revision of electrostatics (week 1) vector calculus (week 2-4)

http://www.cee.hw.ac.uk/~arharvey/22.3MB1electromagnetism.zip Dr Andy Harvey

1

Revision of Electrostatics (week 1)

Vector Calculus (week 2-4)

Maxwell’s equations in integral and differential form (week 5)

Electromagnetic wave behaviour from Maxwell’s equations (week 6-7.5)–Heuristic description of the generation of EM waves

–Mathematical derivation of EM waves in free space

–Relationship between E and H

–EM waves in materials

–Boundary conditions for EM waves

Fourier description of EM waves (week 7.5-8)

Reflection and transmission of EM waves from materials (week 9-10)

22.3MB1 ElectromagnetismDr Andy Harvey

e-mail [email protected]

lecture notes are at:http://www.cee.hw.ac.uk/~arharvey/22.3MB1electromagnetism.zip


2

Maxwell’s equations in differential form

continuity of Equationt

law s Faradaydt

law s Amperedt

tics magnetosta for law Gauss

tics electrosta for law Gauss

J

BE

DJ H

B

D

.

'

'

' 0 .

' .

HB

ED

• Varying E and H fields are coupled

NB Equations brought from elsewhere, or to be carried on to next page, highlighted in this colour

NB Important results highlighted in this colour


3

Fields at large distances from charges and current sources

• For a straight conductor the magnetic field is given by Ampere’s law

dt

HE

dt

EJH

I H

• dH/dt produces E

• dE/dt term produces H

• etc.

• How do the mixed-up E and H fields spread out from a modulated current ?– eg current loop, antenna etc

• At large distances or high frequencies H(t,d) lags I(t,d=0) due to propagation time– Transmission of field is not instantaneous

– Actually H(t,d) is due to I(t-d/c,d=0)

• Modulation of I(t) produces a dH/dt term


4

A moving point charge:• A static charge produces radial

field lines

• Constant velocity, acceleration and finite propagation speed distorts the field line

• Propagation of kinks in E field lines which produces kinks of and

• Changes in E couple into H & v.v– Fields due to are short range

– Fields due to propagate

• Accelerating charges produce travelling waves consisting coupled modulations of E and H

tE22 t E

tE22 t E


5

Depiction of fields propagating from an accelerating point charge

• Fields propagate at c=1/xm/sec

E in plane of page

H in and out of page


6

Examples of EM waves due to accelerating charges

• Bremstrahlung - “breaking radiation”

• Synchrotron emission

• Magnetron

• Modulated current in antennas– sinusoidal v.important

• Blackbody radiation


7

2.2 Electromagnetic waves in lossless media - Maxwell’s equations

SI Units• J Amp/ metre2 • D Coulomb/metre2

• H Amps/metre• B Tesla

Weber/metre2

Volt-Second/metre2

• E Volt/metre• Farad/metre• Henry/metre• Siemen/metre

dt

BE

dt

DJH

D.

0. B

EJ

HHB

EED

or

or

t

J.

Maxwell

Equation of continuity

Constitutive relations


8

2.6 Wave equations in free space• In free space

– =0 J=0

– Hence:

dt

BE

dtdt

DDJH

2

2

t

tt

ttt

o

o

o

EE

D

HBB

E

– Taking curl of both sides of latter equation:


9

Wave equations in free space cont.

• It has been shown (last week) that for any vector A

where is the Laplacian operator

Thus:

2

2

to

E

E

AAA 2.

2

2

2

2

2

22

zyx

2

22.

to

E

EE

2

22

to

E

E

• There are no free charges in free space so .E==0 and we get

A three dimensional wave equation


10

• Both E and H obey second order partial differential wave equations:

2

22

to

E

E


2

22

to

H

H

22 seconds

eVolts/metr

metre

eVolts/metr o

• What does this mean – dimensional analysis ?

– has units of velocity-2

– Why is this a wave with velocity 1/ ?


11

The wave equation

2

22

to

A

A

)( vtzfAA o

vtzfvAvtzfA ooo 2

ov 1cv oo m/sec109979.21 8

v

Substitute back into the above EM 3D wave equation

This is a travelling wave if

• In free space

• Why is this a travelling wave ?

• A 1D travelling wave has a solution of the form:

Constant for a travelling wave

vtzfvAt

vtzfAz oo

2

2

2

2

2 AA

z


12

• Substitute this 1D expression into 3D ‘wave equation’ (Ey=Ez=0):


vtzEx 2

sin

o

o

v

vtxvvtx

vtzvvtzt

vvtzt

vtzvtzz

vtzz

1

sinsin

sincossin

sincossin

222

222

22

2

22

to

E

E

• Sinusoidal variation in E or H is a solution ton the wave equation


13

• Maxwells equations lead to the three-dimensional wave equation to describe the propagation of E and H fields

• Plane sinusoidal waves are a solution to the 3D wave equation

• Wave equations are linear

– All temporal field variations can be decomposed into Fourier components• wave equation describes the propagation of an arbitrary disturbance

– All waves can be written as a superposition of plane waves (again by Fourier analysis)

• wave equation describes the propagation of arbitrary wave fronts in free space.

Summary of wave equations in free space cont.


14

Summary of the generation of travelling waves

24 r

qE

o

022 tE• We see that travelling waves are set up when – accelerating charges

– but there is also a field due to Coulomb’s law:

– For a spherical travelling wave, the power carried by the travelling wave obeys an inverse square law (conservation of energy)

• P E2 1/r2

– to be discussed later in the course

• E/r

– Coulomb field decays more rapidly than travelling field– At large distances the field due to the travelling wave is much larger than the

‘near-field’


15

Heuristic summary of the generation of travelling waves

E due to stationary charge (1/r2)

due to charge constant velocity (1/ r2)

tE

22 t E kinks due to charge

acceleration (1/r)

22 t H due to E (1/r)


16

• Consider a uniform plane wave, propagating in the z direction. E is independent of x and y

2.8 Uniform plane waves - transverse relation of E and H

0.

zyx

zyx EEEE

00

yx

EE

)waveanotisconst(000,0

zzzyx EE

z

E

y

E

x

E

In a source free region, .D= =0 (Gauss’ law) :

E is independent of x and y, so

• So for a plane wave, E has no component in the direction of propagation. Similarly for H.

• Plane waves have only transverse E and H components.


17

Orthogonal relationship between E and H:• For a plane z-directed wave there are no variations along x and y:

y

A

x

A

x

A

z

A

z

A

y

AA

xyz

zxy

yzx

a

a

a

z

H

z

Hx

yy

x

aaH

t

E

z

H

t

E

z

H

yx

xy

t

H

z

E

t

H

z

E

yo

x

xo

y

to HE • Equating terms:dt

DJH

• and likewise for :

t

E

t

E

t

E

t

zy

yy

xx aaa

D

• Spatial rate of change of H is proportionate to the temporal rate of change of the orthogonal component of E & v.v. at the same point in space


18

t

E

z

H

t

E

z

H

yx

xy

• Consider a linearly polarised wave that has a transverse component in (say) the y direction only:

vtzfvEt

E

vtzfEE

oy

oy

t

H

z

E

t

H

z

E

yo

x

xo

y

xo

y EH

• Similarly

yo

x

y

oox

EH

vE

vtzfvEconstzvtzfvEH

d

z

H x

Orthogonal and phase relationship between E and H:

• H and E are in phase and orthogonal


19

• The ratio of the magnetic to electric fields strengths is:

which has units of impedance

yo

x EH

xo

y EH

o

yx

yx

H

E

HH

EE

22

22

metreamps

metreVolts

/

/

37712010

36

1104

9

7

o

o

cH

E

B

E

ooo

1

Note:

• and the impedance of free space is:


20

Orientation of E and H• For any medium the intrinsic impedance is denoted by

and taking the scalar product

so E and H are mutually orthogonal

y

x

x

y

H

E

H

E

0

.

yxxy

yyxx

HHHH

HEHE

HE

2

22

H

HH

HEHE

z

xyz

xyyxz

aHE

a

aHE

xyyxz

zxxzy

yzzyx

BABA

BABA

BABA

a

a

aBA

• Taking the cross product of E and H we get the direction of wave propagation


21

A ‘horizontally’ polarised wave

Hy Ex

xo

y EH

HE

• Sinusoidal variation of E and H

• E and H in phase and orthogonal


22

A block of space containing an EM plane wave• Every point in 3D space is characterised by

– Ex, Ey, Ez

– Which determine• Hx, Hy, Hz

• and vice versa– 3 degrees of freedom

HEHy

Ex

yo

x

xo

y

EH

EH


23

An example application of Maxwell’s equations:

The Magnetron


24

Poynting vector S

Displacement current, D

Current, J

The magnetron

Lorentz force F=qvB

• Locate features:


25

Power flow of EM radiation• Intuitively power flows in the direction of

propagation– in the direction of EH

2

2

E

H

HEHE

z

z

xyyxz

a

a

aHE

• What are the units of EH?– H2=.(Amps/metre)2 =Watts/metre2 (cf I2R)

– E2/=(Volts/metre)2 / =Watts/metre2 (cf I2R)

• Is this really power flow ?


26

• Energy stored in the EM field in the thin box is:

Power flow of EM radiation cont.

HEHy

Ex

dx

Area A

xAuuUUU HEHE dddd

2

22

2

Hu

Eu

oH

E

xAE

xAHE

U o

d

d22

d

2

22

xo

y EH

• Power transmitted through the box is dU/dt=dU/(dx/c)....


27

• This is the instantaneous power flow– Half is contained in the electric component

– Half is contained in the magnetic component

• E varies sinusoidal, so the average value of S is obtained as:

2

22

W/mddd

d

E

xAcxA

E

tA

US

o

xAEU dd 2

Power flow of EM radiation cont.

2sinRMS

sin

2sinE

222

2

22

oo

o

o

o

EvtzE

ES

vtzES

vtzE

• S is the Poynting vector and indicates the direction and magnitude of power flow in the EM field.


28

Example problem• The door of a microwave oven is left open

– estimate the peak E and H strengths in the aperture of the door.

– Which plane contains both E and H vectors ?

– What parameters and equations are required?

222

W/mHE

S

• Power-750 W

• Area of aperture - 0.3 x 0.2 m

• impedance of free space - 377 • Poynting vector:


29

Solution

μTesla2.775.5104B

A/m75.5377

2170H

V/m171,22.0.3.0

750377

WattsA

7

22

H

E

A

PowerE

HAE

SAPower

o


30

• Suppose the microwave source is omnidirectional and displaced horizontally at a displacement of 100 km. Neglecting the effect of the ground:

• Is the E-fielda) vertical

b) horizontal

c) radially outwards

d) radially inwards

e) either a) or b)

• Does the Poynting vector pointa) radially outwards

b) radially inwards

c) at right angles to a vector from the observer to the source

• To calculate the strength of the E-field should onea) Apply the inverse square law to the power generated

b) Apply a 1/r law to the E field generated

c) Employ Coulomb’s 1/r2 law


31

Field due to a 1 kW omnidirectional generator (cont.)

mV/m5.1104

750377

4

nW/m97.5104

10

4

252

225

3

2

r

Power

A

PowerE

r

PSPower


32

4.1 Polarisation

• In treating Maxwells equations we referred to components of E and H along the x,y,z directions– Ex, Ey, Ez and Hx, Hy, Hz

• For a plane (single frequency) EM wave – Ez=Hz =0

– the wave can be fully described by either its E components or its H components

yo

x

xo

y

EH

EH

– It is more usual to describe a wave in terms of its E components

• It is more easily measured

– A wave that has the E-vector in the x-directiononly is said to be polarised in the x direction

• similarly for the y direction


33

Polarisation cont..• Normally the cardinal axes are Earth-referenced

– Refer to horizontally or vertically polarised

– The field oscillates in one plane only and is referred to as linear polarisation• Generated by simple antennas, some lasers, reflections off dielectrics

– Polarised receivers must be correctly aligned to receive a specific polarisation

A horizontal polarised wave generated by a horizontal dipole and incident upon horizontal and vertical dipoles


34

Horizontal and vertical linear polarisation


35

Linear polarisation• If both Ex and Ey are present and in phase then components

add linearly to form a wave that is linearly polarised signal at angle

x

y

E

E1tan

Horizontal polarisation

Verticalpolarisation

Co-phased vertical+horizontal=slant Polarisation


36

Slant linear polarisation

Slant polarised waves generated by co-phased horizontal and vertical dipoles and incident upon horizontal and vertical dipoles


37

Circular polarisation

vtzEE ohh sin

vtzE

vtzEE

ov

ovv

cos

2sin

vtzEE ohh sin

vtzE

vtzEEv

ov

ov

cos

2sin

RHC polarisation

LHC polarisation NB viewed as approaching wave


38

Circular polarisation

LHC & RHC polarised waves generated by quadrature -phased horizontal and vertical dipoles and incident upon horizontal and vertical dipoles

LHC

RHC


39

Elliptical polarisation - an example

3sin5.1

vtzEv

vtzEh sin0.1


40

Constitutive relations• permittivity of free space =8.85 x 10-12 F/m

• permeability of free space o=4x10-7 H/m

• Normally r(dielectric constant) andr

– vary with material

– are frequency dependant• For non-magnetic materials r ~1 and for Fe is ~200,000

• ris normally a few ~2.25 for glass at optical frequencies

– are normally simple scalars (i.e. for isotropic materials) so that D and E are parallel and B and H are parallel

• For ferroelectrics and ferromagnetics r and r depend on the relative orientation of the material and the applied field:

EJ

HHB

EED

or

or

z

y

x

zzzyzx

yzyyyx

xzxyxx

z

y

x

H

H

H

B

B

B

o

ij

00

0j

0jAt microwavefrequencies:


41

Constitutive relations cont...• What is the relationship between and refractive index for non

magnetic materials ?– v=c/n is the speed of light in a material of refractive index n

– For glass and many plastics at optical frequencies• n~1.5

• r~2.25

• Impedance is lower within a dielectric

What happens at the boundary between materials of different n,r,r

?

r

roo

n

n

cv

1

ro

ro


42

Why are boundary conditions important ?• When a free-space electromagnetic wave is incident upon a medium secondary waves are

– transmitted wave

– reflected wave

• The transmitted wave is due to the E and H fields at the boundary as seen from the incident side

• The reflected wave is due to the E and H fields at the boundary as seen from the transmitted side

• To calculate the transmitted and reflected fields we need to know the fields at the boundary– These are determined by the boundary conditions


43

Boundary Conditions cont.

• At a boundary between two media, r,r are different on either side.

• An abrupt change in these values changes the characteristic impedance experienced by propagating waves

• Discontinuities results in partial reflection and transmission of EM waves

• The characteristics of the reflected and transmitted waves can be determined from a solution of Maxwells equations along the boundary

2,22

1,11


44

2.3 Boundary conditions• The tangential component of E is continuous

at a surface of discontinuity

– E1t,= E2t

• Except for a perfect conductor, the tangential component of H is continuous at a surface of discontinuity

– H1t,= H2t

E1t, H1t

2,22

1,11

E2t, H2t

2,22

1,11D1n, B1n

D2n, B2n

• The normal component of D is continuous at the surface of a discontinuity if there is no surface charge density. If there is surface charge density D is discontinuous by an amount equal to the surface charge density.

– D1n,= D2n+s

• The normal component of B is continuous at the surface of discontinuity

– B1n,= B2n


45

Proof of boundary conditions - continuity of Et

• Integral form of Faraday’s law:A

BsE d

td

A..

yxt

BxE

yE

yExE

yE

yE z

xyyxyy

243112 2222

21

21 0xx

xxEE

xExE

00 00

0That is, the tangential

component of E is continuous

0,0As yxtBy z

2,22

1,11y

x1yE

2yE

2xE

1xE 3yE

4yE


46

• Ampere’s law

Proof of boundary conditions - continuity of Ht

AJD

sH dt

dA

..

yxJt

DxH

yH

yHxH

yH

yH z

zxyyxyy

243112 2222

0,0As yxJtDy zz

21

21 0xx

xxHH

xHxH

That is, the tangential component of H is

continuous

00 00

0

2,22

1,11y

x1yH

2yH

2xH

1xH 3yH

4yH


47

• The integral form of Gauss’ law for electrostatics is:

Proof of boundary conditions - Dn

V

dVAD d.

applied to the box gives

yxyxDyxD snn edge21

0,0dAs edge z hence

snn DD 21 The change in the normal component of D at a boundary is equal to the surface charge density

y

2,22

1,11

1nD

2nD

z

x


48

• For an insulator with no static electric charge s=0snn DD 21

Proof of boundary conditions - Dn cont.

21 nn DD

• For a conductor all charge flows to the surface and for an infinite, plane surface is uniformly distributed with area charge density s

In a good conductor, is large, D=E0 hence if medium 2 is a good conductor

snD 1


49

Proof of boundary conditions - Bn

• Proof follows same argument as for Dn on page 47,

• The integral form of Gauss’ law for magnetostatics is

– there are no isolated magnetic poles 0d. AB

21

edge21 0

nn

nn

BB

yxByxB

The normal component of B at a boundary is always continuous at a boundary


50

2.6 Conditions at a perfect conductor• In a perfect conductor is infinite• Practical conductors (copper, aluminium silver) have

very large and field solutions assuming infinite can be accurate enough for many applications– Finite values of conductivity are important in calculating

Ohmic loss

• For a conducting medium– J=E

• infinite infinite J

• More practically, is very large, E is very small (0) and J is finite


51

• It will be shown that at high frequencies J is confined to a surface layer with a depth known as the skin depth

• With increasing frequency and conductivity the skin depth, x becomes thinner

2.6 Conditions at a perfect conductor

Lower frequencies, smaller

Higher frequencies, larger

xx

Current sheet

• It becomes more appropriate to consider the current density in terms of current per unit with:

0A/mlim

xx s

JJ


52

• Ampere’s law:AJ

DsH d

td

A..

yxJt

DxH

yH

yHxH

yH

yH z

zxyyxyy

243112 2222

szzz xJyxJyxtDy ,0,0As

szxx JHH 21

That is, the tangential component of H is discontinuous by an amount equal to the

surface current density

Jszx0 00

0

2,22

1,11y

x1yH

2yH

2xH

1xH 3yH

4yH

Conditions at a perfect conductor cont. (page 47 revisited)


53

• From Maxwell’s equations:– If in a conductor E=0 then dE/dT=0– Since

Conditions at a perfect conductor cont. (page 47 revisited)cont.

dt

HE

szx JH 1

Hx2=0 (it has no time-varying component and also cannot be established from zero)

The current per unit width, Js, along the surface of a perfect conductor is equal to the magnetic field just outside the surface:

• H and J and the surface normal, n, are mutually perpendicular:

HnJ s


54

Summary of Boundary conditions

21

21

21

21

nn

nn

tt

tt

BB

DD

HH

EE

0.

0.

0

0

21

21

21

21

BB

DD

HH

EE

n

n

n

n

At a boundary between non-conducting media

0.

.

0

0

21

21

21

21

BB

DD

HH

EE

n

n

n

n

s

At a metallic boundary (large )

At a perfectly conducting boundary

0.

.

0

1

1

1

1

B

D

JH

E

n

n

n

n

s

s


55

2.6.1 The wave equation for a conducting medium• Revisit page 8 derivation of the wave equation with J0

dt

BE

dt

DJH

As on page 8:HB

BE

ttt o

2

2

2

2

tt

tt

tt

oo

oo

o

EE

DJ

DJE

ED

EJ


56

In the absence of sources hence:

2

22

2

2

.tt

EEE

tt

E

oo

oo

E

EE

2.6.1 The wave equation for a conducting mediumcont.

0. E

2

22

tt oo

EE

E

• This is the wave equation for a decaying wave– to be continued...


57

Reflection and refraction of plane waves

• At a discontinuity the change in and results in partial reflection and transmission of a wave

• For example, consider normal incidence: ztj

ieE waveIncident ztj

reE waveReflected

• Where Er is a complex number determined by the boundary conditions


58

Reflection at a perfect conductor• Tangential E is continuous across the

boundary – see page 45

• For a perfect conductor E just inside the surface is zero– E just outside the conductor must be zero

ri

ri

EE

EE

0

• Amplitude of reflected wave is equal to amplitude of incident wave, but reversed in phase


59

Standing waves• Resultant wave at a distance -z from the interface

is the sum of the incident and reflected waves

tj

i

tjzjzji

ztjr

ztji

T

ezjE

eeeE

eEeE

tzE

sin2

wavereflectedwaveincident,

and if Ei is chosen to be real

tzE

tjtzjEtzE

i

iT

sinsin2

sincossin2Re,

j

ee jj

2sin


60

Standing waves cont...

• Incident and reflected wave combine to produce a standing wave whose amplitude varies as a function (sin z) of displacement from the interface

• Maximum amplitude is twice that of incident fields

tzEtzE iT sinsin2,


61

Reflection from a perfect conductor


62

Reflection from a perfect conductor• Direction of propagation is given by EH

If the incident wave is polarised along the y axis:

xixi

yiyi

HH

EE

a

a

xiyiz

xiyixy

HE

HE

a

aaHE

then

From page 18

That is, a z-directed wave.

xiyiz HEaHΕ For the reflected wave and yiyr EE aSo and the magnetic field is reflected without change in phase

ixixr HHH a


63

• Given that

derive (using a similar method that used for ET(z,t) on p59) the form for HT(z,t)


2cos

jj ee

tj

i

tjzjzji

ztjr

ztjiT

ezH

eeeH

eHeHtzH

cos2

,

As for Ei, Hi is real (they are in phase), therefore

tzHtjtzHtzH iiT coscos2sincoscos2Re,


64

• Resultant magnetic field strength also has a standing-wave distribution

• In contrast to E, H has a maximum at the surface and zeros at (2n+1)/4 from the surface:

tzHtzH iT coscos2,


free space silver

resultant wave

z = 0

z [m]

E [V/m]

free space silver

resultant wave

z = 0

z [m]

H [A/m]


65


• ET and HT are /2 out of phase( )

• No net power flow as expected– power flow in +z direction is equal to power flow in - z direction

tzHtzH iT coscos2,

tzEtzE iT sinsin2,

2/cossin tt


66

Reflection by a perfect dielectric

• Reflection by a perfect dielectric (J=E=0)– no loss

• Wave is incident normally– E and H parallel to surface

• There are incident, reflected (in medium 1)and transmitted waves (in medium 2):


67

Reflection from a lossless dielectric


68

Reflection by a lossless dielectric

• Continuity of E and H at boundary requires:

tri

tri

HHH

EEE

tt

rr

ii

HE

HE

HE

2

1

1

Which can be combined to give

rittriri EEEHEEHH 221

111

1212

12

21

11

ri

riri

riri

EE

EEEE

EEEE

12

12

i

rE E

E

The reflection coefficient

roj

j


69

• Similarly

Reflection by a lossless dielectric

tri

tri

HHH

EEE

12

2

12

12

12

12 21

i

r

i

ir

i

tE E

E

E

EE

E

E

12

22

E

The transmission coefficient


70

• Furthermore:Reflection by a lossless dielectric

Hi

t

i

t

Hi

r

i

r

E

E

H

H

E

E

H

H

12

1

12

2

2

1

2

1 22

And because =o for all low-loss dielectrics

21

2

21

2

21

1

21

1

21

21

21

21

22

22

nn

n

nn

n

E

E

nn

nn

E

E

H

i

rE

Hi

rE


71

The End

revision of electrostatics (week 1) vector calculus (week 2-4)

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