

Displacement current density, :

Current flow in the dielectric material of the capacitor

dJ

(1)

(1) is also a Maxwell’s equation for time varying field.

+ + + + + + + + + + + + + +

- - - - - - - - - - - - - - - -

Ic

dJ

For a perfect dielectric:0=J

dJH =×∇∴

Maxwell’s Equations

Time Varying FieldStatic Field

vD ρ=•∇ vD ρ=•∇

0=•∇ B 0=•∇ B

0=×∇ EdtBdE −=×∇

JH =×∇ dtDdJH +=×∇

Example 1

A coaxial cable has a dielectric with an εr=4. The inner conductor has a radius of 1.0 mm and the inside radius of the outer conductor is 5.0 mm. Find the displacement current between the two conductors per meter length of the cable for an applied voltage, V = 100 cos (12π ×106t) V.

Hint: Use ⎟⎟⎠

⎞⎜⎜⎝

⎛=

a

b

rrln

LC πε2

Example 2Find the amplitude of the displacement current density:

(a) In the air near a car antenna where the field strength of an FM signal is:

(b) In an air space within a large power transformer where ( ) ( )m/Vzy.t.cosE 09221027768 8 −×=

( )( ) ( )m/Ayz.tcosH 66 102566137710 −×+=

Example 3

Find the amplitude of the displacement current density in a metallic conductor at 60 Hz, if ε=εo, µ=µo, =5.8×107 S/m and ( ) ( )21117377 m/MAxz.tsinJ −=

Example 4

Consider the region defined by and .

If : Find;

(a)

(b)

(c)

(d) Numerical value of b

y,x 1<z

( ) 28105120 m/Aybxt.cosJ d µ−×=

E;D

H;B

dJ

Example 5

Let the internal dimension of a coaxial capacitor be a =1.2cm, b=4cm and l=40 cm. The homogeneous material inside the capacitor has the parameters ε=10-11 F/m, µ=10-5 H/m and =10-5 S/m. If the electric field intensity is

Find:

(a) J

(b)The total conduction current through the capacitor

(c) The total displacement current through the capacitor.

(d)The quality factor of the capacitor.

( )m/Vrtcosr

E 56

1010=

Lossy Dielectrics

Recall;dtDdJH +=×∇

For a perfect (lossless, non-conducting) dielectrics, 00 == σ;J

In reality, 0≠J Lossy dielectrics0≠σ;

(1)

Consider a lossy dielectric and from (1), it can be written:

DjJH ω+=×∇

( )E'jH

E'jEH

ωεσ

ωεσ

+=×∇⇒

+=×∇⇒(2)

Consider a lossless dielectric and from (1), it can be written:

EjH ωε=×∇ (1)

Lossy Dielectrics

To investigate the relation of dielectric permittivity in lossless (ε) and lossy dielectric (ε’ ), we shall then equate (1) and (2):

'jj ωεσωε +=

⎟⎠⎞

⎜⎝⎛ −=⇒

−=⇒

+=⇒

'j'

j'

j'

ωεσεε

ωσεε

ωσεε

1

Loss tangent;

For a lossless dielectric, loss tangent = 0

⎟⎟⎠

⎞⎜⎜⎝

⎛

d

c

II

Lossy Dielectric

For lossless dielectric in parallel plate capacitor:

For lossy dielectric in parallel plate capacitor:

+

--

+++++

--- -ve

cJtDJd ∂∂

→

Leakage current

R CcI dI

+

--

+++++

--- -ve

tDJd ∂∂

→ C dI

Lossy Dielectrics

Loss tangent concept can be used to classify the material:

From, loss tangent = 'ωεσ

Good conductor

Bad conductor

Loss tangent high

Loss tangent small

For material with small conductivity, to be a good conductor, frequency used must be small to increase the loss tangent.

Example

Consider a parallel plate capacitor having a plate area of 1.0 cm2 each and where the plates are separated by a distance of 1.0mm by a dielectric having the following properties at 1GHz.

εr’ =2 ; = 10-7 S/m

Find the equivalent circuit for this capacitor and calculate the conduction current, displacement current and the loss tangent if 1 V at 1 GHz is applied across the capacitor.

Hint: Use and dSC ε

=S

dRσ

=

Next Lecture


1) Propagation of Plane Waves in Free Space


Lecture #22


i) Wave Propagation in Free Space/ Lossless Dielectrics


Objectives

1. To use the Maxwell’s equations in time varying field in describing the wave propagation behaviour.

2. To study the wave propagation behaviour in free space and lossless dielectrics in terms of wave velocity, wavenumber (propagation constant) and relation between propagating electric field and magnetic field.


(Time Varying Field)

Wave Propagation in Free Space /Lossless Dielectrics

Wave Propagation in Lossy Dielectrics

Wave Propagation in Good Conductors

Wave Propagation

CommTower

Portable devices

The propagating wave is a combination of time varying electrical field, and magnetic field, .

( )tE( )tH

y

Hy

But how to determine the both field in the propagating wave?

Consider the simplest wave propagation, i.e. plane wave

z

x Ex

Direction of z


Recall:

dtEdJH;

dtHdE

B;D

oo

v

εµ

ρ

+=×∇−=×∇

=•∇=•∇ 0

Wave propagating medium (free space and lossless dielectric) is assume to be sourceless ρv =0 ; J =0

Hence; the Maxwell’s equations become:

dtEdH;

dtHdE

B;D

oo εµ =×∇−=×∇

=•∇=•∇ 00

Wave EquationConsider the plane wave propagation in the direction of z, and electric field is polarized along x xEE x=

From the diagram:

The magnitude of change only with respect to z coordinate.

Ez

xxEE x=

From Maxwell’s:

dtHdy

dzdE

Edzd

dyd

dxd

zyx

E ox

x

µ−===×∇

00

ydt

dH yoµ−=

E and are orthogonal

H

z

y yHH y=

(1)

Wave Equation

Knowing that magnetic field is polarized along y direction

and the magnitude change only with respect to z (diagram)

yHH y=

From Maxwell’s:

xdt

dEdtEdx

dzdH

Hdzd

dyd

dxd

zyx

H xoo

y

y

εε ==−==×∇

00

H

(2)

From (1) and (2):

dtdH

dzdE y

ox µ−=

dtdE

dzdH x

oy ε−=

(3)

(4)

Wave Equation (Electric Field)

To find relation between (3) and (4); two steps need to be worked out:(i) Differentiate (3) with respect to z

(ii) Differentiate (4) with respect to t

Hence:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

dzdH

dtd

dzEd y

ox µ2

2

2

2

dtEd

dtdH

dzd x

oy ε−=⎟⎟⎠

⎞⎜⎜⎝

⎛(5) (6)

Substituting (6) into (5):

2

2

2

2

dtEd

dzEd x

oox εµ= Wave equation for x-

polarised electric field, propagating along z direction.

2

2

2

2

dtEd

dzEd xx µε=

(Free space)

(Lossless dielectric)

Wave Equation (Magnetic Field)

To find relation between (3) and (4); two steps need to be worked out:(i) Differentiate (3) with respect to t

(ii) Differentiate (4) with respect to z

Hence:

2

2

dtHd

dtdE

dzd y

ox µ=⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

dzdE

dtd

dzHd x

oy ε2

2

(7) (8)

Substituting (7) into (8):

2

2

2

2

dtHd

dzHd y

ooy εµ=

Wave equation for y-polarised magnetic field, propagating along z direction.

2

2

2

2

dtHd

dzHd yy µε=

(Free space)

(Lossless dielectric)

Wave EquationConsider again, the wave equation for x-polarised electric field in free space:

2

2

2

2

dtEd

dzEd x

oox εµ=

In phasor form:

xoox E

dzEd εµω 22

2

−= (9)

Alternatively, (9) can be written as:

xox Ek

dzEd 22

2

−=

ko : propagation/phase constant (rad/m) = cooωεµω =

ω = 2πf : Frequency in rad/s ; f is the wave frequency

( )s/mcoo

81031×==

εµ

Noted that:

(light velocity)

(10)

Wave Equation

From (10):xo

x Ekdz

Ed 22

2

−=

Mathematically, solution for (10) will be:

( ) zjkx

zjkxx

oo eEeEzE +−−+ +=

z

x

( ) zjkxx

oeEzE −+=

( ) zjkxx

oeEzE +−=

(11)

In real form; (11) can be written as: (ignoring the complex quantity)

( ) ( ) ( )zktcosEzktcosEzE oxoxx ++−= −+ ωω (12)

Wave Equation

Recall, from (1):

dtdH

dzdE y

ox µ−=

( ) ( ) ( )zHjeEjkeEjk yozjk

xozjk

xooo ωµ−=+− −−+

Using (11):

( ) zjkx

o

ozjkx

o

oy

oo eEkeEkzH −−+⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ωµωµ

( ) zjk

o

ox

zjk

o

oxy

oo eEeEzH ⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛= −−+

µε

µε

( ) zjky

zjkyy

oo eHeHzH −−+ −= (13)

(Wave Equation)

From (13); the following relation between E field and H field in wave propagating in free space can be deduced:

Ω==−= −

−

+

+

377o

o

y

x

y

x

HE

HE

εµ (14)

(14) is called intrinsic impedance of free space.

In lossless dielectric:

εµ

=−= −

−

+

+

y

x

y

x

HE

HE (15)

(15) is called intrinsic impedance of dielectric.

Example

An E field in free space is given as:

( ) mVzytE /ˆ10cos800 8 β−=

Find:

a) β

b) λ

c) H at P (0.1,1.5,0.4) at t=8ns

Example

In a certain lossless dielectric, µ=µo; ε=3.4εo and electric field is presented by:

( ) mVxztE /ˆ103sin10 8 βπ −×=

( )( )( )( )( )( )( ) λ

η

β

g

Hfe

Ud

cfbEa

y

x

+

+

Next Lecture


1) Propagation of Plane Waves in Lossy Dielectrics


Lecture #23


(i) Wave Propagation in Lossy Dielectrics

(ii)Wave Power – Poynting Vector


1. To study the wave propagation behaviour in lossydielectrics in terms of intrinsic impedance, wave velocity, wavenumber (propagation constant) and relation between propagating electric field and magnetic field.

2. To calculate the wave power by utilizing the PoyntingTheorem

Objectives

Recall: Lossy Dielectrics

The concept has been discussed in the topic of displacement current.

The lossy dielectric permittivity is complex

( )m/Fj'j' or ωσεε

ωσεε −=−= (1)

Where;

ωσε ' = dielectric permittivity

= dielectric conductivity

= frequency in rad/s

Plane Wave in Lossy DielectricsRecall: For wave propagation in lossless dielectric;

Propagation constant,

µεω=k (rad/m) (2)

However, in lossy dielectric;

ε complex

∴ k complex

Thus, from (2):

⎟⎠⎞

⎜⎝⎛ −=

ωσεµω jk '

'1'

ωεσµεω jk −= (3)

Plane Wave in Lossy Dielectrics

Let:βα jjk +=

( )22 βα jk +=−⇒

( ) αββα 2222 jk −−−=⇒ (4)

From (3):ωµσµεω jk −= '22 (5)

From (4) and (5):

( ) '222 µεωβα −=− (real)

ωµσαβ =2 (imag)

(6)

(7)


(6) and (7) are used to determine the and β:

)8( / 1-'

12

'22

2

mNp⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

εωσµεωα

)9( /1'

12

'22

2

mrad⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

εωσµεωβ

Please prove (8) and (9) on your own !!!

is an attenuation constant: A measure of wave attenuation while travelling in a medium.

β is a phase constant. A measure of phase change while travelling in a medium.

Recall: From the Lecture #22; the wave propagation equation can be written as:


( ) zjkx

zjkxx

oo eEeEzE +−−+ +=

Where ko is a wave propagation constant in free space.

In lossy dielectrics, replace jko with . (10) becomes: βα j+

(10)

( ) ( ) ( )zjx

zjxx eEeEzE βαβα ++−+−+ +=

Consider a wave propagation in +z direction:

( ) zjzxx eeEzE βα −−+=

(11)

(12)

In real form:( ) ( )ztcoseEzE z

xx βωα −= −+ (13)


From (13); wave will be attenuated by e-z when propagate in the lossy dielectrics.

( ) ( )ztcoseEzE zxx βωα −= −+

( ) ( )zktcosEzE oxx −= + ω ( ) ( )kztcosEzE xx −= + ω(free space) (lossless dielectrics)

(lossy dielectrics)

z

x

z

x

e-z


In lossless dielectric; the intrinsic impedance:

εµη = (Ω)

In lossy dielectric; the intrinsic impedance:

ηθη ηθη

ωεσεµ

ωεσε

µη jejj

=∠=⎟⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛ −

=

'1

'

'1'

(Ω)

Where:

412

1/

'

'/

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

=

ωεσ

εµη

'tan

ωεσθη =2

From:

ηθη

η

∠=⇒

=

++

+

+

xy

y

x

EH

HE


Thus, consider only the +z propagation of the magnetic field wave:

( ) ( )ηα θβω

η−−= −

+

zteEzH zxy cos

ηθη

η

∠−=⇒

−=

−−

−

−

xy

y

x

EH

HE

Ex

Hy

x

z

y


Previously, in lossless dielectric; the wave velocity:

kων = (m/s)

In lossy dielectric; the wave velocity:

βων = (m/s)

In conclusion: for wave propagation in lossy dielectrics, two important observations can be made:

(i) Both electric and magnetic field waves will be attenuated by e-z

(ii) E leading H by θη

Example

A lossy dielectric has an intrinsic impedance of at the particular frequency. If at that particular frequency a plane wave that propagate in a medium has a magnetic field given by :

Ωo30∠200

./ˆ)x/2-cos(10 - mAyteH x ωα=

Find and .E α

Wave Power Calculation

From previous lecture, the plane wave and plane wave were found to be perpendicular to each other.

E H

Hence the wave power:

HEP ×= (W/m2) (14)

Equation (14) can also provide the wave propagation direction.

Equation (14) Poynting vector

(i) For lossless dielectrics:

( )

( ) ( )ykztEykztHH

xkztEE

xy

x

ˆcosˆcos

ˆcos

−=−=

−=+

+

+

ωη

ω

ω


From (14); The wave power:

( ) ( ) 222

/ˆcos mWzkztEP x −=+

ωη

To find the time average power density:

( ) ( )

( )η

ωη

2

cos1

1

2

2

0

20

+

+

=⇒

−=⇒

=

∫

∫

xavg

Tx

avg

T

avg

EP

dtkztET

P

dtPT

P

(15)


(ii) For lossy dielectrics:

( )

( ) ( )yzteEyztHH

xzteEE

zxy

zx

ˆcosˆcos

ˆcos

ηα

η

α

θβωη

θβω

βω

−−=−−=

−=

−+

+

−+

( ) ( ) ( ) zztzteEP zx ˆcoscos22

ηα θβωβω

η−−−= −

+

( ) ( ) ( )η

α

θη

ωη

cos2

cos1

1

222

0

20

zx

Tx

avg

T

avg

eEdtkztET

P

dtPT

P

−++

=−=⇒

=

∫

∫

The time average power density:

(16)

Example

At frequencies of 1, 100 and 3000 MHz, the dielectric constant of ice made from pure water has values of 4.15, 3.45 and 3.20 respectively, while the loss tangent is 0.12, 0.035 and 0.0009, also respectively. If a uniform plane wave with an amplitude of 100 V/m at z=0 is propagating through such ice, find the time average power density at z=0 and z=10 m for each frequency.

Next Lecture


1) Propagation of Plane Waves in Good Conductors


Lecture #24


(i) Wave Propagation in Good Conductors


1. To study the wave propagation behaviour in good conductors in terms of intrinsic impedance, wave velocity, wavenumber (propagation constant) and relation between propagating electric field and magnetic field.

Objective

Propagation of Plane Wave in Good Conductors

Recall: From our discussion in lossy dielectrics;

Good conductors: 1>>'ωε

σ

Hence: the wave properties previously obtained for lossydielectrics can be used in describing wave propagation in good conductors.

The attenuation constant and phase constant can be written as:

( )2 1-12 22

2

m/Np'

'

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

εωσµεωα

( )3 112 22

2

m/rad'

'

⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

εωσµεωβ

(1)


To determine any changes to wave properties in conductor: (1) is applied into (2) and (3):

( )42

m/Npf µσπωµσα ==

( )52

m/radfµσπωµσβ ==

Also; for intrinsic impedance:

⎟⎠⎞

⎜⎝⎛ −

=⎟⎠⎞

⎜⎝⎛ −

=

'j

'

'j'

ωεσεµ

ωεσε

µη11

(6)


It can be shown that for good conductors, (6) can be altered to be:

o45∠=σωµη (7)

From (7), it can be deduced that the electric field wave will lead the magnetic field wave by 45o.

( ) ( )zftcoseEzE zfxx µσπωµσπ −= −+

( ) ( )ozfyy zftcoseEzH 45−−= −+ µσπωµσπ

(8)

(9)


An important concept, related to the wave propagation in good conductors is skin depth.

( ) ( )zftcoseEzE zfxx µσπωµσπ −= −+

Considering the electric field wave as in (8):

z

Consider the wave magnitude:

At z=0, the magnitude is Ex+

At , the magnitude is µσπf

z 1= ++

xx E.@Ee

36801

µσπfz 1=0=z

δ

+xE+xE.3680

Skin depth @ Depth of penetration:

µσπβαδ

f111

===

The distance is denoted by and is called skin depth

Propagation distance (in conducting medium) that reduce the amplitude of propagating wave to e-1 or 37% of its initial value.

ExampleFind the phase and amplitude of the E field at a depth of 0.1 mm into a copper sheet relative to that entering the surface for a 1 GHz wave directed normal to the sheet.

Given: σcopper=5.8x107 ; εr=1; µr=1

ExampleCalculate the skin depth of copper at 60 Hz and at 6 GHz

Given: σcopper=5.8x107 ; εr=1; µr=1

THANK YOU

Education

Sem5 em skee 2523 electromagnetic field theory lect 01-24