ch1

1FUNDAMENTALS

1.1 Maxwell Equations

A. Spatial Frequency k

B. Vector Analysis and Boundary Conditions

1.2 Polarization

Topic 1.2A Stokes Parameters and Poincare Sphere

1.3 Lorentz Force Law

A. Lenz Law and Electromotive Force (EMF)

B. Poyntings Theorem and Poynting Vector

C. Momentum Conservation Theorem

1.4 Hertzian Waves

Topic 1.4A Electric Field Pattern

1.5 Reection and Transmission

A. Reection and Transmission of TE Waves

B. Reection and Transmission of TM Waves

C. Brewster Angle and Zero Reection

1.6 Phase Matching

A. Critical Angle and Total Reection

B. Guided Waves in a Dielectric Waveguide

1.7 Wave Guidance

A. Guidance by Conducting Parallel Plates

B. Guidance by Rectangular Waveguides

1

2 1. Fundamentals

C. Rectangular Cavity Resonators

1.8 Constitutive Relations

A. Anisotropic and Bianisotropic Media

Topic 1.8A Constitutive Matrices

1.9 Boundary Conditions

Topic 1.9A Derivation of Boundary Conditions

Topic 1.9B Boundary Conditions for Moving Boundaries

Answers

1.1 Maxwell Equations 3

1.1 Maxwell Equations

The laws of electricity and magnetism were established in 1873 by JamesClerk Maxwell (18311879). In three-dimensional vector notation, the Max-well equations are

H (r, t) = t

D (r, t) + J (r, t) (1.1.1)

E (r, t) = t

B (r, t) (1.1.2)

D (r, t) = (r, t) (1.1.3) B (r, t) = 0 (1.1.4)

where E, B, H, D, J, and are real functions of position and time.

E (r, t) = electric eld strength (volts/m)

B (r, t) = magnetic ux density (webers/m2)

H (r, t) = magnetic eld strength (amperes/m)

D (r, t) = electric displacement (coulombs/m2)

J (r, t) = electric current density (amperes/m2)

(r, t) = electric charge density (coulombs/m3)

Equation (1.1.1) is Ampe`res law or the generalized Ampe`re circuit law.Equation (1.1.2) is Faradays law or Faradays magnetic induction law.Equation (1.1.3) is Coulombs law or Gauss law for electric elds.Equation (1.1.4) is Gauss law or Gauss law for magnetic elds.Maxwells contribution to the laws of electricity and magnetism is the addi-tion of the displacement term D/t in Ampe`res law (1.1.1).

The electric current density J (r, t) and the electric charge density (r, t) are governed by the continuity law

J (r, t) = t

(r, t) (1.1.5)

which states that the electric current and charge densities at r are conserved.The divergence of current J from an innitesimal volume surrounding r isequal to the decreasing of electric charge density with time t .

4 1. Fundamentals

Constitutive Relations for Free SpaceThe Maxwell equations are fundamental laws governing the behavior of

electromagnetic elds in free space and in media. Free space is characterizedby the following constitutive relations:

D = oE (1.1.6a)B = oH (1.1.6b)

where

o 8.85 1012 farad/metero= 4 107 henry/meter

are, respectively, the permittivity and the permeability of free space. Givingthe velocity of light in free space being c = 3 108 m/s, the permittivity

o = 1/(oc2) , which follows from the dispersion relation as derived below.

Wave EquationThe Maxwell equations in dierential form are valid at all times for

every point in space. First we shall investigate solutions to the Maxwellequations in regions devoid of source, namely in regions where J = 0 and = 0 . This of course does not mean that there is no source anywhere in allspace. Sources must exist outside the regions of interest in order to produceelds in these regions. Thus in source-free regions in free space, the Maxwellequations become

H = o t

E (1.1.7)

E = o t

H (1.1.8)

E = 0 (1.1.9) H = 0 (1.1.10)

In the form of scalar partial dierential equations, we have

yHz

zHy = o

tEx (1.1.11a)

zHx

xHz = o

tEy (1.1.11b)

xHy

yHx = o

tEz (1.1.11c)


yEz

zEy = o

tHx (1.1.12a)

zEx

xEz = o

tHy (1.1.12b)

xEy

yEx = o

tHz (1.1.12c)

xEx +

yEy +

zEz = 0 (1.1.13a)

xHx +

yHy +

zHz = 0 (1.1.13b)

A wave equation for E can be derived by eliminating H from (1.1.11) and(1.1.12). Taking time derivatives of (1.1.11a) and substituting (1.1.12c) and(1.1.12b), we have

oo2

t2Ex =

y

(

xEy

yEx

)+

z

(

zEx

xEz

)

=(2

x2+

2

y2+

2

z2

)Ex

where use is made of (1.1.13). Thus we obtain the following equations forthe three components of E :(

2

x2+

2

y2+

2

z2 oo

2

t2

)Ex = 0 (1.1.14a)(

2

x2+

2

y2+

2

z2 oo

2

t2

)Ey = 0 (1.1.14b)(

2

x2+

2

y2+

2

z2 oo

2

t2

)Ez = 0 (1.1.14c)

Making use of the Laplacian operator 2 in rectangular coordinate system

2 = 2

x2+

2

y2+

2

z2(1.1.15)

we have

2E oo 2

t2E = 0 (1.1.16)

This is known as the Helmholtz wave equation.

6 1. Fundamentals

Wave SolutionSolutions to the wave equation (1.1.16) that satisfy all Maxwell equa-

tions are electromagnetic waves. We shall now study a solution to (1.1.14a)assuming Ey = Ez = 0 . Let Ex be a function only of z and t and inde-pendent of x and y . The electric eld vector can be written as

E = xEx(z, t)

The wave equation it satises follows from (1.1.16) which becomes

2

z2Ex oo

2

t2Ex = 0 (1.1.17)

The simplest solution to (1.1.17) takes the form

E = xEx(z, t) = xE0 cos(kz t) (1.1.18)

Substituting (1.1.18) in (1.1.17) we nd that the following equation,called the dispersion relation, must be satised:

k2 = 2oo (1.1.19)

The dispersion relation provides an important connection between the spatialfrequency k and the temporal frequency .

There are two points of view useful in the study of a space-time varyingquantity such as Ex(z, t) . The temporal view point is to examine the timevariation at xed points in space. The spatial view point is to examine spatialvariation at xed times, a process that amounts to taking a series of pictures.

From the temporal view point, we rst x our attention on one par-ticular point in space, say z = 0 . We then have the electric eld Ex(z =0, t) = E0 cost . Plotted as a function of time in Fig. 1.1.1, we nd thatthe waveform repeats itself in time as t = 2m for any integer m. Theperiod is dened as the time T for which T = 2 . The number of periodsin a time of one second is the frequency f dened as f = 1/T , which gives

f =

2(1.1.20)

The unit for frequency f is Hertz (Hz) with 1 Hz = 1 s1 , which is equal tothe number of cycles per second. Since = 2f , is the angular frequencyof the wave.


t

Ex(z = 0, t) = E0 cost

2

3

T =

Figure 1.1.1 Electric eld strength as a function of t at z = 0.

1 1 1

Ex = E0 cost Ex = E0 cos 2t Ex = E0 cos 3t

t t t

sec sec sec

a. = o = 2 Hz b. = 2o = 4 Hz c. = 3o = 6 Hz

Figure 1.1.2 Electric eld strength vs. t for dierent frequencies .

In this book, we often refer to as the frequency, simply because is more commonly encountered than f . The temporal frequency charac-terizes the variation of the wave in time. We plot in Fig. 1.1.2a Ex(z = 0, t)as a function of t instead of t . Let there be one period within the timeinterval of 1 second. Thus f = fo = 1 Hz , and we let = o = 2 rad/s.In Fig. 1.1.2b, we plot = 2o ; there are two periods in a time interval ofone second and the period in time is 0.5 seconds. In Fig. 1.1.2c, = 3oand there are three periods in one second.

8 1. Fundamentals

A. Spatial Frequency k

To examine wave behavior from the spatial view point, we let t = 0 andplot Ex(z, t = 0) in Fig. 1.1.3. The waveform repeats itself in space whenkz = 2m for integer values of m. The spatial frequency k characterizesthe variation of the wave in space. The wavelength is dened as thedistance for which k = 2 . Thus = 2/k , or

k =2

(1.1.21)

We call k the spatial frequency or the wavenumber which is equal to thenumber of wavelengths in a distance of 2 and has the dimension of inverselength.

2

3

kz

k =

Ez(z, t = 0) = E0 cos kz

Figure 1.1.3 Electric eld strength as a function of kz at t = 0.

To further understand the meaning of k as a spatial frequency, we plotin Fig. 1.1.4a Ex(z, t = 0) as a function of z instead of kz . Let there beone period within the wavelength of 1 meter. We dene Ko = 2 rad/m .Thus k = 1Ko = 2 rad/m . In Fig. 1.1.4b, we plot k = 2 Ko ; there are twoperiods in a spatial distance of one meter and the wavelength is 0.5 meters.In Fig. 1.1.4c, k = 3 Ko and there are three periods in one meter.

We dene for the spatial frequency k a fundamental unit Ko :

1Ko = 2 rad/m (1.1.22)

Similar to the unit Hz which is cycles per second in temporal variation, Ko iscycles per meter in spatial variation. For a wave that has a spatial frequency


Ex = E0 cos Koz Ex = E0 cos 2Koz Ex = E0 cos 3Koz

a. k = Ko b. k = 2Ko c. k = 3Ko

z z z

1m 1m 1m

Figure 1.1.4 Electric eld strength vs. distance z with dierent spatial fre-quency k.

of one period in one meter distance, we have k = 1 Ko . An electromagneticwave in free space with k = 5 Ko has ve spatial periods in a distance of onemeter. From the dispersion relation for electromagnetic waves, the spatialfrequency and the temporal frequency are related by the velocity of light.In free space, the conversion factor is c = (oo)1/2 = 3 108 m/s . Thusfor a spatial frequency of 1 Ko , the corresponding temporal frequency isf = 300 MHz .

Within the spatial frequency range of 0.01 Ko to 100 Ko electromag-netic waves are used for microwave heating, radar, navigation, and carryingsignals from radio, television, and satellite communications. The visible lighthas a spatial frequency band between 1.4106 2.6106 Ko . In Fig. 1.1.5we illustrate the electromagnetic wave spectrum according to the spatial fre-quency in Ko and corresponding wavelength in meters, frequency in Hz, andenergy in electron-volts (eV).

In this book we shall place great emphasis on the use of k , whichis of more fundamental importance in electromagnetic wave theory thanboth of the more popular concepts of wavelength and frequency f .The corresponding values of wavelength and frequency are, for k = AKo , = 2/k = 2/(AKo) = 1/Am and f = ck/2 = cAKo/2 = 3108AHz .The photon energy in electron-volts (eV) is calculated from h Joule =hck/q eV , where h = 1.05 1034 Joule-second is Plancks constant di-vided by 2 and q = 1.6 1019 coulombs is the electron charge. Thush = (hcAKo/q) eV 1.24 106A eV .

10 1. Fundamentals

k

Ko

m

h f

eV

Hz

108

106

104

102

Ko

102

104

MKo

108

1010

1012

106

km

m

cm

mm

104

108

A

1012

Hz

kHz

MHz

GHz

THz

1014

1016

1018

1020

eV

keV

MeVKo = 2m1k = AKo = 1/Amf = 300AMHzh = 1.24 106A eV

-Ray

X-Ray

Ultraviolet

Visible (1.4 2.6 MKo; 417 789 )Near Infrared (0.77 1.4 MKo;Thermal Infrared (0.067 0.14 MKo;

Millimeter WaveEHF

SHF Radar C, X, Ku, K bands (13.3

470

GHz )

UHF

4

Microwave Oven (8.17Ko; 2.45 GHz )

VHF Television 890

FM

HF Short Wave Radio

MF AM Radio (.00178 .00535 K 535 1605 kHz)

LF (104 103 Ko

;

30 300 kHz)VLF (105 104 Ko; 3 30 kHz)ULF (106 105 Ko; 0.3 3 kHz)SLF (107 106 Ko; 30 300Hz)ELF (108 107 Ko; 3 30 Hz)

10 8

o;

(.01 .001 K 3 30 MHz )o ;

Radio (.29 .36 K 88 108 MHz )o ;

90 K

MHz )Television (1.57 2.97 K

27

o ;(.18 .72 K 54 216 MHz )o ;

o

Frequency Frequency( )) (Spatial Temporal

THz231 417 )THz

20 43 )THz

Figure 1.1.5 Electromagnetic wave spectrum.


Phase Velocity and Phase Delay

In Figs. 1.1.6b and 1.1.6c we plot Ex(z, t) at two progressive timest = /2 and t = . We observe that the electric eld vector at Aappears to be propagating along the z direction as time progresses. Thevelocity of propagation Vp is determined from kz t = constant whichgives

Vp =dz

dt=

k(1.1.23)

We call Vp the phase velocity. By virtue of the dispersion relation (1.1.19),we see that Vp = (oo)1/2 , which is equal to the velocity of light in freespace c .

Ex = E0 cos kz Ex = E0 sin kz Ex = E0 cos kz

a. t = 0 b. t =

2c. t =

kz kz kz

A

A

A

Ex Ex Ex

2 2 2

3 3 3

Figure 1.1.6 Electric eld strength vs. kz at dierent times.

The spatial frequency k is, according to the dispersion relation, directlyrelated to the temporal frequency by the phase delay

p =k

=oo (1.1.24)

which determines how much time it takes for the wave to propagate a unitdistance. In free space p = 108/3 s/m or it takes 3.33 nanoseconds for anelectromagnetic wave to travel the distance of one meter.

12 1. Fundamentals

Example 1.1.1 Operating frequencies of common devices:

Device Temporal frequency (Hz) Spatial frequency (Ko)AM Radio 535 1605 kHz 0.00178 0.00535 KoShortwave Radio 3 30 MHz 0.01 0.1 KoFM Radio 88 108 MHz 0.293 0.36 KoAirport ILS 108 112 MHz 0.35 0.373 KoCommercial Television

Channels 2-4 54 72 MHz 0.18 0.24 KoChannels 5-6 76 88 MHz 0.253 0.293 KoChannels 7-13 174 216 MHz 0.58 0.72 KoChannels 14-83 470 890 MHz 1.57 2.97 Ko

Microwave Oven 2.45 GHz 8.17 KoCommunication Satellite

Downlink 3.70 4.20 GHz 12.3 14 KoUplink 5.925 6.425 GHz 19.75 21.4 Ko

End of Example 1.1.1

Electric Field Vector E and Magnetic Field Vector HFor the wave solution in (1.1.18) for the vector electric eld E ,

E = xEx(z, t) = xE0 cos(kz t) (1.1.25)the vector magnetic eld H can be determined from (1.1.8). We nd

o

tH = E =

x y z

x

y

z

Ex 0 0

= y kEo sin(kz t)

The magnetic eld vector H is then

H = yk

oE0 cos(kz t) (1.1.26)

Equations (1.1.25) and (1.1.26) are seen to satisfy all Maxwell equations

H = o t

E (1.1.27)

E = o t

H (1.1.28)

E = 0 (1.1.29) H = 0 (1.1.30)


z

Magne

ticel

d H

Electriceld E

Figure 1.1.7 Electric and magnetic eld vectors of an electromagnetic wave.

Write the amplitude of the magnetic eld vector H as H0

H = yHy(z, t) = yH0 cos(kz t) (1.1.31)where H0 = E0/ and =

o/o is called the free-space impedance. The

electromagnetic wave is propagating in the positive z direction because astime t increases, z must increase in order to maintain a constant phasekzt . The eld vectors of the electromagnetic wave are transversal to thedirection of propagation and lie in the xy -plane, on which the phase kztof the wave is a constant. Since the phase front of the wave is the xy -plane,we call the electromagnetic wave as represented by (1.1.25) and (1.1.31) aplane wave.

Exercise 1.1.1An electromagnetic wave propagating in the negative z direction

E = xEx(z, t) = xE0 cos(kz + t) (Ex1.1.1.1)

because as time t increases, z must decrease in order to maintain kz + t aconstant. The associated magnetic eld vector H takes the form

H = yHy(z, t) = yH0 cos(kz + t) (Ex1.1.1.2)

where H0 = E0/ and =o/o is the free-space impedance.

14 1. Fundamentals

Personalities in Electromagnetics

James Clerk Maxwell (13 June 1831 5 November 1879)James Clerk Maxwell attended University of Edinburgh (18471850),

and studied under William Hopkins at Cambridge University (18501854).He was a fellow of Trinity (18551856), Professor of Natural Philosophy atMarischal College of the University of Aberdeen (18561860), and at KingsCollege (18601865). He was the rst Cavendish Professor of ExperimentalPhysics at Cambridge University to build and direct the Cavendish Labo-ratory (18711879). He published four books and about 100 papers startingage 14. Inspired by Faradays concept of lines of force, Maxwell publishedhis papers On Faradays Lines of Forces in 1855, On Physical Lines ofForce in 1861, and A Dynamical Theory of the Electromagnetic Field inDecember 1864. In 1865, at age 33, he retired to his country home estate andspent six years to write his monumental book A Treatise of Electricity andMagnetism (Constable and Company, London, 1873; Dopver Publications,New York, 1006 pages, 1954).

Originally written in Cartesian component form, the Maxwell equationswere cast in the vector form by Oliver Heaviside (18 May 1850 3 February1925). Heinrich Rudolf Hertz (22 February 1857 1 January 1894) experi-mentally veried Maxwells theory in 1888. Since then, electromagnetic the-ory has played a central role in the development of radio, television, wirelesscommunications, radar, microwave heating, remote sensing, and numerousother practical applications. The special theory of relativity developed by Al-bert Einstein (14 March 1879 18 April 1955) in 1905 further asserted therigorousness and elegance of Maxwells theory. As a well-established scienticdiscipline, this sophisticated theoretical structure embodies many principlesand concepts which serve as fundamental rules of nature and vital links toother scientic disciplines.

Michael Faraday (22 September 1791 25 August 1867)Faraday became an assistant to Sir Humphry Davy at the Royal Insti-

tution on 1 March 1813. In September 1821, his experimentation demon-strated electro-magnetic rotation, initiated the concept of electric motor. InAugust 1831, he discovered electro-magnetic induction, and that magnetismproduced electricity through movement, the principle behind the electrictransformer and generator. He became professor of chemistry in 1833. Oneof his most important contributions to physics was his development of theconcept of lines of force leading to the development of the concept of eldsby Maxwell. Faraday published many of his results in the three-volume Ex-perimental Researches in Electricity (18391855).


Johann Carl Friedrich Gauss (30 April 1777 23 February 1855)Gauss studied mathematics at the University of Gottingen from 1795 to

1798, and received his doctoral degree from the University of Helmstedt in1799. In 1807 he took the position of director of the Gottingen Observatory.In 1832 he presented a systematic use of absolute units (length, mass, time)to measure nonmechanical quantities. From 1831 to 1837 he worked closelywith Weber on terrestrial magnetism and organized a system of stations forsystematic observations.

Andre-Marie Ampe`re ( 20 January 1775 10 June 1836)Ampe`re was appointed professor at Bourg Ecole Centrale in 1802, at

the Ecole Polytechnique in 1809, and at Universite de France in 1826. InSeptember 1820, Ampe`re showed that two parallel conductors attract eachother if they carry currents that ow in the same direction and repel if thecurrents ow in opposite directions. In 18231826, he completed his Memoiron the Mathematical Theory of Electrodynamic Phenomena, Uniquely De-duced from Experience, which contains the description of his experimentsand a mathematical formulation of the laws that govern the interaction ofelectric currents with magnetic elds.

Charles-Augustin de Coulomb (14 June 1736 23 August 1806)Coulomb worked in the Corps du Genie until he retired in 1791. In

1777 he invented the torsion balance, which enabled him to establish thefundamental laws of electricity by measuring the force between two smallspheres charged with electricity. Between 1785 and 1791, he published seventreatises on electricity and magnetism. Using the torsion balance method,he established laws of attraction and repulsion, the electric point charges,magnetic poles, distribution of electricity on the surface of charged bodiesand others.

Hermann Ludwig Ferdinand von Helmholtz (31 August 1821 8 September 1894)

Hermann von Helmholtz was a professor of anatomy and physiology atthe University of Bonn in 1858, then became a professor of physics at theUniversity of Berlin in 1871, and the rst director of the Physico-TechnicalInstitute of Berlin in 1888. His 3-volume Handbook of Physiological Opticsappeared between 1856 and 1867. Helmholtz, who was born ten years earlierthan Maxwell and died eight months later than his student Hertz death, thuswitnessed the whole development and triumphant verication of Maxwellselectromagnetic wave theory.

16 1. Fundamentals

B. Vector Analysis and Boundary Conditions

A vector has a magnitude and a direction. We write vector A as

A = aA

where A is the magnitude of A and a is a dimensionless vector with a unitmagnitude pointing in the direction of A . The vector A can be representedgraphically by a directed straight-line element of length A and pointing inthe direction of a .

Two vectors A and B , when they are not in the same direction or inopposite directions, determine a plane. The addition and subtraction of Aand B are illustrated graphically in Fig. 1.1.8.

A

B

A+B

a

(a)

A

BAB

AB

B

(b)

Figure 1.1.8 Addition and subtraction of A and B.

The scalar or dot product of A and B , denoted by A B , is a scalarnumber:

A B = AB cos ABwhere AB is the angle between A and B .

AB

AB

B

A

Figure 1.1.9 Cross product AB.


A

x

y

z

Ax

Ay

Az

x

y

z

Figure 1.1.10 Projection of A in rectangular coordinate system.

The vector or cross product of two vectors A and B , denoted by AB ,is a vector perpendicular to the plane containing A and B . Thus A Bis perpendicular to both A and B [Fig. 1.1.9]. The magnitude of A Bis |AB sin AB| , which is equal to the area of the parallelogram formed byA and B . Its direction follows the right-hand rule, i.e., when the ngers ofthe right hand rotate from A to B , the thumb of the right hand points inthe direction of AB . Division by a vector is not dened; thus B/A , 1/Aare meaningless expressions.

If none of the operations of addition, subtraction, dot product, or crossproduct is imposed on A and B , the entity AB is called a dyad. In thelanguage of tensor analysis, a dyad is a tensor of second rank, while allvectors are tensors of rst rank.

Any vector can be represented by three components projected onto aCartesian coordinate system (also called the rectangular coordinate system)with three orthogonal unit vectors x, y, and z where x = y z, y =z x, z = x y, x x = y y = z z = 1, and x y = y z = z x = 0. Wewrite Ax, Ay, Az as projections of A onto the x, y, z axes [Fig. 1.1.10].

Position Vector r = xx+ yy + zz , whose three components are (x, y, z) .

18 1. Fundamentals

For the two vectors A and B , we write

A = xAx + yAy + zAz

B = xBx + yBy + zBz

Thus,A B = AxBx +AyBy +AzBz

AB = x(AyBz AzBy) + y(AzBx AxBz) + z(AxBy AyBx)

=

x y zAx Ay AzBx By Bz

C (AB) = Cx(AyBz AzBy) + Cy(AzBx AxBz)

+ Cz(AxBy AyBx)

=

Cx Cy CzAx Ay AzBx By Bz

It is useful to prove the vector identities

C (AB) = A (B C) = B (C A) (1.1.32)

C (AB) = A(C B) (C A)B (1.1.33)Both identities will be used frequently later on.

Gradient of a ScalarIn the Cartesian coordinate system, the del operator is a vector

dierential operator expressed as

= x x

+ y

y+ z

z

When operating on a scalar function , the result is a vector

= x x

+ y

y + z

z (1.1.34)

called the gradient of .


Example 1.1.2Consider the function = x+ y . The gradient of the function is

= x+ y

For 2 = x2 + y2 > 1 = x1 + y1 , we see that is pointing in the direction ofincreasing .


Example 1.1.3The function = x2 + 2y2 describes an ellipse. Its gradient is

= xx+ y2y

For the ellipse with equal to a constant, d = dr = 0 , where dr is tangentto the ellipse. Thus the gradient is normal to the ellipse and pointing in thedirections of an expanding ellipse.


Example 1.1.4 Electric eld vector as gradient of a potential function.When there is no time variation, we may write the electric eld vector E as

E = (E1.1.4.1)

and call a potential function. The picture is that E points from high potentialtowards low potential, similar to water owing from a high altitude to lower ground.


Divergence of a VectorThe divergence of a vector function is a scalar, dened as

D =(x

x+ y

y+ z

z

) (xDx + yDy + zDz)

=

xDx +

yDy +

zDz (1.1.35)

20 1. Fundamentals

x

y

z

xy

z

(x0, y0, z0)

Figure 1.1.11 Dierential volume xyz.

Consider a dierential volume of sides x,y,z centered around apoint (x0, y0, z0) [Fig. 1.1.11]. The divergence as dened states that

D = limx0y0z0

1xyz

{yz

[Dx(x0 +

x2, y0, z0)Dx(x0 x2 , y0, z0)

]

+ zx[Dy(x0, y0 +

yz, z0)Dy(x0, y0 y

z, z0)

]

+ xy[Dz(x0, y0, z0 +

z2

)Dz(x0, y0, z0 z2 )]}

(1.1.36)

The rst term in the braces is equal to the eld component Dx at thesurface at x = x0 + x2 multiplied by the surface area yz . We dene asurface normal vector dS pointing outward of the volume such that at thesurface at x = x0 + x2 , dS = xyz and at the surface at x = x0 x2 ,dS = xyz . Then the negative sign in the second term is due to Ddot multiplied by dS . All six terms account for the six dierential areasbounding the dierential volume V = xyz with a surface normaldS . We thus express the divergence of D as

D = limV0

dS DV

(1.1.37)


Derivation of Boundary Conditions for D and BWhen there is a plane boundary at z = z0 , and the D eld values

are nite above and below the plane boundary, we can derive the boundarycondition for D by using (1.1.36) with a small pill-box [Fig. 1.1.12] andletting z go the zero rst. We nd

D = limz0

1z

[Dz(x0, y0, z0 +

z2

)Dz(x0, y0, z0 z2 )](1.1.38)

Making use of (1.1.3), we nd

limz0

[Dz(x0, y0, z0 +

z2

)Dz(x0, y0, z0 z2 )]

= limz0

z

(1.1.39)

Since Dz(x0, y0, z0 + z2 ) is in region 1 and Dz(x0, y0, z0 z2 ) in region 2,we write Dz(x0, y0, z0+ z2 ) = D1z and Dz(x0, y0, z0z2 ) = D2z . The righthand side of (1.1.39) becomes zero when the volume charge density withunit coulomb/m3 is nite. However, if we assume is innite contained in azero thickness, we may dene a surface charge density s = lim

z0z which

is nite and has dimension coulombs/m2 . The concept of surface chargedensity will have very practical usefulness. Equation (1.1.39) becomes

Region 1

Region 2

x

y12z

dS = sdS

Figure 1.1.12 Small pill-box volume.

D1z D2z = s (1.1.40)Letting the surface element dS = sxy = sdS we can write (1.1.40) as

s (D1 D2) = s (1.1.41)

22 1. Fundamentals

Thus the dierence between the D eld components normal to the boundarysurface is equal to the surface charge density at the boundary surface.

When there is no surface charge density at the boundary surface, wehave

D1z D2z = 0 (1.1.42)or

s (D1 D2) = 0 (1.1.43)Thus the normal D component is continuous across the boundary. By thesame token, we conclude from (1.1.4) that

B1z B2z = 0 (1.1.44)or

s (B1 B2) = 0 (1.1.45)Thus the B eld component normal to the boundary surface is continuoussince there is no surface charge density at the boundary surface.

Divergence TheoremApplying the above result to a large volume V containing an innite

number of such innitesimal dierential volumes [Fig. 1.1.13], we note thatwhen integrating the divergence over the volume surfaces shared by adjacentdierential volumes will have no contribution because the surface normalspoint in opposite directions and thus cancel. The result is the divergence orGauss theorem

VdV D =

SdS D (1.1.46)

The divergence theorem states that the volume integral of the divergence ofthe vector eld D is equal to the total outward ux D through the surfaceS enclosing the volume.

Example 1.1.5 Interpretation of Gauss law for electric elds.Applying (1.1.46) to Gauss or Coulombs law for electric elds (1.1.3), we nd

S

dS D =

V

dV D =

V

dV = q (E1.1.5.1)

Thus the divergence of the vector eld D , also called electric ux, out of an enclosedsurface S , is equal to the total charge q in the volume V enclosed by the surface.



V

S

Figure 1.1.13 Derivation of divergence theorem.

Curl of a VectorThe curl of a vector eld H is a vector, dened as

H =(x

x+ y

y+ z

z

)H (1.1.47)

Consider a dierential volume of sides x,y,z centered around a point(x0, y0, z0) . In the Cartesian coordinate system, the curl of a vector H asdened states that

H = limx0y0z0

1xyz

{yz

[x

(H(x0 +

x2, y0, z0)H(x0 x2 , y0, z0)

)]

+ zx[y

(H(x0, y0 +

y2, z0)H(x0, y0 y2 , z0)

)]

+ xy[z

(H(x0, y0, z0 +

z2

)H(x0, y0, z0 z2 ))]}(1.1.48)

The six terms in the above equation are associated with the six dierentialsurfaces bounding (x0, y0, z0) . For the rst term, the surface normal is in thex direction; we write dS = xyz . For the second term dS = xyz .

24 1. Fundamentals

For the third term dS = yzx , etc. Thus we write (1.1.48) as

H = limV0

dS sHV

(1.1.49)

Applying the above result to a large V containing an innite number ofsuch dierential volumes, we nd the curl theorem

VdV H =

SdS sH (1.1.50)

This is similar to the divergence theorem except that now the result is invector form.

Derivation of Boundary Conditions for E and HWhen there is a plane boundary at z = z0 , and the E eld values

are nite above and below the plane boundary, we can derive the boundarycondition for E by using (1.1.48) and letting z go to zero rst [Fig. 1.1.12].We nd

H = limz0

1z

{x

[Hy(x0, y0, z0 +

z2

)Hy(x0, y0, z0 z2 )]

+ y[Hx(x0, y0, z0 +

z2

)Hx(x0, y0, z0 z2 )]}

From (1.1.1), we nd

limz0

[Hy(x0, y0, z0 + z2 ) +Hy(x0, y0, z0

z2

)]

= limz0

z[Dxt

+ Jx

]

limz0

[Hx(x0, y0, z0 +

z2

)Hx(x0, y0, z0 z2 )]

= limz0

z[Dyt

+ Jy

](1.1.51)

On the right hand sides of the above two equations, the time derivativesDx/t and Dy/t are nite but we may assume Jx and Jy to be inniteto create a surface current density Js when z 0 :

Js = limz0J

J z (1.1.52)


Notice that Hy(x0, y0, z0 + z2 ) and Hx(x0, y0, z0 +z2 ) are in region 1, and

Hy(x0, y0, z0 z2 ) and Hx(x0, y0, z0 z2 ) are in region 2, we obtain from(1.1.51)

H1y +H2y = JsxH1x H2x = Jsy (1.1.53)

Since the dierential surface dS = sdS = zxy we can write (1.1.53) invector form as

s (H1 H2) = Js (1.1.54)Thus the discontinuity in the tangential components of H is equal to thesurface current at the boundary surface.

When there is no surface current density at the surface boundary, wehave

H1y = H2yH1x = H2x (1.1.55)

ors (H1 H2) = 0 (1.1.56)

Thus the tangential H components are continuous across the boundarysurface. By the same token, we conclude from (1.1.2) that

E1y = E2yE1x = E2x (1.1.57)

ors (E1 E2) = 0 (1.1.58)

Thus the tangential components of E are continuous across the boundarysurface.

Stokes TheoremThe curl of H is dened in the Cartesian coordinate system as the

vector

H =(x

x+ y

y+ z

z

)H =

x y z

x

y

z

Hx Hy Hz

= x

(

yHz

zHy

)+ y

(

zHx

xHz

)+ z

(

xHy

yHx

)(1.1.59)

26 1. Fundamentals

Writing in dierential form, we have

H = limx0y0z0

1xyz

{x

[xz

(Hz(x0, y0 +

y2, z0)Hz(x0, y0 y2 , z0)

)

xy(Hy(x0, y0, z0 +

z2

)Hy(x0, y0, z0 z2 ))]

+ y[xy

(Hx(x0, y0, z0 +

z2

)Hx(x0, y0, z0 z2 ))

yz(Hz(x0 +

x2, y0, z0)Hz(x0 x2 , y0, z0)

)]

+ z[yz

(Hy(x0 +

x2, y0, z0)Hy(x0 x2 , y0, z0)

)

xz(Hx(x0, y0 +

y2, z0)Hx(x0, y0 y2 , z0)

)]}(1.1.60)

The 12 terms in the above equation are associated with the 6 dierentialsurfaces bounding (x0, y0, z0) .

Applying to open surfaces, the curl theorem becomes the well-knownStokes theorem in vector calculus. We have, for the z component,

(H)z = limx0y0

1xy

{

y[Hy(x0 +

x2, y0, z0)Hy(x0 x2 , y0, z0)

]

x[Hx(x0, y0 +

y2, z0)Hx(x0, y0 y2 , z0)

]}

The rst term in the bracket is equal to the component Hy at x = x0 + x2multiplied by the dierential length y . We dene a vector dierentiallength dl [Fig. 1.1.14] such that for the side y at x = x0 + x2 , dl = ydy ;for the side x at y0 + y2 , dl = xdx ; for the side y at x = x0 x2 ,dl = ydy ; and for the side x at y = y0 y2 , dl = xdx . If we use


z

C

y

x

(x0, y0)

dl = xdx

dl = ydy

Figure 1.1.14 Derivation of z-component of the curl of a vector eld.

the ngers of the right hand to trace the direction of dl along the loop, theright-hand thumb points in the surface normal direction z . Thus

z (H) = limx0y0

1S

Cdl H

where C denotes the contour circulating the area S = xy . Similarresults are derivable for the x and y components of H . Consequentlyfor a dierential area Sj bounded by a contour Cj and with a surfacenormal an , we have Sj = anSj and

Sj (H)j =Cj

dl H

For an open surface S , we can subdivide it into N such as dierential areas[Fig. 1.1.15]. Adding the contributions of all N dierential areas, we have

limSj0N

Nj=1

Sj (H)j =Cdl H

Since the common part of the contours in two adjacent elements is traversedin opposite direction by the two contours, the net contribution of all thecommon parts in the interior sums to zero and only the contribution fromthe external contour C bounding the open surface S remains in the lineintegral on the right-hand side. The left-hand side becomes a surface integral,and the result is Stokes theorem:

28 1. Fundamentals

S

C

C

S

Figure 1.1.15 Derivation of Stokes theorem.

dS (H) =

Cdl H (1.1.61)

Stokes theorem states that the surface integral of the curl of the vector eldH over an open surface S is equal to the closed line integral of the vectoralong the contour enclosing the open surface.

Example 1.1.6 Electromotive force (EMF) and Lenz law.Applying Stokes theorem (1.1.61) to Faradays law (1.1.2), we have

C

dl E = t

A

dS B (E1.1.6.1)

The line integral of E is dened as the electromotive force (EMF):

EMF =C

dl E = t

(E1.1.6.2)

where

=

A

dS B (E1.1.6.3)

is the magnetic ux linking a loop with area A bounded by the closed contour C .Equation (E1.1.6.2) states that the EMF is equal to the negative time derivative ofthe magnetic ux linking the loop. Thus the EMF always produces a ux in the loop


to oppose the direction of change of the ux linking the loop; if is increasing,the EMF decreases the ux, and vice versa. This is known as the Lenz law.

The integration of E along a line segment of l from point a to point b isdened as the voltage drop between a and b :

Vab = ba

dl E (E1.1.6.4)

Note that the EMF has unit of voltage (Volt) and not unit of force. The voltageVab between points a and b is the dierence of potential at the two points. Forpositive Vab , the electric eld vector points from a to b . Thus point a is at a higherpotential a and point b is at a lower potential b < a and Vab = a b .

Consider a closed loop C . According to (E1.1.6.1), the sum of the voltagearound the closed loop is equal to the magnetic ux linking the area formed by theloop. Kirchhos voltage law (KVL) states that the voltage around a closed loop isequal to zero. Thus KVL is correct only when E = 0 . It is important to notethat if there is magnetic eld linking the loop, then KVL is incorrect.


Example 1.1.7 Electromotive force (EMF).

Consider the loop in Fig. E1.1.7.1 consisting of two resistors with resistancesR1 = 2.5 ohm and R2 = 7.5 ohm . Let the magnetic ux linking the loop be in-creasing at the rate of 10 Wb/s. According to (E1.1.6.1), an EMF of 10 V isinduced to counter the increase. The direction of the induced current is as shown soas to produce a magnetic eld in the opposite direction of the increasing magneticeld. The voltage across R1 is V1 = 2.5 V , which can be obtained by taking theclosed loop consisting of the voltmeter and R1 yielding 0 = 2.5 V1 , or by takingthe loop consisting of the voltmeter and R2 which includes the time varying mag-netic eld and yielding 10 = 7.5 + V1 . Likewise, the voltage readings for the othertwo voltmeters are V2 = 7.5 V and V3 = 2.5 V . It is noted that although thevoltmeters for V2 and V3 are connected to the same two nodes, the two readingsare drastically dierent, a clear violation of Kirchhos voltage law (KVL). Thisdemonstrates that KVL is correct only when there is no magnetic ux linking theloop where it is applied, namely E = 0 .

Consider another case as shown in Fig. E1.1.7.2 where the induced counter EMFis 20 V. Following the same analysis, we nd V1 = 5 V , V2 = 15 V , V3 = 5 V ,V4 = 10 V , and V5 = 20 V .

30 1. Fundamentals

V1V2 V3+

7.5V 2.5V

+

+

+

+

R1R2

Figure E1.1.7.1 EMF of the loop is 10 volts.

V1V2 V3

+

15V 5V+

+

+

+

R1R2

V4+

V5+

Figure E1.1.7.2 EMF of the loop is 20 volts.


Example 1.1.8 Derive boundary conditions for E and H with Stokes theorem.Consider a ribbon-like surface as shown in Fig. E1.1.8.1. Integrating over the

surface of the ribbon area, Faradays law and Ampe`res law become

dl E = d

dt

dS s B

dl H = ddt

dS s D +

dS s J

Let the ribbon area approach zero in such a manner that goes to zero rst andthe terms involving are discarded. To relate E1, H1 in region 1 to E2, H2 inregion 2, we proceed as follows.

The integral forms of Faradays law and Ampe`res law as applied to the ribbon


dl

dl

region 2

2

dS = s dS

region 1

n

Figure E1.1.8.1

area in Fig. E1.1.8.1 yield, as 0,

ddt

dS s B = 0

d

dt

dS s D = 0

because ddt s B and ddt s D remain nite while the ribbon area approaches zero.Therefore

dl (E1 E2) = 0dl (H1 H2) = s J dl

The electric eld E in the dl direction is tangential to the surface and can bewritten as dl E = dls n E = dls n E for all dls along the interface andsimilarly for H . We thus have

n (E1 E2) = 0n (H1 H2) = lim

0J Js


Useful Vector IdentitiesSeveral vector identities that are used later on are listed below:

( E) = ( E)2E (1.1.62)

32 1. Fundamentals

(E H) = H ( E) E (H) (1.1.63) (A) = 0 (1.1.64) () = 0 (1.1.65)

The proofs can be carried out in the Cartesian coordinate system or ingeneral vector notation.

Exercise 1.1.2 To prove (1.1.62), we use the vector identity C (A B) =A(C B) (C A)B and identify C and A with and B with E . We can alsouse the denition for curl to show that

( E) =

x y zx

y

z

Ezy

Eyz

Exz

Ezx

Eyx

Exy

= x

(2Eyxy

+2Ezxz

2Exy2

2Exz2

)

y(2Eyx2

+2Eyz2

2Ex

xy

2Ezyz

)

+ z(2Exxz

+2Eyzy

2Ez2x

2Ezy2

)

= x(

x( E)2Ex

)+ y

(

y( E)2Ey

)

+ z(

z( E)2Ez

)= ( E)2E

where

2 = = 2

x2+

2

y2+

2

z2

is called the Laplacian operator, and in the x component we added and subtractedthe term

2Exx2

and similarly for the y and z components.

To prove (1.1.63), we write

(E H)=

x(EyHz EzHy) +

y(EzHx ExHz) +

z(ExHy EyHx)


= Ey

xHz +Hz

xEy Ez

xHy Hy

xEz

+ Ez

yHx +Hx

yEz Ex

yHz Hz

yEx

+ Ex

zHy +Hy

zEx Ey

zHx Hx

zEy

= Hx

(

yEz

zEy

)+Hy

(

zEx

xEz

)+Hz

(

xEy

yEx

)

Ex(

yHz

zHy

) Ey

(

zHx

xHz

) Ez

(

xHy

yHx

)= H ( E) E (H)

To prove (1.1.64), we write

(A) = x

(

yAz

zAy

)+

y

(

zAx

xAz

)

+

z

(

xAy

yAx

)= 0

To prove (1.1.65), we nd

() =

x y zx

y

z

x

y

z

= 0

Example 1.1.9 Poisson equation and Laplace equation.In (E1.1.4.1), we wrote the electric eld vector as the gradient of a potential

function :

E = (E1.1.9.1)

By virtue of (1.1.65), we see that E = 0 . Thus the above denition for theelectric eld is true only when the term B/t in Faradays law can be neglected,i.e., when there is no time variation. We may refer to the above electric eld as thestatic electric eld.

The Coulomb law (or Gauss law for electricity) in free space is E = /o .In terms of the potential function, we have

2 = /o (E1.1.9.2)

34 1. Fundamentals

This is known as the Poisson equation. In places where there is no charge density, = 0 , we have

2 = 0 (E1.1.9.3)This is known as the Laplace equation.


Exercise 1.1.3 Show that the potential function

=C

x2 + y2 + z2

where C is a constant, satises the Laplace equation (E1.1.9.3).

Example 1.1.10 Static electric eld vector as gradient of a potential function.With the potential function = C/

x2 + y2 + z2 , the associated static elec-

tric eld vector is

E = = C[x

x

(x2 + y2 + z2)3/2+ y

y

(x2 + y2 + z2)3/2+ z

z

(x2 + y2 + z2)3/2

]

In terms of the position vector r = xx + yy + zz , and the length of the positionvector r =

x2 + y2 + z2 , the electric eld vector takes the form

E = rC

r3= r

C

r2

where r is pointing in the direction of r with unit length. Thus the electric eldvector is pointing away from the origin along the direction of the position vectorr , and its magnitude decreases as the squared inverse distance r .


Index NotationA vector in the Cartesian coordinate system can be represented by its

three components. Thus Aj with j = 1, 2, 3 represents A1, A2, A3 of thevector A . The dot product A B is written as AjBj where the repeatedindex j implies summation over j from 1 to 3:

AjBj =3

j=1

AjBj = A1B1 +A2B2 +A3B3


To express cross products in index notation we need to dene a Levi-Cevitasymbol ijk where i, j, k take values from 1 to 3. When any of the twoindices are equal the Levi-Cevita symbol is zero. Otherwise, ijk is either+1 or 1 . It is +1 if ijk is an even permutation of 1,2,3; 1 if ijk is anodd permutation of 1,2,3. Thus 123 = 231 = 312 = 1 and 213 = 132 =321 = 1 and all others equal to 0. Let C = AB . In index notation, wewrite Ci = ijkAjBk . Thus, C1 = 123A2B3 + 132A3B2 = A2B3 A3B2 .The dyad AB is then AjBk , no summation implied because no index isrepeated. The identities (1.1.32) and (1.1.33) are

CiijkAjBk = AjjkiBkCi = BkkijCiAj

and

ijkCjklmAlBm = (ijkklm)CjAlBm = (iljm imlj)CjAlBm= AiCmBm ClAlBi

where ij = 1 when i = j and ij = 0 when i = j .In index notation, divergence of Dj , Dj , is jDj .In index notation, is represented by i and by i .In index notation, curl of Hi , Hi , is written as ijkjHk.The identities (1.1.62)(1.1.65) are, in index notation

ijkjklmlEm = (iljm imjl)jlEm= miEm jjEi

i(ijkEjHk) = ijkHkiEj + ijkEjiHk= HkkijiEj EjjikiHk

iijkjAk = jikijAk = 0ijkjk = ikjjk = ikjkj = 0

Cylindrical and Spherical Coordinate SystemsAlthough we have emphasized the use of the rectangular Cartesian co-

ordinate system, expressions written in vector notation are not coordinatedependent. In addition to the coordinates with unit vectors x, y, z , the cylin-drical coordinate system with unit vectors , , z , and the spherical coordi-nate system with unit vectors , , are often used in this book.

36 1. Fundamentals

x

y

dz

ddzddz

dd

d

z

Figure 1.1.16 Cylindrical coordinate system.

In the cylindrical coordinate system [Fig. 1.1.16], the vector dierentiallength is

dl = d+ d+ zdz

the dierential area is

dS = ddz + ddz + zdd

and the dierential volume is

dV = dddz

In the spherical coordinate system [Fig. 1.1.17], the vector dierentiallength is

dl = rdr + rd + r sin d

the dierential area is

dS = rr2 sin dd+ r sin drd+ rdrd

the dierential volume is

dV = r2 sin drdd


r

d

r sin x

y

z

d

dr

Figure 1.1.17 Spherical coordinate system.

In a general orthogonal coordinate system, we use ui (i = 1, 2, 3) todenote the three basis vectors, dli = hidui to denote a dierential length,where hi is called a metric coecient. The basis vectors are perpendicularto one another ui uj = 0 for i = j but they are not necessarily of unitlength or even with the dimension of length. In Table 1.1.1 we summarize thebasis vectors and the metric coecients for the rectangular (or Cartesian),cylindrical, and spherical coordinate systems.

OrthogonalCoordinate System

RectangularCoordinates

(x, y, z)

CylindricalCoordinates

(, , z)

SphericalCoordinates

(r, , )Base Vectors(u1, u2, u3) x, y, z , , z r, ,

MetricCoecients(h1, h2, h3) 1, 1, 1 1, , 1 1, r, r sin

Dierential Volume(h1h2h3du1du2du3) dxdydz dddz r

2 sin drdd

Table 1.1.1 Orthogonal coordinate systems.

In terms of the general orthogonal coordinate system, the gradient, the

38 1. Fundamentals

divergence, and the curl are dened as

= u1 h1u1

+ u2

h2u2+ u3

h3u3

D = 1h1h2h3

[

u1(h2h3D1) +

u2(h3h1D2) +

u3(h1h2D3)

]

H = 1h1h2h3

h1u1 h2u2 h3u3u1

u2

u3

h1H1 h2H2 h3H3

The Laplacian operator is

2 = = 1h1h2h3

[

u1h2h3

h1u1

+

u2h3h1

h2u2+

u3h1h2

h3u3

]

Identifying the metrics h1, h2, h3 with those as listed in Table 1.1.1, wereadily obtain the expressions in cylindrical and spherical coordinates.

Example 1.1.11 Static electric eld due to a charged particle.In the spherical coordinate system, the static electric eld associated with the

potential function = C/r is

E = = r r

= rC

r2

The corresponding electric displacement eld vector is D = oE . Applying theCoulomb law D = to the divergence theorem (1.1.46)

V

dV D =

S

dS D (E1.1.11.1)

we nd V

dV =

S

dS D (E1.1.11.2)

Assuming the electric eld is due to a charged particle q situated at the origin,we can integrate over a small spherical volume with radius r = surroundingthe origin. The left hand side, integrating over the charge density over the volumecontaining the charged particle, gives rise to the total charge q . In the spherical


coordinate system, the surface integration element at a small distance r = isrr2 sin dd . We thus have

q =

S

dS D =

0

20

dd2 sin

oC

2= 4oC

Thus we have determined the constant C = q/4o and obtained the static electriceld

E = rq

4or2(E1.1.11.3)

due to a charged particle situated at the origin.End of Example 1.1.11

Problems

P1.1.1

Maxwells equations were originally written in the form of scalar partial dier-ential equations. For all eld components, write (1.1.1), (1.1.2), (1.1.3), (1.1.4), and(1.1.5) in terms of partial derivatives. Derive the continuity equation from (1.1.1)to (1.1.4). Prove that given the continuity law, Coulombs law can be derived fromAmpe`res law. Likewise, show that Gauss law can be derived from Faradays law,and that (1.1.3) and (1.1.4) are not independent scalar equations.

P1.1.2

An electromagnetic wave has spatial frequency ko = 100 Ko. Determine thewavelength in meters and the temporal frequency in GHz.

Determine the spatial frequency in units of Ko for a laser light at wavelength = 0.6328m .

Determine the spatial frequency in units of Ko for a microwave oven at fre-quency 2.4 GHz.

P1.1.3

Electromagnetic waves satisfy all of the Maxwell equations. Consider, in freespace, the following electric eld vectors:

E1 = x cos(t kz)E2 = z cos(t kz)E3 = (x+ z) cos

(t+ k|x z|/

2)

E4 = (x+ z) cos(t+ k|x+ z|/

2)

E5 = (x+ z) cos(t+ ky)

40 1. Fundamentals

Do these electric eld vectors satisfy the wave equation and all Maxwell equations?Which of the ve elds qualify as electromagnetic waves? For those not qualiedas electromagnetic waves, you should state which of the Maxwell equations areviolated.

P1.1.4

The known spectrum of electromagnetic waves covers a wide range of frequen-cies. Electromagnetic phenomena are all described by Maxwells equations and, byconvention, are generally classied according to wavelengths or frequencies. Radiowaves, television signals, radar beams, visible light, X rays, and gamma rays areexamples of electromagnetic waves.

(a) Give in meters the wavelengths corresponding to the following frequencies:(i) 60 Hz(ii) AM radio (5351605 kHz)(iii) FM radio (88108 MHz)(iv) L- band (12 GHz)(v) Visible light ( 1014 Hz)(vi) X-rays ( 1018 Hz)

(b) Give in Hertz the temporal frequencies corresponding to the following wave-lengths:(i) 1 km, (ii) 1 m, (iii) 1 mm, (iv) 1 m , (v) 1 A.

(c) Give in unit Ko the spatial frequencies corresponding to the wavelengths in(b).

P1.1.5

Prove the following identities: ( E) = ( E)2E (E H) = H ( E) E (H) (A) = 0 () = 0

P1.1.6

Three vectors A, B , and C drawn in a head-to-tail fashion form the threesides of a triangle. What is A+B + C and what is A+B C ?P1.1.7

A position vector r = x

2+ y+ z2 . Determine its spherical components r, , and its cylindrical components , , z .

P1.1.8

Find a unit vector c that is perpendicular to both A = x4 + y5 z3 andB = x2 y7 z1.5 .P1.1.9

For the vector A = 2 + z2z , verify the divergence theorem for the circularcylindrical region enclosed by = 5, z = 0 , and z = 3 .


P1.1.10Derive the boundary conditions for E by applying the curl theorem to a small

pill-box volume on the x-y plane which has an area A and an innitesimal thicknessz .

P1.1.11What is the result if the surface integral of H is carried out over a closed

surface? Compare with the curl theorem we obtained for the curl integrated over avolume V enclosed by a surface S in (1.1.50).

42 1. Fundamentals

1.2 Polarization

The polarization of a wave is conventionally dened by the time variation ofthe tip of the electric eld vector E at a xed point in space. If the tip movesalong a straight line, the wave is linearly polarized. When the locus of thetip is a circle, the wave is circularly polarized. For an elliptically polarizedwave, the tip of E describes an ellipse. If the right-hand thumb points inthe direction of propagation while the ngers point in the direction of the tipmotion, the wave is dened as right-hand polarized. The wave is left-handpolarized when it is described by the left-hand thumb and ngers.

Consider the following wave solution:

E(z, t) = xEx + yEy= x cos(kz t) + yA cos(kz t+ ) (1.2.1)

The wave propagates in the +z direction. From the temporal view point,

E(t) = x cos(t) + yA cos(t )

We now study polarization for the following special cases:Case 1) = 2m , where m = 0, 1, 2, ... is an integer. We have

E(t) = x cos(t) + yA cos(t)

The tip of the electric eld vector moves along a line as shown in Fig. 1.2.1a.The wave is linearly polarized.Case 2) = (2m+ 1) , we have

E(t) = x cos(t) yA cos(t)

The tip of the electric eld vector moves along a line as shown in Fig. 1.2.1b.The wave is linearly polarized.Case 3) = /2 and A = 1 , we have

E(t) = h cos(t) + y sin(t) (1.2.2)

It can be seen that while the x component is at its maximum the y com-ponent is zero. As time progresses, the y component increases and the xcomponent decreases. The tip of E rotates from the positive Ex axis to thepositive Ey axis [Fig. 1.2.1c]. Elimination of t from the x and y compo-nents in (1.2.2) yields a circle of radius 1 , E2x + E

2y = 1 . Thus the wave is

right-hand circularly polarized.

1.2 Polarization 43

Ey

Ey

Ey

Ey

Ey

Ex

Ex

Ex

Ex

Ex

a. Linear polarization b. Linear polarization

c. Right-handcircular polarization

d. Left-handcircular polarization

e. Right-handelliptical polarization

f. Left-handelliptical polarization

AA

Ey

Ex

AA

Figure 1.2.1 Polarizations.

44 1. Fundamentals

Case 4) = /2 and A = 1 , we have

E(t) = x cos(t) y sin(t) (1.2.3)

As time progresses, the y component increases and the x component de-creases. The tip of E rotates from the positive Ex axis to the negative Eyaxis. Thus the wave is left-hand circularly polarized [Fig. 1.2.1d].Case 5) = /2 , we have

E(t) = x cos(t) yA sin(t) (1.2.4)

The wave is right-hand elliptically polarized for = /2 [Fig. 1.2.1e] andleft-hand elliptically polarized for = /2 [Fig. 1.2.1f].

A

12 1 22

Figure 1.2.2 Polarizations for various values of and A.

The above discussion can be summarized in Fig. 1.2.2 where we illustratethe polarization for dierent values of A and . On the horizontal axis, = 0, or , the wave is linearly polarized. If A = 1 and = /2, thewave is right-hand circularly polarized. For A = 1 and = /2, the waveis left-hand circularly polarized. Otherwise the wave is elliptically polarized.The polarization is right-handed if the phase dierence is between zero and, and left-handed if is between and 2.

1.2 Polarization 45

Example 1.2.1 Polarization from the spatial view point.Wave polarization can be viewed by either taking a series of still pictures at

several xed times, called the spatial view point or by making observations at a xedpoint in space, called the temporal view point. The denition of polarization so farhas been discussed from the temporal view point. Let us now look at polarizationfrom the spatial view point.

x

y

z

z = zo E|t=to

E|t=t+k

Figure E1.2.1.1 Spatial view of polarization.

Consider the right-hand circularly polarized wave with = /2 and A = 1in case 3), setting t = 0 in wave solution (1.2.1), we have

E(z = 0, t) = x cos(kz) y sin(kz) = xEx(z) yEy(z)This is a left-handed helix as shown below.

Ex = E0 cos(2z)

Ey = E0 sin(2z)

The parametric equation of a helix is

x = R cos(

2pz

)y = R sin

(2pz

)r

where p is the pitch of the helix. Thus the locus of the tip point of the electric eldvector measured along the z axis is a left-handed helix with the pitch p = . Thehelix advances along +z without rotating. At z = z0 = 3/4 , electric eld vectoris at E|t=to when to = 0 , it is shown as E|t=t+ when t+ = /4 .


46 1. Fundamentals

Topic 1.2A Stokes Parameters and Poincare Sphere

To facilitate a mathematical discussion of polarization, we decompose theE vector of a wave into two components perpendicular to the direction ofpropagation. For a specic point in space, we write

E(t) = hEh + vEv = heh cos(t h) + vev cos(t v) (1.2A.1)where h , v , and the direction of propagation are mutually perpendicularand thus form an orthogonal system. We assume the amplitudes ev and ehare both positive. The locus of the tip E(t) is determined by eliminatingthe time t dependence between the two components Eh and Ev .

When h and v dier by an integral multiple of 2 , vh = 2n ,the two components are in phase. We have Eh/eh = Ev/ev . The wave is lin-early polarized, and the straight-line locus traverses the rst and third quad-rants [Fig. 1.2.1a]. When h and v dier by an odd integral multiple of ,the two components are 180 out of phase. We have Eh/eh = Ev/ev , andthe straight-line locus traverses the second and fourth quadrants [Fig. 1.2.1b].When the magnitudes of the two components are equal, eh = ev = e0 , andv h = /2 , the wave is right-hand circularly polarized [Fig. 1.2.1c].When eh = ev = e0 , and v h = /2 , the wave is left-hand circularlypolarized [Fig. 1.2.1d].

Eh

Ev

e1

e2

Figure 1.2A.1 Elliptical polarization.

In general, a polarized wave has elliptical polarization; that is, whentime is eliminated from the two components of E , the resultant equationdescribes an ellipse. Consider the case h = 0, vh = /2 in (1.2A.1)and let eh = e1 > e2 = ev. We have

E(t) = he1 cos(t 0) ve2 sin(t 0) (1.2A.2)

1.2 Polarization 47

Elimination of time yields an ellipse

(Ehe1

)2+

(Eve2

)2= 1

We see that e1 is the major axis of the ellipse and e2 the minor axis. Withthe plus sign in (1.2A.2), we have a right-hand elliptically polarized waveand with the minus sign a left-hand elliptically polarized wave. The shape ofthe ellipse can be specied by an ellipticity angle . Referring to (1.2A.2),with e1 denoting the major axis and e2 the minor axis, we write

tan = e2e1

(1.2A.3)

where the plus sign corresponds to right-hand polarization for which 0 /4 and the negative sign to left-hand polarization for which /4 0.

Eh

Ev

e1

e2

Figure 1.2A.2 Elliptical polarization.

The general polarization states are more popularly described with thePoincare sphere as discussed below. Consider the elliptical polarization asgiven by (1.2A.1), which describes a tilted ellipse as plotted in Fig. 1.2A.2.The major axis of the ellipse described in (1.2A.2) is rotated and makes theangle with the Eh axis with 0 . We call the orientation angle.

48 1. Fundamentals

Example 1.2A.1 Coordinate transformation.Consider two rectangular coordinate systems represented by unit vectors (x, y)

and (x, y) . In terms of the unprimed coordinates, the unit vectors for the primedcoordinate system are represented as

x = x cos + y sin

y = x sin + y cosA position vector r can be represented as r = xx + yy = xx + yy. The trans-formation between coordinates (x, y) and (x, y) is obtained as follows:

x = x r = (x x)x+ (x y)y = x cos + y siny = y r = (y x)x+ (y y)y = x sin + y cos

End of Example 1.2A.1

In view of (1.2A.2) and Fig. 1.2A.2, we have from coordinate transfor-mation

e1 cos(t 0) = Eh cos + Ev sin (1.2A.4a)e2 sin(t 0) = Eh sin + Ev cos (1.2A.4b)

Substituting the components Eh and Ev of (1.2A.1) in (1.2A.4) and com-paring the coecients of cost and sint , we obtain

e1 cos0 = eh cosh cos + ev cosv sin (1.2A.5a)e1 sin0 = eh sinh cos + ev sinv sin (1.2A.5b)e2 cos0 = eh sinh sin + ev sinv cos (1.2A.5c)e2 sin0 = eh cosh sin ev cosv cos (1.2A.5d)

Eliminating 0 from (1.2A.5a) and (1.2A.5b) by squaring and adding, wend

e21 = e2h cos

2 + e2v sin2 + ehev sin 2 cos (1.2A.6a)

Similarly from (1.2A.5c) and (1.2A.5d) , we have

e22 = e2h sin

2 + e2v cos2 ehev sin 2 cos (1.2A.6b)

Multiplying (1.2A.5a) by (1.2A.5c) , (1.2A.5b) by (1.2A.5d) and adding,we again eliminate 0 and obtain

e1e2 = ehev sin (1.2A.6c)

1.2 Polarization 49

Finally we multiply (1.2A.5a) by (1.2A.5d) and subtract from (1.2A.5b)multiplied by (1.2A.5c) , which yields

2ehev cos = (e2h e2v) tan 2 (1.2A.6d)

Equation (1.2A.6) will be used in the following discussion on Stokes param-eters and the Poincare sphere.

To facilitate the discussion of polarization states of electromagneticwaves, the four Stokes parameters pertaining to E(t) given in (1.2A.1) aredened as follows :

I =1

(e2h + e

2v

)(1.2A.7a)

Q =1

(e2h e2v

)(1.2A.7b)

U =2ehev cos (1.2A.7c)

V =2ehev sin (1.2A.7d)

Notice that I2 = Q2 + U2 + V 2.Adding (1.2A.6a) and (1.2A.6b) yields e21+e

22 = e

2h+e

2v = I . Making

use of (1.2A.3), we havee21 = I cos

2 (1.2A.8)

Subtracting (1.2A.6b) from (1.2A.6a) and making use of (1.2A.6d) , wend e21 e22 = (e2h e2v)/ cos 2 . Making use of (1.2A.3) and (1.2A.8) , wend

Q =1(e2h e2v) = I cos 2 cos 2 (1.2A.9a)

In terms of I , we nd from (1.2A.7c) , (1.2A.6d) and (1.2A.9a)

U = I cos 2 sin 2 (1.2A.9b)

and from (1.2A.7d) , (1.2A.6c) and (1.2A.8)

V = I sin 2 (1.2A.9c)

Again we see from (1.2A.9) that I2 = Q2 + U2 + V 2.Equation (1.2A.9) suggests a simple geometrical representation of all

states of polarization by recognizing that Q, U, and V can be regarded as

50 1. Fundamentals

Q

V

U

Figure 1.2A.3 Poincare sphere.

the rectangular components of a point on a sphere with radius I [Fig. 1.2A.3].We dene, in the spherical coordinate system, = /2 2 and = 2.As seen from (1.2A.3), positive is for right-hand polarization which isrepresented by points on the upper hemisphere. On the lower hemisphere, thepoints correspond to left-hand polarization. The north pole represents right-hand circular polarization and the south pole represents left-hand circularpolarization. The sphere is called the Poincare sphere. Fig. 1.2.2 is seen tobe a planar projection of the Poincare sphere with the plane and the spheretouching each other at Q = I. The equator is mapped into the horizontalaxis.

For polarized waves,

I = Ih + IvQ = Ih Iv = I cos 2 cos 2U = I cos 2 sin 2V = I sin 2

When the wave is right-hand circularly polarized Q = U = 0 , V = I , as = /4 . When the wave is left-hand circularly polarized, Q = U = 0 ,V = I , as = /4 . When the wave is linearly polarized, V = 0 , as = 0 .

1.2 Polarization 51

Partial PolarizationRadiation from many natural and man-made sources consists of eld

components that uctuate with time. We write

Eh = eh(t) cos(t h(t)

)Ev = ev(t) cos

(t v(t)

)

The wave is quasi-monochromatic when eh(t), ev(t), h(t), and v(t) areslowly varying compared with cost. The Stokes parameters are denedby a time-average procedure over a large time interval T , denoted with thebrackets :

=1T

T0

dt [Eh(t)]2

The Stokes parameters are

I = Ih + Iv =1

( +

)Q = Ih Iv = 1

(

)= I

U =2= I

V =2= I

For completely unpolarized waves, Eh and Ev are uncorrelated and we haveI = total Poynting power and Q = U = V = 0. For completely polarizedwaves we have I2 = Q2 + U2 + V 2. For partially polarized waves it can beshown that I2 Q2 +U2 +V 2 [Example 1.2A.2]. With the Poincare sphereof radius I , the partially polarized waves correspond to points inside thesphere.

In concluding this section on wave polarization, we remark that thepolarization is dened according to the time variations of the E vector. Aswe shall see in Chapter 3, it is imperative that we dene polarization interms of D when anisotropic and bianisotropic media are involved. This isbecause in isotropic media E is perpendicular to k , k E = 0, while innon-isotropic media k D = 0. This also suggests that wave polarization canbe dened in terms of the eld vector B .

52 1. Fundamentals

Example 1.2A.2

(a) Assume ev(t), eh(t) and (t) remain constant for a fractional time interval tnand let T be subdivided into t1, t2, . . . tN . The time-averages can be writtenin a summation form such that

Ih =1T

n

tne2hn

Iv =1T

n

tne2vn

U =2T

n

tne2vnAn cosn

V =2T

n

tne2vnAn sinn

where

An =ehnevn

denotes the ratio of eh and ev in the time interval tn . Assume that both ehand ev are positive.Show that

4IvIh U2 + V 2

or

I2 = (Ih + Iv)2 Q2 + U2 + V 2

The equal sign holds when n = m and An = Am for all n and m ,which means that the amplitude ratio and phase dierence of Ev and Ehstay constant. This is the case for elliptical polarization.

(b) For polarized waves

I = Ih + IvQ = Ih Iv = I cos 2 cos 2U = I cos 2 sin 2

V = I sin 2

Show that when the wave is right-handed circularly polarized Q = U = 0 andV = I, when it is left-hand circularly polarized, Q = U = 0 and V = I,and when the wave is linearly polarized, V = 0.

1.2 Polarization 53

Answer:

(a)

4IhIv =4

2T 2

[n

m

tntmA2ne

2vne

2vm

]

=4

2T 2

[n

t2nA2ne

4vn +

n =m

tntmA2ne

2vne

2vm

]

=4

2T 2

[n

t2nA2ne

4vn +

n>m

tntm(A2n +A2m)e

2vne

2vm

]

U2 =4

2T 2

[n

m

tntmAnAme2vne

2vm cosn cosm

]

=4

2T 2

[n

t2nA2ne

4vn cos

2 n + 2n>m

tntmAnAme2vne

2vm cosn cosm

]

V 2 =4

2T 2

[n

m

tntmAnAme2vne

2vm sinn sinm

]

=4

2T 2

[n

t2nA2ne

4vn sin

2 n + 2n>m

tntmAnAme2vne

2vm sinn sinm

]

Thus

U2 + V 2 =4

2T 2

[n

t2nA2ne

4vn + 2

n>m

tntme2vne

2vmAnAm cos(n m)

]

We nd

4IvIh (U2 + V 2)=

42T 2

n>m

tntme2vne

2vm[A

2m +A

2n 2AmAn cos(n m)]

42T 2

n>m

tntme2vne

2vm[A

2m +A

2n 2AmAn]

=4

2T 2

n>m

tntme2vne

2vm(Am An)2

The right-hand side is always non-negative. Hence 4IvIh U2 + V 2 .

54 1. Fundamentals

(b) For a right-handed circularly polarized wave = /4 , then

Q = I cos(2/4) cos(2) = 0U = I cos(2/4) sin(2) = 0V = I sin(2/4) = I

For a left-handed circularly polarized wave = /4 , then

Q = I cos(2/4) cos(2) = 0U = I cos(2/4) sin(2) = 0V = I sin(2/4) = I

For linearly polarized wave = 0 , then

V = I sin 0 = 0

End of Example 1.2A.2

Problems

P1.2.1Consider an electromagnetic wave propagating in the z-direction with

E = xex cos(kz t+ x) + yey cos(kz t+ y)

where ex , ey , x , and y are all real numbers.(a) Let ex = 2, ey = 1, x = /2, y = /4. What is the polarization?(b) Let ex = 1, ey = x = 0. This is a linearly polarized wave. Prove that it can

be expressed as the superposition of a right-hand circularly polarized wave anda left-hand circularly polarized wave.

(c) Let ex = 1, x = /4, y = /4, ey = 1. This is a circularly polarized wave.Prove that it can be decomposed into two linearly polarized waves.

P1.2.2From E(t) = hEh + vEv = heh cos(t h) + vev cos(t v) show that

Eheh

sinv Evev

sinh = cost sin

Eheh

cosv Evev

cosh = sint sin

Eliminating the time dependence t to obtain the equation

(Eheh

)2+

(Evev

)2 2EhEv

ehevcos = sin2

1.2 Polarization 55

show that this is a tilted ellipse.

P1.2.3Wave polarization can be viewed by either taking a series of still pictures at

several xed times, called the spatial view point or by making observations at axed point in space, called the temporal view point. We dene polarization fromthe temporal view point. Let us now look at polarization from the spatial viewpoint.

Consider an electromagnetic wave with k = 100 Ko propagating in the zdirection.

E(r, t) = E0[x cos(kz t) y sin(kz t)]

What are the wavelength and the polarization of this wave?From the spatial point of view, by taking a picture at t = 0 , the tips of the

electric eld vectors form a helix. Is the helix right-handed or left-handed? What isthe pitch of this helix?

Observing at a xed point in space, show that the tip of the electric eld de-scribes the same polarization as in the temporal view point when the helix advanceswithout turning.

P1.2.4Sun navigation was rst observed in 1911. It was found that some species of

ants, horseshoe crabs, honeybees, etc., are sensitive to polarized light. These crea-tures can navigate as long as there is a small patch of blue sky. The sky polarizationdepends upon the angle between the suns rays to a particular point in the skyand an observers line of sight to the same point. The sunlight, which is unpolarized,or randomly polarized, excites air molecules which behave like small dipole antennaswhen irradiated by the incident electric elds of the sunlight. The scattered electriceld Es for each excited dipole antenna is linearly polarized in planes perpendic-ular to the sunlight path; and looking along the sun ray path the scattered wave isunpolarized, or randomly polarized.

At sunset, if an ant looks directly at the sun ( = 0 ), what is the polarization?What is the polarization if the ant looks at the zenith ( = 90 ) perpendicular tothe sun ray path? Show that the sky light appears to be partially linearly polarizedwhen it looks at other parts of the sky [Scientic American, July 1955].

56 1. Fundamentals

1.3 Lorentz Force Law

The interaction of the electric and magnetic elds with the current andcharge densities are governed by the Lorentz force law

f = E + J B (1.3.1)where f is the force density. The Lorentz force law relates electromagnetismto mechanics. The manifestation of the electric eld vector E and the mag-netic eld vector B can be demonstrated with the forces exerted on thecharge density and the current density J . It can thus be used to denethe elds E and B .

Example 1.3.1 Coulombs law.For static electric elds in the absence of magnetic elds, the Lorentz force law

becomes f = E. Acting on a charged particle q , the total force is F = qE .Assuming the electric eld E is generated by another charged particle Q situatedat the origin, we have from (E1.1.11.3)

E = rQ

4or2

Thus the total force acting on the charged particle q is

F = rqQ

4or2

which is proportional to the squared inverse distance. This is known as Coulombslaw.


Example 1.3.2 Cyclotron frequency.Consider a particle with charge q and mass m moving with velocity v in a

uniform static magnetic eld in the z direction, B = zB0 . In the absence ofelectric elds, if the velocity v has no component in the z direction, the Lorentzforce is perpendicular to the direction of the velocity and the charge particle movesin the x-y plane. Let v = xvx + yvy , we have

F = qv B = xqvyB0 + yqvxB0Equating to Newtons law

F = mdv

dt= x

mdvxdt

+ ymdvydt

1.3 Lorentz Force Law 57

we nddvxdt

= cvy (E1.3.2.1a)dvydt

= cvx (E1.3.2.1b)

wherec =

qB0m

(E1.3.2.2)

is called the cyclotron frequency, which is proportional to the magnitude of themagnetic eld and is independent of the velocity of the particle.

x

y

v

Figure E1.3.2.1 Cyclotron frequency.

The solution to (E1.3.2.1) can be written as

vx =dx

dt= v cosct (E1.3.2.3a)

vy =dy

dt= v sinct (E1.3.2.3b)

To nd the trajectory of the particle, we write the solution of (E1.3.2.3) as

x =v

csinct (E1.3.2.4a)

y = vc

cosct (E1.3.2.4b)

The trajectory of the particle is thus a circle with radius

R = (x2 + y2)1/2 =v

c=

mv

qB0(E1.3.2.5)

in view of (E1.3.2.2). It is seen that the larger the magnetic eld, the smaller theradius. If the charged particle has a velocity component in the z direction, thetrajectory of the particle will follow a helical path.


58 1. Fundamentals

Example 1.3.3 Cyclotron.A cyclotron [Fig. E1.3.3.1] is an accelerator for charged particles. The a.c.

source provides an alternating voltages at the cyclotron frequency and a chargedparticle is repeatedly accelerated every time it passes through the voltage drop.

Uniform B eld

a.c. source

Figure E1.3.3.1 Cyclotron.


Example 1.3.4 Isotope separation.To separate the isotope Uranium 235 from Uranium 238, the isotopes are rst

vaporized and then ionized by electric discharge. Accelerated through a voltage dropV , they acquire a kinetic energy qV = mv2/2 . Passing through [Fig. E1.3.4.1] auniform magnetic eld, the isotopes move along circular paths of dierent radii.

Uniform B eld

m235m238 +

V

Figure E1.3.4.1 Isotope separation.

R235R238

=m235v235m238v238

=m235m238

m238m235

=

m235m238

Thus Uranium 235 can be obtained in a collector with a smaller radius.End of Example 1.3.4


Example 1.3.5The two rods attract each other when their currents are in the same direction

and are repulsive when their currents are in the opposite directions.

I1 I2

FF

I1 I2

F F

Figure E1.3.5.1 Attractive and repulsive forces.


Example 1.3.6 Linear motor.In Fig. E1.3.6.1, we show a sliding bar with length l moving perpendicular to

a DC magnetic eld B = zB0 in the z direction. According to the Lorentz forcelaw, a force

Fm = yIl zB0 = xIlB0is produced that moves the sliding bar in the x direction.

sliding bar

Fm

x

y

V

Figure E1.3.6.1 Linear motor.

If a force is applied to move the sliding bar with velocity v = x , an inducedvoltage V = vlB0 will be generated across the resistor.


60 1. Fundamentals

Example 1.3.7 Magnetic moment and magnetic torque.A rectangular loop [Figure E1.3.7.1] carrying a static current I is placed in

a static magnetic eld B = xB0 . The magnetic moment of the current loop isM = mM . Its direction m follows from the right-hand rule: with the ngerspointing in the direction of the current, the thumb of the right hand is pointing inthe direction of m . Its magnitude M is equal to the area of the loop A times thecurrent I , M = AI . If the rectangular loop has lengths lb and ly , the area of theloop is A = lbly .

B = x

x

l

ly

B0

I

x

F

F

B

M

z

x

Figure E1.3.7.1 Torque on a loop.

The loop is on the x-z plane with two sides aligned with the x-axis and twosides aligned with the y-axis. Since the static magnetic eld is in the x direction,there is no force acting on the two sides with length lb aligned with the x-axis.The forces acting on the two sides with length ly aligned with the y axis are inthe positive and negative y directions. Thus the loop is rotating around the y-axis following the right-hand rule; with the ngers pointing in the direction of therotation, the thumb of the right hand is pointing in the y direction.

The torque acting on the loop is calculated as

T =12lbx (y xIlzB0) 12 lbx (y xIlzB0) = yIAB0

For the current conguration, M = zIA and B = xB0 . In general, the magnetictorque is

T = M B (E1.3.7.1)

Thus there is no torque acting on the component of M in the direction of themagnetic eld.



Example 1.3.8A simple DC motor [Fig. E1.3.8.1] consists of a loop of area A with N turns,

called an armature, which is immersed in a uniform magnetic eld, either producedby a permanent magnet or an electromagnet. The armature is connected to a com-mutator which is a divided slip ring. A DC current I is supplied through a pair ofbrushes resting against the commutator such that the torque

T = NBoIA sinproduced by the current on the armature always acts in the same direction.

N S

Bo

Brush

Commutator

Armature+

I

B =

F

F

B

M

z

x

Figure E1.3.8.1a DC motor.

N S

Brush

Figure E1.3.8.1b Side view of a DC motor.


62 1. Fundamentals

A. Lenz Law and Electromotive Force (EMF)

We apply Stokes theorem to Faradays law and dene the line integral of Eas the electromotive force (EMF):

EMF =Cd l E =

t

dS B

= t

(1.3.2)

where =

AdS B (1.3.3)

is the magnetic ux linking a loop with area A bounded by a closed contourC [Fig. 1.3.1]. Equation (1.3.2) states that the EMF is equal to the negativetime derivative of the magnetic ux linking the loop. Thus the EMF alwaysproduces a ux in the loop to oppose the direction of change of the uxlinking the loop; if is increasing, the EMF decreases the ux, and viceversa. This is known as Lenz law.

+

V

C

Figure 1.3.1 Flux linking a loop.

Notice that the EMF has unit of voltage (Volt) and not unit of force. Thevoltage drop across the loop V is equal to the negative of the induced EMF.

V = EMF = ddt

(1.3.4)

Thus in the presence of a time varying magnetic eld linking a loop, a voltageis generated to oppose the time change of the magnetic eld. The voltagegenerated across the loop V is equal to the negative of the induced EMF.


The voltage Vab is dened as the integration of E along a line segmentof ? from point a to point b .

Vab = bad? E (1.3.5)

Thus Vab is the potential dierence between points a and b . For positiveVab , the electric eld vector points from a to b . Point a is at a higherpotential a than b at point b , b < a and Vab = a b .

LeChateliers Principleis the chemists version of Lenz law, which states thatwhen an external stress (pressure, concentration, or temperature change) isapplied to a chemical system that is in a state of equilibrium, the system willautomatically respond so as to undo the stress applied externally. In Physics,this same phenomenon is embodied in the Third Law of Motion,that is, forevery action there is an equal and opposite reaction. In biology, a conditionin an organism known as homeostasis means that when a stress is appliedto an organism, the organisms bodily functions automatically respond so asto remove the stress.

Example 1.3.9 Linear generator.If a force is applied to move the sliding bar with velocity v = dx/dt as

shown in Fig. E1.3.9.1, the total magnetic eld = xlBo linking the loop will bedecreasing at the rate of vlB0 . According to Lenz law, a current in the bar mustbe produced to oppose the decreasing of the magnetic ux. Thus an induced voltageV = vlB0 is generated across the resistor.

vV

x

y

+

sliding bar

v

x

y

+

sliding bar

EMF

Figure E1.3.9.1 Linear generator.


64 1. Fundamentals

Example 1.3.10 AC generator.An AC generator can be made of the DC motor by replacing the DC current

source with a load resistance R and providing an external rotatory force on thearmature. Applying a torque that makes the loop turn in the direction as shown inFig. E1.3.10.1, a motional EMF

V =

dl E =

dl F/q =

dl v B = AB sin

+

slip rings

B

V

M

F

F z

x

Figure E1.3.10.1 AC generator.

is produced. For the armature rotating with an angular frequency , we havev B = lBA sin and = t .

The same result can be derived by using Lenz law

EMF = d/dt

where

=

A

dS B = AB cos (E1.3.10.1)

We nd the generated AC voltage

V = EMF = AB sintEnd of Example 1.3.10


B. Poyntings Theorem and Poynting Vector

Energy conservation immediately follows from the Maxwell equations. Dot-multiply Faradays law (1.1.2) by H , Amperes law (1.1.1) by E and sub-tract. By making use of the vector identity (EH) = H EE H ,we obtain Poyntings theorem

(E H) +H Bt

+ E Dt

= E J (1.3.6)

The Poynting vectorS = E H (1.3.7)

is interpreted as the power ow density with the dimension of watts/m2 ,and H (B/t) + E (D/t) represents the time rate of change of thestored electric and magnetic energy density. On the right-hand side of (1.3.6),E J is the power supplied by the current J .

Example 1.3.11Consider the simple wave solution

E = xE0 cos(kz t) (E1.3.11.1a)H = yH0 cos(kz t) (E1.3.11.1b)

where H0 = E0/o and o =o/o is called the characteristic impedance of free

space. Substituting (E1.3.11.1) in (1.3.6) we see that Poyntings theorem is satised.The Poyntings vector is calculated to be

S = E H = z

oo

E20 cos2(kz t) (E1.3.11.2)

In free space, we nd

H t

(oH) =

t

[12oH H

]=

tWm

and

E t

(oE) =

t

[12

oE E

]=

tWe

In the source-free region we also have J = 0 . Poyntings theorem becomes

(E H) + t

(We +Wm) = 0 (E1.3.11.3)

66 1. Fundamentals

where

We =12

o

E2 = 12

oE

20 cos

2(kz t) (E1.3.11.4)

is the stored electric energy density and

Wm =12o

H2 = 12oH

20 cos

2(kz t) (E1.3.11.5)

is the stored magnetic energy density. It is seen that the stored electric energy isequal to the stored magnetic energy, We = Wm .


Example 1.3.12 Power, energy, force, and radiation pressure.The time-average Poynting vector power density is given by

=1T

T0

dt S = zE202o

= z12oH

20 = zP (E1.3.12.1)

where

P =E202o

=12oH

20

is the power density of the wave with units of Watts/m2 . The total time-averageelectromagnetic energy density (with units J/m3 ) is equal to the sum of the electricenergy density and the magnetic energy density,

W = + =12

oE

20 =

12oH

20 (E1.3.12.2)

We may dene an energy velocity ve equal to the ratio of power density to energydensity. We nd P/W = ve = 1/

oo which is the velocity of light.

Radiation pressure is force per unit area F = P/ve (with unit N/m2 ). Thusthe radiation pressure of the wave p is

p = F = P/ve = W =12

oE

20 =

12oH

20 (E1.3.12.3)

which is equal to the time-average total energy density in the wave and acts in thedirection of propagation of the wave. The radiation pressure, although generallyvery small, can lead to large scale eects. For example, comet tails are forced topoint away from the Sun due to the radiation pressure from the Sun.



Example 1.3.13 Consider an electromagnetic wave with

Ei = xE0 cos(kz t) (E1.3.13.1a)Hi = yH0 cos(kz t) (E1.3.13.1b)

impinging upon the surface of a perfectly conducting surface [Fig. E1.3.13.1]. Theboundary condition at the surface of the boundary requires that

n (E1 E2) = 0 (E1.3.13.2a)n (H1 H2) = Js (E1.3.13.2b)

where n = z is the normal to the surface. A perfect conductor is dened to haveelds zero inside, thus E1 = H1 = 0 .

z

Js

n = z E = H = 0

H

Perfect conductor

F

x

y

z 0Ei

Hi

Er

Hr

incident reected

Figure E1.3.13.1 Reection by a perfect conductor.

The reected wave that satises the boundary conditions (E1.3.13.2) is

Er = xE0 cos(kz + t) (E1.3.13.3a)Hr = yH0 cos(kz + t) (E1.3.13.3b)

which is propagating in the z direction. The surface current Js at z = 0 isfound to be

Js = n [(Hi +Hr) 0]z=0 = x2H0 costFrom the Lorentz force law, the force density acting on Js is

F =12Js B = 2zoH20 cos2 t

The factor 1/2 is due to the fact that there is magnetic eld only on one side ofthe current sheet. The time-average value is thus

F = oH20which is twice the value of that in Example 1.3.12. This is because the reectedwave exerts a recoil force on the conductor when it launches the reected wave.


68 1. Fundamentals

Example 1.3.14For an electromagnetic wave with electric eld with E0 = 3 106 V/m (which

is the breakdown electric eld strength for air), nd the power density and radiationpressure. What is the area required in order to supply the electric power of 2.4 1011 W for use by a nation?

Answer:

The power density is P = 1.2 1010 W/m2 .The radiation pressure is p = 40 N/m2 .The area required is 20 m2 .


Applying the divergence theorem to Poyntings theorem (1.3.6), we write

SdS E H =

t

VdV

(12

oE

2 +12oH

2

)

VdV E J (1.3.8)

The left-hand side represents power ow out of the surface enclosing thevolume V . The rst term on the right-hand side represents the depletionof the electric energy and the magnetic energy inside the volume in orderto supply the outow of the Poynting power. The last term represents thepower generated by the source J inside the volume V .

C. Momentum Conservation Theorem

Substituting the Maxwell equations for and J in the Lorentz force law

f = E + J B (1.3.9)

we nd that

f = t

(D B) [12(D E +B H)I DE BH

](1.3.10)

where I is a unit dyad with diagonal elements equal to 1 and all o-diagonalelements equal to zero.

The interpretation of the terms is

G = D B = momentum density vector (1.3.11)T =

12(D E +B H)I DE BH

= Maxwell stress tensor (1.3.12)


Thus we have the theorem

T + Gt

= f (1.3.13)which expresses conservation of momentum. This is in a form similar toPoyntings theorem in (1.3.6) except that it is now a vector equation. Infact, (1.3.6) and (1.3.13) combine to become a four-dimensional conservationtheorem in relativity.

Problems

P1.3.1According to the Bohr model of an atom, electrons revolve around the nucleus

in quantized orbits with radii R = nh/mv where n is an integer, m is the electronmass and v is the electron velocity. Letting the nucleus be a positive charge of Ze ,calculate R by equating the centrifugal force with the Lorentz force. Estimate theradius for a hydrogen atom with Z = 1 .

P1.3.2The Earth receives over all frequency bands about 1.5 kW/m2 of power from

the Sun.(a) The Earth-Sun distance is 150 109 m . How long does it take the sunlight to

reach the Earth?(b) The Earth radius is 6400 km. What is the

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