# Electromagnetic Fields and Waves

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More on MaxwellWave EquationsBoundary ConditionsPoynting VectorTransmission Line

A. Nassiri - ANL

Lecture 2Review 2

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2

Maxwells equations in differential form

HBED

== Varying E and H fields are coupled

continuityofEquationt

lawsFaradaydt

lawsAmperedt

ticsmagnetostaforlawGaussticselectrostaforlawGauss

=

=

+=

==

J

BE

DJH

0BD

.

'

'

'.'.

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3

Electromagnetic waves in lossless media - Maxwells equations

SI Units J Amp/ metre2 D Coulomb/metre2 H Amps/metre B Tesla

Weber/metre2Volt-Second/metre2

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

dtBE =dtDJH +=

= D.0. = B

EJHHB

EED

===

==

or

or

t

=J.

Maxwell

Equation of continuity

Constitutive relations

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4

Wave equations in free space

In free space =0 J=0 Hence:

dtBE =

dtdtDDJH =+=

2

2

t

tt

ttt

o

o

o

=

=

=

=

=

EE

D

HBBE

Taking curl of both sides of latter equation:

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 5

Wave equations in free space cont.

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

where is the Laplacian operatorThus:

2

2

to

=EE

AAA 2. =

2

2

2

2

2

22

zyx

+

+

=

2

22.

to

=EEE

2

22

to

=EE

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

A three dimensional wave equation

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6

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

2

22

to

=EE

Wave equations in free space cont.

2

22

to

=HH

22 secondseVolts/metr

metreeVolts/metr

= o

What does this mean dimensional analysis ?

has units of velocity-2

Why is this a wave with velocity 1/ ?

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 7

Uniform plane waves - transverse relation of E and H

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

0. =

+

+

=zyx

zyx EEEE

00 =

=

yxEE

)waveanotisconst(000,0 ===

=

=

zzzyx EE

zE

yE

xE

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.

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8

Orthogonal relationship between E and H:

For a plane z-directed wave there are no variations along x and y:

+

+

=

yA

xA

xA

zA

zA

yAA

xyz

zxy

yzx

a

a

a

zH

zH x

yy

x

+

= aaH

tE

zH

tE

zH

yx

xy

=

=

tH

zE

tH

zE

yo

x

xo

y

=

=

to = HE Equating terms:dtDJH +=

and likewise for :

+

+

=

=

tE

tE

tE

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 9

Orthogonal and phase relationship between E and H:

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

tE

zH

tE

zH

yx

xy

=

=

( )

( )vtzfvEt

EvtzfEE

oy

oy

=

=

tH

zE

tH

zE

yo

x

xo

y

=

=

xo

y EH

=

Similarly

( ) ( )

yo

x

y

oox

EH

vE

vtzfvEconstzvtzfvEH

=

=

=+= dz

H x

=

H and E are in phase and orthogonal

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 10

The ratio of the magnetic to electric fields strengths is:

which has units of impedance

yo

x EH

= xo

y EH

=

===

+

+o

yx

yx

HE

HH

EE22

22

=metreampsmetreVolts

//

==

=

37712010

361

1049

7

o

o

cH

EBE

ooo===

1

Note:

and the impedance of free space is:

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 11

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

HE

HE

==

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 12

A horizontally polarised wave

Sinusoidal variation of E and H E and H in phase and orthogonal

Hy Ex

xo

y EH

=

HE

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 13

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

=

=

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 14

Power flow of EM radiation

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

HEHy

Exdx

Area A

( ) xAuuUUU HEHE dddd +=+=

2

22

2

Hu

Eu

oH

E

=

=

xAE

xAHEU o

d

d22

d

2

22

=

+=

xo

y EH

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 15

Power flow of EM radiation cont.

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

22W/md

ddd

ExA

cxAE

tAUS

o====

xAEU dd 2=

( )

( )

( )( )

2sinRMS

sin

2sinE

222

2

22

oo

o

o

o

EvtzEES

vtzES

vtzE

==

=

=

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

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16

Example

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/mHES

==

Power-750 W Area of aperture - 0.3 x 0.2 m impedance of free space - 377 Poynting vector:

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17

Tesla2.775.5104B

A/m75.53772170H

V/m171,22.0.3.0

750377

WattsA

7

22

===

===

===

===

H

E

APowerE

HAESAPower

o

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18

Constitutive relations

permittivity of free space 0=8.85 x 10-12 F/m permeability of free space o=4x10-7 H/m Normally r (dielectric constant) and r

vary with material are frequency dependant

For non-magnetic materials mr ~1 and for Fe is ~200,000 er is normally a few ~2.25 for glass at optical frequencies

a