Time-Varying Fields and Maxwell’s Equations

Chapter 6

Overview

6 - 1

2006/9/18 3

Constitutive relations(linear and isotropic media)

Governing equations

MagnetostaticModel

ElectrostaticModel

FundamentalRelations

Fundamental Relations for Electrostatic and Magnetostatic Models

0∇ × =E 0∇ ⋅ =B=∇ × H J

=εD E 1=µ

H B

vρ∇ ⋅ =D

7-1

2006/9/18 4

Introduction

In the static (non-time-varying) case, (E , D) and (B , H) form separate and independent pairs

In a conducting medium, static electric and magnetic fields may both exist and form an electromagnetostatic field.

The magnetic fields is a consequence, it does not affect the calculation of the electric field.

(E , D) (B , H)not related

X

σ Esteady current J = σE static Hconducting

mediumE

E

E

6-1

2006/9/18 5

Time-Varying Fields

When the source producing the fields are changing with time (time-varying)

Therefore, the fundamental relations for electrostatic and magnetostatic models need to be modified to show the mutual dependencebetween the E and H fields in the time-varying case.

changing magnetic field electric field

changing electric field magnetic field

6-1

Faraday’s Law of Electromagnetic Induction

6 - 2

electromagnetic inductiontransformerstransformer emfeddy currentflux-cutting emf

2006/9/18 7

Electromagnetic Induction

Michael Faraday discovered experimentally in 1831 that a current was induced in a conducting loop when the magnetic flux linking the loop changed.The quantitative relationship between the induced electromotive force and the rate of change of flux linkage is the Faraday’s law.

6-2

2006/9/18 8

Fundamental Postulate for Electromagnetic Induction

=t

∂∇× −

∂BE

( , , ; )( , , ; )= x y z tx y z tt

∂∇× −

∂BE

point-function relationship, applies to every point in space, whether in free space or in a material

S Sd d

t∂

∇× ⋅ = − ⋅∂∫ ∫BE s s

C Sd d

t∂

⋅ = − ⋅∂∫ ∫BE s

take surface integral on both sides over an open surface Swith a bounding contour C

can be real or virtualStoke’s theorem

S

C

6-2

2006/9/18 9

Electromagnetic Induction

changing magnetic field induces electric fieldthree different cases to consider the induction phenomena

a stationary circuit in a time-varying magnetic fielda moving conductor in a static magnetic fielda moving circuit in a time-varying magnetic field

6-2

2006/9/18 10

A Stationary Circuit in a Time-Varying Field

C S

dd ddt

⋅ = − ⋅∫ ∫E B s

C Sd d

t∂

⋅ = − ⋅∂∫ ∫BE s

take ∂/∂t outside the integral

Cd= ⋅∫ EV

SdΦ = ⋅∫ B s

ddtΦ

= −V

: emf induced in circuit with contour C

: magnetic flux crossing surface S

6-2.1

2006/9/18 11

Faraday’s Law of Electromagnetic Induction

the electromotive force induced in a stationary closed circuit is equal to the negative rate of increase of the magnetic flux linking the circuitLenz’s law : the negative sign asserts that the induced emf will cause a current to flow in the closed loop in such a direction as to oppose the change in the linking magnetic fluxtransformer emf : emf induced in a stationary loop caused by a time-varying magnetic field

ddtΦ

= −V

6-2.1

2006/9/18 12

Example 6-1

A circular loop of N turns of conducting wire lies in the xy-plane with its center at the origin of a magnetic field specified by , where b is the radius of the loop and ω is the angular frequency. Find the emf induced in the loop.Solution :

0 cos( / 2 )sinzB r b tπ ω=B a

( )b

00

2

0

= cos sin 22

8 1 sin2

S

z x

d

rB t rdrb

b B t

π ω π

π ωπ

Φ = ⋅

⎡ ⎤ ⋅⎢ ⎥⎣ ⎦

⎛ ⎞= −⎜ ⎟⎝ ⎠

∫

∫

B s

a a

20

8 1 cos2

d NN b B tdt

π ω ωπ

Φ ⎛ ⎞= − = − −⎜ ⎟⎝ ⎠

V

N turns

2006/9/18 13

Transformers6-2.2

ferromagnetic core

an a-c device for transforming voltages, currents, and impedances

1 1 2 2N i N i− =ℜΦ

1 1 2 2 SN i N i

µ− = Φ

Sµℜ = : reluctance

2006/9/18 14

Ideal Transformer (1)

1 1 2 2 SN i N i

µ− = Φ

1 2

2 1

i Ni N=

µ →∞ratio of the currents of an ideal transformer is equal to the inverse ratio of the numbers of turns

1 1dv NdtΦ

= 2 2dv NdtΦ

=

1 1

2 2

v Nv N

=

,

ratio of the voltages of an ideal transformer is equal to the ratio of the numbers of turns

6-2.2

2006/9/18 15

Ideal Transformer (2)when the secondary winding is terminated in a load resistance RL

1 1 2 2 1 eff

1 2 1 2

( / )( )( / )

v N N vRi N N i

= =

21

1 eff2

( ) LNR RN

⎛ ⎞= ⎜ ⎟⎝ ⎠

21

1 eff2

( ) LNZ ZN

⎛ ⎞= ⎜ ⎟⎝ ⎠

6-2.2

2006/9/18 16

Real Transformers

existence of leakage flux ( k < 1 )non-infinite inductancenonzero winding resistancepresence of hysteresiseddy current lossnonlinearity of the ferromagnetic core ( µ(H) )

6-2.2

2006/9/18 17

Eddy Currents

When time-varying magnetic flux flows in the ferromagnetic core, the induced emf will produce local currents in the conducting core normal tothe magnetic flux. These currents are called eddy currents.Eddy currents produce ohmic power loss and cause local heating. As a matter of fact, this is the principle of induction heating.Eddy current power loss is undesirable for transformers and can be reduced by using core materials with high µ and low σ.

Ferrite is one such material.

6-2.2

2006/9/18 18

Reduction of Eddy Currents

For low-frequency, high-power applications an economical way for reducing eddy-current power loss is to use laminated cores.

use stacked ferromagnetic (iron) sheets to make transformer core, each sheet is separated by thin varnish or oxide coating

The insulating coatings are parallel to the direction of the magnetic flux so that eddy currents normal to the flux are restricted to the laminated sheets.

The total eddy-current power loss decreases as the number of laminations increases.

6-2.2

2006/9/18 19

Laminated Cores6-2.2

2006/9/18 20

A Moving Conductor in a Static Magnetic Field

6-2.3

++

--

m q= ×F u B

↑ and ⊕ ↓

Coulomb force of attraction between and ⊕

equilibrium state when electric and magnetic forces balance

At equilibrium, which is reached very rapidly, the net forceon the free charges in the moving conductor is zero.

2006/9/18 21

Motional emf

To an observer moving with the conductor, there is no apparent motion

/m q = ×F u B has the same effect as an electric field

2

21 1( )V d= × ⋅∫ u B

a voltage is created along the conductor

( )C

d′ = × ⋅∫ u BV

if the moving conductor is part of a closed circuit C

flux-cutting emf, motional emf

6-2.3

2006/9/18 22

Example 6-2 (1)

A metal bar slides over a pair of conducting rails in a uniform magnetic field B = azB0 with a constant velocity u.a) determine V0 across the terminals 1 and 2b) find electric power dissipated in Rc) show that the electric power dissipated in R is equal to the mechanical

power required to move the sliding bar with a velocity uSolution :

++

--

2006/9/18 23

a) the moving bar generates a flux-cutting emf

b) a current I = uB0h/R with flow from 2 to 1 when R is connected

c) the mechanical force F must balance the magnetic force Fm to move the sliding bar at a constant velocity

0 1 2

1

0 02

( )

( ) ( )

C

x z y

V V V d

u B d uB h′

′

= − = × ⋅

= × ⋅ = −

∫∫

u B

a a a

22 0( )

euB hP I R

R= =

m eP P= ⋅ =F u

1 2 20 02

/m x xI d IB h uB h R′

′= − = − × = =∫F F B a a

Example 6-2 (2)

2006/9/18 24

Example 6-3

A circular metal disk with radius b rotates with a constant angular velocity ω in a uniform and constant B = azB0. Brush contacts are provided at the axis and on the rim of the disk. Determine the open-circuit voltage V0.Solution :

Faraday Disk Generator

B = azB0

consider the circuit 122'341'1

only 2'34 moves

only 34 “cuts” the magnetic flux

0

4

03

20 00

( )

[( ) ] ( )

2

z r

b

V d

r B dr

B bB rdr

φ ω

ωω

= × ⋅

= × ⋅

= = −

∫∫

∫

u B

a a a

2006/9/18 25

A Moving Circuit in a Time-Varying Magnetic Field (1)

When a charge q moves with velocity u in a region where both E and B exist, the electromagnetic (EM) force F on q is given by

To an observer moving with q, there is no apparent motion, and the force on q can be regarded as caused by an equivalent electric field E'

( )q= + ×F E u B Lorentz’s force equation

6-2.4

′ = + ×E E u B

( )C S C

d d dt

∂′ ⋅ = − ⋅ + × ⋅∂∫ ∫ ∫BE s u B

SC

u

BE

take surface integral on both sides over an open surface Swith a bounding contour C

induced emf in the moving frame of reference

transformer emf motional emf

2006/9/18 26

A Moving Circuit in a Time-Varying Magnetic Field (2)

Cd′ ′≡ ⋅∫ EV

S

d dddt dt

Φ′ = − ⋅ = −∫ B sV

emf induced in C measured in the moving frame

6-2.4

( )C S C

d d dt

∂′ ⋅ = − ⋅ + × ⋅∂∫ ∫ ∫BE s u B

designate the left-hand side by

it can be proved that,

( )S C S

dd d dt dt

∂− ⋅ + × ⋅ = − ⋅

∂∫ ∫ ∫B s u B B s

2006/9/18 27

Division of the Induced emf

The division of the induced emf between the transformer and the motional parts depends on the chosen frame of reference.The transformer and motional emf’s can be combined together.

calculation of the induced emf needs not separate into transformer part and motional partonly one equation is necessary

6-2.4

2006/9/18 28

Applicability of Faraday’s Law

Faraday law that induced emf in a closed circuit equals the negative time-rate of increase of the magnetic flux linking a circuit applies to a stationary circuit as well as a moving one.

If the circuit is not moving, V' reduces to V

6-2.4

2006/9/18 29

Example 6-4

Determine the open-circuit voltage of the Faraday disk generator by using

Solution :

Faraday Disk Generator

B = azB0

Φ linking the circuit 122'341'1

flux pass through the wedge area 2'342'

2

0 00 0( )

2

S

b t

d

bB rd dr B tω

φ ω

Φ = ⋅

= =

∫

∫ ∫

B s

S

d dddt dt

Φ′ = − ⋅ = −∫ B sV

20

0 2d B bVdt

ωΦ′= − = − = −V

2006/9/18 30

Example 6-5 (1)

An h by w rectangular conducting loop is situated in a changing magnetic field B = ayB0sinωt. The normal of the loop initially makes an angle α with ay. Find the induced emf in the loop : a) when the loop is at restb) when the loop rotates with an angular velocity ω about the x-axis

Solution :

α

2006/9/18 31

a) when the loop is at rest

b) when the loop rotates about the x-axis

0 0( sin ) ( ) sin cosy nSd B t hw B hw tω ω αΦ = ⋅ = ⋅ =∫ B s a a

0 cos cosad B S tdt

ω ω αΦ= − = −V

S hw=

1

02

3

04

0

( ) ( sin ) ( )2

( sin ) ( )2

2 sin sin2

a n y xC

n y x

wd B t dx

w B t dx

w B t h

ω ω

ω ω

ω ω α

⎛ ⎞′= × ⋅ = × ⋅⎜ ⎟⎝ ⎠⎛ ⎞+ − × ⋅⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

∫ ∫

∫

u B a a a

a a a

V

sides 23 and 41 do not contribute to Va'

Example 6-5 (2)

2006/9/18 32

02 sin sin2aw B t hω ω α⎛ ⎞′= ⎜ ⎟

⎝ ⎠V

0 sin sina B S t tω ω ω′=V

if α = 0 at t = 0, then α = ωt

2 20 0(cos sin ) cos2t a a B S t t B S tω ω ω ω ω′ ′= + = − − = −V V V

using directly

[ ] 0 0 01( ) ( ) ( ) sin cos sin cos sin 22nt t t S B S t B S t t B S tω α ω ω ωΦ = ⋅ = = =B a

sS

d dddt dt

Φ′ = − ⋅ = −∫ BV

0 01 sin 2 cos22t

d d B S t B S tdt dt

ω ω ωΦ ⎛ ⎞′= − = − = −⎜ ⎟⎝ ⎠

V

Example 6-5 (3)

Maxwell’s Equations

6 - 3

Faraday’s lawAmpere’s circuital lawGauss’s lawno isolated magnetic chargeintegral form of Maxwell’s equations

2006/9/18 34

Maxwell’s Equations (1)

The fundamental postulate for electromagnetic induction assures that a time-varying magnetic field induces an electric field.The ∇×E = 0 equation is replaced by Faraday’s law

The relationship between the sources charge and current is the continuity equation representing the principle of conservation of charge

6-3

t∂

∇× = −∂BE ∇× =H J

vρ∇⋅ =D 0∇⋅ =B

v

tρ∂

∇⋅ = −∂

J

(6-40)

2006/9/18 35

Maxwell’s Equations (2)

Certain equation (one, or maybe more) in (6-40) needs to be modified further in order to be consistent with continuity equation.

because continuity equation is a conservation lawWhich equation needs to be modified?

( ) 0∇⋅ ∇× ≡ = ∇⋅H J v

tρ∂

∇⋅ = −∂

J

∇× =H J

contradicts

( ) 0 v

tρ∂

∇ ⋅ ∇× = = ∇⋅ +∂

H J

modified

6-3

2006/9/18 36

Maxwell’s Equations (3)

vρ∇ ⋅ =D

( ) 0 v

tρ∂

∇ ⋅ ∇× = = ∇⋅ +∂

H J

( ) ( )t

∂∇ ⋅ ∇× = ∇⋅ +

∂DH J

t∂

∇× = +∂DH J

t∂∂D

: displacement current density (A/m2)

adding the ∂D/∂t term in the ∇×H = 0 equation is one of the major contribution of James Clerk Maxwell (1831-1879)

a time-varying electric field will give rise to a magnetic field even when the current density is zero

6-3

2006/9/18 37

Maxwell’s equations

Continuity equation

Lorentz’s force equation

t∂

∇× = −∂BE

t∂

∇× = +∂DH J

vρ∇⋅ =D 0∇⋅ =B

( )q= + ×F E u B

v

tρ∂

∇⋅ = −∂

J

Foundation of Electromagnetic Theory

ρv : volume density of free charge

convection current ρvu

conduction current σEJ : density of free current

explain and predict ALL macroscopic EM phenomena

6-3

2006/9/18 38

Solving Maxwell’s Equations

The 4 Maxwell’s equations are consistent, but they are not all independent.In fact, the two divergence equations can be derived from the two curl equations using the continuity equation.

t∂

∇× = −∂BE

t∂

∇× = +∂DH J

only two equations but four unknowns

, , ,E D B H

constitutive relations

, ε µ= =D E B H

sufficient for the solution

6-3

2006/9/18 39

Integral Form of Maxwell’s Equations

The four Maxwell’s equations are differentialequations that are valid at every point in space.The differential form of equations represent the microscopic behavior of EM fields.In dealing with EM phenomena in a large scale, it is often necessary to integrate the differential equations into integral form to indicate the macroscopic behavior of EM fields.

Take surface integrals of the two curl equations and apply the Stoke’s theorem.Take volume integrals of the two divergence equationsand apply the divergence theorem.

6-3.1

2006/9/18 40

Maxwell’s Equations in Integral Form

sC S

d dt

∂⋅ = − ⋅

∂∫ ∫BE

( ) sC S

d dt

∂⋅ = + ⋅

∂∫ ∫DH J

s vS Vd dvρ⋅ = ⋅∫ ∫D

s 0S

d⋅ =∫ B

Faraday’s law of EM induction

Ampere’s circuital law

Gauss’s law

no name

6-3.1

2006/9/18 41

no isolated magnetic charge

Gauss’s law

Ampere’s circuital law

Faraday’s law

SignificanceIntegral FormDifferential Form

Maxwell’s Equations

t∂

∇ × = −∂BE

t∂

∇ × = +∂DH J

vρ∇⋅ =D

Cd

t∂Φ

⋅ = −∂∫ E

C Sd I d

t∂

⋅ = + ⋅∂∫ ∫DH s

Sd Q⋅ =∫ D s

0S

d⋅ =∫ B s0∇⋅ =B

vVdv Qρ ⋅ =∫ s

Sd I⋅ =∫ J

6-3.1

2006/9/18 42

Example 6-6 (1)

An a-c voltage source vc = V0sinωt is connected across a parallel-plate capacitor C1.a) Verify that the displacement current in the capacitor is the same as the

conduction current in the wires.b) Determine the magnetic field intensity at a distance r from the wire.

Solution :a) conduction current in the connecting wire

1 1 0 cosCC

dvi C C V tdt

ω ω= =

capacitance of a parallel-plate capacitor

1ACd

ε=

0 sincv VD E td d

ε ε ε ω= = =

S2

S1

2006/9/18 43

a) displacement current

b) H at a distance r can be found from Ampere’s circuit law to contour C

Two typical open surfaces with rim C may be chosen

0 1 0s cos cosD CA

Ai d V t C V t it d

ε ω ω ω ω∂ ⎛ ⎞= ⋅ = = =⎜ ⎟∂ ⎝ ⎠∫D

( ) sC S

d dt

∂⋅ = + ⋅

∂∫ ∫DH J

(1) planar disk surface S1

11 02 cosCC S

d rH ds i C V tφπ ω ω⋅ = = ⋅ = =∫ ∫H J

(2) curved surface S2 passing through the plates

21 02 cosDC S

d rH ds i C V ttφπ ω ω∂

⋅ = = ⋅ = =∂∫ ∫DH

Example 6-6 (2)

2006/9/18 44

b) therefore,

If not for the presence of the displacement current density in Ampere’s circuit law, the results would be dependent on the choice of surface S, which is a contradiction.

1 cos2

C HH t

rφ

φ ω ωπ

=

Example 6-6 (3)

2006/9/18 45

Electromagnetic Boundary Conditions

In order to solve EM problems involving contiguous regions of different materials, it is necessary to know the boundary conditions that the field vectors must satisfy at the interface.Boundary conditions are derived by applying the integral form of Maxwell’s equations to a small region at an interface of two media.

6-3.2

2006/9/18 46

Tangential Components

1 2 ( )abcda S

d dt

∂⋅ = ⋅∆ + ⋅ −∆ = − ⋅

∂∫ ∫BE E w E w S

0 0 0S

h S dt

∂∆ → ⇒ → ⇒ ⋅ =

∂∫D S

1 2t tE E=

1 2 ( )abcda S S

d d dt

∂⋅ = ⋅∆ + ⋅ −∆ = ⋅ + ⋅

∂∫ ∫ ∫DH H w H w J S S

0 0 0S

h S dt

∂∆ → ⇒ → ⇒ ⋅ =

∂∫B S

1 2t t snH w H w J w∆ − ∆ = ∆

2 1 2( )n s× − =a H H J

a

cd

b∆h

∆w

6-3.2

2006/9/18 47

Normal Components

0 0 0S

h S dt

∂∆ → ⇒ → ⇒ ⋅ =

∂∫D S

0 0 0S

h S dt

∂∆ → ⇒ → ⇒ ⋅ =

∂∫B S

2 1 2( )n sρ⋅ − =a D D

1 2 2 1( )n n s sS Sd S Q ds Sρ ρ⋅ = ⋅ + ⋅ ∆ = = = ∆∫ ∫D s D a D a

1 2 2 1( ) 0n nSd S⋅ = ⋅ + ⋅ ∆ =∫ B s B a B a

1 2n nB B=

E

∆S

anEn

ρs

6-3.2

2006/9/18 48

General Statements About BC

tangential component of an E field is continuousacross an interfacetangential component of an H field is discontinuousacross an interface where a surface current exists

the amount of discontinuity is equal to the surface current density

normal component of a D field is discontinuousacross an interface where a surface charge exists

the amount of discontinuity is equal to the surface charge density

normal component of a B field is continuous across an interface

6-3.2

2006/9/18 49

Equivalence Between BC

As noted previously, the two divergence equations can be derived from the two curl equations and the continuity equation.Therefore, the two conditions for normal components cannot be independent from the two conditions for tangential components.As a matter of fact, in time-varying case,

simultaneous specification of Et and Bn would be redundant and could result in contradiction if not careful

1 2 1 2t t n nE E B B= ⇔ =

2 1 2 2 1 2( ) ( )n s n sρ× − = ⇔ ⋅ − =a H H J a D Dequivalent

6-3.2

2006/9/18 50

Interface Between Two Lossless Media

1 11 2

2 2

1 11 2

2 2

1 2 1 1 2 2

1 2 1 1 2 2

tt t

t

tt t

t

n n n n

n n n n

DE EDBH HB

D D E EB B H H

εεµµ

ε εµ µ

= =

= =→

→== =

→

→ =

lossless σ = 0 ρs = 0 and Js = 0

6-3.2

2006/9/18 51

Interface Between a Dielectric and a Perfect Conductor

Good conductors are often considered perfect electric conductor (PEC) in regard to BC.

σ ∞ for PECIn the interior of a PEC, the E field is zero, or else it would produce an infinite current density.

B and H are also zero through Maxwell’s equations.

2 1n s× =a H J

on the dielectric side (1) on the conductor side (2)

1 0tE =

2 1n sρ⋅ =a D

1 0nB =

2 0tE =

2 0nB =

2 0tH =

2 0nD =

outward normal from medium 2

6-3.2

2006/9/18 52

Additional Comments

When the materials have finite conductivities, currents flowing in the material are expressed in terms of volume current density.

surface current density defined for current flowing through an infinitesimal thickness is zero

Therefore, the tangential components of H is continuous across an interface with a conductor having finite conductivity.

6-3.2

2006/9/18 53

Importance of BC

Boundary conditions are of basic importance in the solution of EM problems.

General solutions of Maxwell’s equations carry little meaning until they are adapted to physical problems each with a given region and associated BC.

Maxwell’s equations are partial differential equations and their solutions will contain integration constants that are determined from the additional information supplied by BC.

Therefore, each solution will be unique for each given problem.

6-3.2

Potential Functions

6 - 4

magnetic vector potentialelectric scalar potentialwave equationsLorentz gauge

2006/9/18 55

Potential Functions

Maxwell’s equations are a set of coupled partial differential equations.If we try to solve for E and H directly from the equations, it would be a very difficult process and the results are very complicated.Therefore, we can make use of some auxiliarypotential functions to simplify the process of obtaining the fields in terms of the sources.

The use of potential functions is purely mathematical.Electric and magnetic fields obtained through the potential functions must still obey the Maxwell’s equations.

6-4

2006/9/18 56

Vector Magnetic Potential

= ∇×B A

0∇⋅ =B

0∇⋅∇× ≡Avector identity

t∂

∇× = −∂BE

from Faraday’s law

( )t∂

∇× = − ∇×∂

E A

( ) 0t

∂∇× + =

∂AE

6-4

2006/9/18 57

Scalar Electric Potential

vector identity

( ) 0t

∂∇× + =

∂AE

0V∇×∇ ≡

Vt

∂+ = −∇∂AE

Vt

∂= −∇ −

∂AE

in static case 0t

∂=

∂A

V= −∇E

6-4

2006/9/18 58

Derivation of Equations for Potentials

Vt

∂= −∇ −

∂AE=∇×B A

t∂

∇× = +∂DH J

tµ µε ∂∇× = +

∂EB J

,

( )Vt t

µ µε ∂ ∂∇×∇× = + −∇ −

∂ ∂AA J

,µ ε= =B H D E

22

2( ) Vt t

µ µε µε∂ ∂⎛ ⎞∇ ∇⋅ −∇ = −∇ −⎜ ⎟∂ ∂⎝ ⎠AA A J

6-4

2006/9/18 59

Lorentz Condition2

22

Vt t

µε µ µε∂ ∂⎛ ⎞∇ − = − +∇ ∇⋅ +⎜ ⎟∂ ∂⎝ ⎠AA J A

a coupled equation between the two unknowns A and V

The definition of a vector A requires the specification of both its curland divergence.

, ?∇= =∇× ⋅B AA

In order to simplify this equation, we can choose whatever the value of ∇·A, as long as the choice will result in a simpler equation form.

0Vt

µε ∂∇ ⋅ + =∂

A

Lorentz condition, Lorentz gauge

6-4

2006/9/18 60

Wave Equation for Vector Potential

22

2

Vt t

µε µ µε∂ ∂⎛ ⎞∇ − = − +∇ ∇⋅ +⎜ ⎟∂ ∂⎝ ⎠AA J A

0Vt

µε ∂∇ ⋅ + =∂

A

22

2tµε µ∂

∇ − = −∂

AA J

6-4

2006/9/18 61

Wave Equation for Scalar Potential

Vt

∂= −∇ −

∂AE

vρ∇⋅ =D

vVt

ε ρ∂⎛ ⎞−∇ ⋅ ∇ + =⎜ ⎟∂⎝ ⎠A

2 ( )Vt

ρε

∂∇ + ∇⋅ = −

∂A

0Vt

µε ∂∇ ⋅ + =∂

A

22

2vVV

tρµεε

∂∇ − = −

∂

6-4

2006/9/18 62

Solutions of Wave Equations

Maxwell’s equations give a complete description of the relation between electromagnetic fieldsand sources (charge and current distributions).

Wave equations for potentials are derived from Maxwell’s equations.

Solutions for wave equations provide the answers to all electromagnetic problems.

However, in most cases the solutions are difficult to obtain.Special analytical and numerical techniques may be devised to aid in the solution procedure; but they do not add to or refine the fundamental structure.

6-4.1

2006/9/18 63

Solutions of Wave Equations for Potentials2

22

vVVt

ρµεε

∂∇ − = −

∂

x

y

z

Rθ

φ

( )v t dvρ ′

for an elemental point charge ρv(t)dv' located at the origin

V depends only on Rand t but not θ and φ

symmetry

0 , 0φ θ∂ ∂

= =∂ ∂

0 , 0φ θ∂ ∂

= =∂ ∂

22

2 2

1 0V VRR R R t

µε∂ ∂ ∂⎛ ⎞ − =⎜ ⎟∂ ∂ ∂⎝ ⎠

change of variable

1( , ) ( , )V R t U R tR

=

2 2

2 2 0U UR t

µε∂ ∂− =

∂ ∂

1-D homogeneous wave equation

6-4.1

2006/9/18 64

Solution of Wave Equation

2 2

2 2 0U UR t

µε∂ ∂− =

∂ ∂

1 2( , ) ( ) ( )U R t C f t R C g t Rµε µε= − + +

C1 , C2 : integration constantf , g : any twice-differentiable function

non-physical, to be proved later

( , ) ( )U R t f t R µε= −a wave traveling in the +R direction with a velocity 1/pu µε=

6-4.1

2006/9/18 65

Proof of the Solution

( ) ( )( )

( )df t R t RU f t R

t td t Rµε µε

µεµε

− ∂ −∂ ′= ⋅ = −∂ ∂−2

2 ( )U f t Rt

µε∂ ′′= −∂

( ) ( )( )

( )df t R t RU f t R

R Rd t Rµε µε

µε µεµε

− ∂ −∂ ′= ⋅ = − −∂ ∂−

2

2 ( )U f t RR

µε µε∂ ′′= −∂

2 2

2 2 0U UR t

µε∂ ∂− =

∂ ∂

6-4.1

2006/9/18 66

Wave Function

( , ) [ ( ) ]

[( ) ( )]

U R R t t f t t R R

f t R t R

µε

µε µε

+ ∆ + ∆ = + ∆ − + ∆

= − − ∆ −∆

( , ) ( ) ( , )U R R t t f t R U R tµε+ ∆ + ∆ = − =

if / pt R R uµε∆ = ∆ = ∆

1/pu µε= : velocity of propagation

U(R, t) : wave functionthe function retains its form if ∆t = ∆R/u

1( , ) ( / )pV R t f t R uR

= −

6-4.1

2006/9/18 67

The Function Form of f

( )( )4v t vV R

Rρπε

′∆∆ =

for a static point charge ρv(t)∆v' at the origin

1( , ) ( / )pV R t f t R uR

= −

( / )( / )

4v p

pt R u v

f t R uρ

πε′− ∆

∆ − =

from comparison

( / )1( , )4

v p

V

t R uV R t dv

Rρ

πε ′

−′= ∫

6-4.1

2006/9/18 68

Retarded Potentials

( / )1( , )4

v p

V

t R uV R t dv

Rρ

πε ′

−′= ∫

V at a distance R from the source at time t depends on the value of ρ at an earlier time t – R /up

V(R, t ) : retarded scalar potential

( )g t R µε+ can not be a physically meaningful solution

following the same procedure

( / )( , )

4p

V

t R uR t dv

Rµπ ′

−′= ∫

JA

6-4.1

Time-Harmonic Fields

6 - 5

use of phasortime-harmonic electromagneticsvector phasorHelmholtz’s equationsimple mediaclassification of materialsthe electromagnetic spectrum

2006/9/18 70

Time-Harmonic FieldsMaxwell’s equations and all the equations derived from them hold for EM quantities with an arbitrary time-dependence.

The actual type of time functions of the field quantities depends on the source quantities ρ and J.

However, for real engineering problems, arbitrary time dependence is very difficult to handle.In engineering, sinusoidal time functions occupy a unique position and they are easy to generate.

periodic time functions can be expanded into Fourier series of harmonic sinusoidal componentstransient non-periodic time functions can be expressed as Fourier integrals

6-5

2006/9/18 71

Principle of Superposition

Since Maxwell’s equations are linear differential equations, sinusoidal time variations of source quantities of a given frequency will produce sinusoidal variations of E and H with the same frequency in the steady state.For source quantities with an arbitrary time dependence, electrodynamic fields can be determined in terms of those caused by the various frequency components of the sources.Application of the principle of superposition will give the total fields.

6-5

2006/9/18 72

The Use of Phasors – A Review

time-harmonic fields sinusoidal time variationFor time-harmonic fields, it is convenient to use a phasor notation.The instantaneous (time-dependent) expression of a sinusoidal scalar quantity, such as a current i, can be written as either a cosine or sine function.

If we choose a cosine function as the reference, then all derived results will refer to the cosine function.

The specification of a sinusoidal quantity requires the knowledge of three parameters : amplitude, frequency, and phase

6-5.1

2006/9/18 73

Inconvenience of cosine Function

To work with an instantaneous expression such as the cosine (or sine) function is inconvenient when differentiation or integration is involved.

combining cos and sin is very tedious

cos sin , sin cosd dt t t tdt dt

ω ω ω ω= − =

( ) cos( )i t I tω φ= +

amplitude phase

angular frequency

6-5.1

2006/9/18 74

Trouble of Dealing Directly with cosine or sine Function

~

R L C

e(t) i(t)1 ( )diL Ri idt e t

dt C+ + =∫

( ) cos( )i t I tω φ= +

1sin( ) cos( ) sin( ) cosI L t R t t E tC

ω ω φ ω φ ω φ ωω

⎡ ⎤− + + + + + =⎢ ⎥⎣ ⎦

complicated!!

6-5.1

2006/9/18 75

Phasors Are Relatively Easy0 0( ) cos [( ) ] ( ) ,j j t j t j

s se t E t Ee e E e E Ee Eω ωω= = ℜ =ℜ = =e e

( ) [( ) ] ( ) ,j j t j t js si t Ie e I e I Ieφ ω ω φ= ℜ =ℜ =e e

( ) , ( )j t j tss

di Ij I e idt edt j

ω ωωω

=ℜ =ℜ∫e e

1 ( )diL Ri idt e tdt C+ + =∫

1s sR j L I E

Cω

ω⎡ ⎤⎛ ⎞+ − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

easy!!

for the series RLC circuit,

6-5.1

2006/9/18 76

Example 6-7

Write the phasor expression Is for the following current function using a cosine reference.a) i(t) = – I0 cos(ωt – 30°)b) i(t) = I0 sin(ωt + 0.2π)

Solution :for cosine reference,

a)

b)

( ) ( )j tsi t I e ω= ℜe

300 0( ) cos( 30 ) [( ) ]j j ti t I t I e e ωω − °= − − ° = ℜ −e

30 / 6 5 / 60 0 0

j j jsI I e I e I eπ π− ° −= − = − = −

0.2 / 20 0( ) sin( 0.2 ) [( ) ]j j j ti t I t I e e eπ π ωω π −= + = ℜ ⋅e

0.2 / 2 0.30 0( )j j j

sI I e e I eπ π π− −= =

2006/9/18 77

Example 6-8

Write the instantaneous expression v(t) for the following phasors using a cosine reference.a)b) Vs = 3 – j4

Solution :a)

b)

/ 40 0( ) ( ) [( ) ] cos( / 4)j t j j t

sv t V e V e e V tω π ω ω π= ℜ =ℜ = +e e

/ 40

jsV V e π=

12 2 tan ( 4 /3) 53.13 4 3 4 5j jsV j e e

− − − °− = + =

53.1( ) ( ) [(5 ) ] 5cos( 53.1 )j t j j tsv t V e e e tω ω ω− °= ℜ =ℜ = − °e e

2006/9/18 78

Time-Harmonic Electromagnetics6-5.2

( , , , ) [ ( , , ) ]j tx y z t x y z e ω= ℜE Ee

( , , , ) ( , , ) , ( , , , ) ( , , ) /x y z t j x y z x y z t dt x y z jt

ω ω∂= =

∂ ∫E E E E

jωµ∇× = −E H

jωε∇× = +H J E

/vρ ε∇ ⋅ =E

0∇⋅ =H

t∂

∇× = −∂BE

t∂

∇× = +∂DH J

vρ∇⋅ =D

0∇⋅ =B

in a simple medium

2006/9/18 79

Helmholtz’s Equations and Solutions

2 2 2 2,vV k V kρ µε

∇ + = − ∇ + = −A A J

2 2p

fku uω π πω µε

λ= = = = : wave number

0j Vωµε∇⋅ + =A

Lorentz condition

Helmholtz’s equations

1( ) , ( )4 4

jkR jkRv

V V

e eV R dv R dvR R

ρ µπε π

− −

′ ′′ ′= =∫ ∫

JA

phasor solutions of Helmholtz’s equations

6-5.2

2006/9/18 80

Complete Solution Procedure

Find phasors V(R) and A(R)

Find phasors E(R) and B(R)

Find instantaneous E(R,t) and B(R, t)

1( ) ( )4 4

,jkR jkR

v

V V

e eV R dv R dvR R

ρ µπε π

− −

′ ′′ ′= =∫ ∫

JA

( ) ( ) ( ) , ( ) ( )R V R j R R Rω= −∇ − = ∇×E A B A

( , ) [ ( ) ] , ( , ) [ ( ) ]j t j tR t R e R t R eω ω= ℜ =ℜE E B Be e

6-5.2

2006/9/18 81

Source-Free Wave Equations

For problems of wave propagation, we are concerned with the behavior of an EM wave in a source-free region where ρ and J are both zero.In other words, we are often not interested at how an EM wave is originated (antenna problem), but how the EM wave propagates (propagation problem).

6-5.3

2006/9/18 82

Homogeneous Wave Equations

in a simple (linear, isotropic, and homogeneous) non-conducting medium

tµ ∂

∇× = −∂HE

tε ∂∇× =∂EH

0∇⋅ =E 0∇⋅ =H

2

2( )t t

µ µε∂ ∂∇×∇× = − ∇× = −

∂ ∂EE H

2 22 2

2 2 2

1 0pt u t

µε ∂ ∂∇ − = ∇ − =

∂ ∂E EE E

2 2( )∇×∇× =∇ ∇ ⋅ −∇ = −∇E E E E0

2( )t t

ε µε∂ ∂∇×∇× = ∇× = −

∂ ∂

2HH E

2 2( )∇×∇× =∇ ∇ ⋅ −∇ = −∇H H H H0

2 22 2

2 2 2

1 0pt u t

µε ∂ ∂∇ − = ∇ − =

∂ ∂H HH H

6-5.3

,

2006/9/18 83

Time-Harmonic Source-Free Fields6-5.3

2 22 2

2 2 2 2

1 10 , 0p pu t u t∂ ∂

∇ − = ∇ − =∂ ∂

E HE H

jt

ω∂⇒

∂2 2

2 22 20 , 0s s s sp pu u

ω ω∇ + = ∇ + =E E H H

p

kuω

=

2 2 2 2,0 0s s s sk k∇ + = ∇ + =E E H H

2006/9/18 84

Electromagnetic Spectrum

Maxwell’ equations, and therefore the wave and Helmholtz’s equations, impose no limit on the frequency of the waves.All electromagnetic waves in whatever frequency range propage in a medium with the same velocity

c ≈ 3 x 108 m/s in air1/u µε=

6-5.3

2006/9/18 85

Electromagnetic Spectrum

6-5.3

2006/9/18 86

Band Designations for Microwaves

0.5 – 1CUHF1 – 2DL2 – 3ES3 – 4FS4 – 6GC6 – 8HC8 – 10IX

10 – 12.4JX12.4 – 18JKu18 – 20JK

20 – 26.5KK26.5 – 40KKa

Frequency Ranges (GHz)NewOld

6-5.3