32-1 Chapter 32 - Physics2000 - Relativity · PDF file32-1 Chapter 32 Maxwell's Equations In...

32-1

Chapter 32Maxwell'sEquations

In electricity theory we have two vector fields E and B,and two equations are needed to define each field.Therefore the total number of equations required mustbe four.

How many of the required equations have we discussedso far? We have Gauss’ law for the divergent part ofE, and Faraday’s law for the solenoidal part. Itappears that we already have a complete theory of theelectric field, and we do. Gauss’ law and Faraday’slaw are two of the four equations needed.

For magnetism, we have Ampere’s law that defines thesolenoidal part of B. But we have not written anequation involving the surface integral of B. We aremissing a Gauss’ law type equation for the magneticfield. It would appear that the missing Gauss’ law forB, plus Ampere’s law make up the remaining twoequations.

This is not quite correct. The missing Gauss’ law is oneof the needed equations for B, and it is easily writtendown because there are no known sources for a diver-gent B field. But Ampere’s law, in the form we havebeen using

B⋅dl = µ0 i (29-18)

has a logical flaw that was discovered by Maxwell.When Maxwell corrected this flaw by adding anothersource term to the right side of Equation (29-18), hethen had the complete, correct set of four equations forE and B.

In 1860 James Clerk Maxwell summarized the entirecontent of the theory of electricity and magnetism in afew short equations. In this chapter we will reviewthese equations and investigate some of the predictionsone can make when the entire theory is available.

What does a complete theory of electricity and magne-tism involve? We have to fully specify the electric fieldE , the magnetic field B, and describe what effect thefields have when they interact with matter.

The interaction is described by the Lorentz force law

F = qE + q v × B (28-18)

which tells us the force exerted on a charge q by the Eand B fields. As long as we stay away from the atomicworld where quantum mechanics dominates, then theLorentz force law combined with Newton’s second lawfully explains the behavior of charges in the presenceof electric and magnetic fields, whatever the origin ofthe fields may be.

To handle the electric and magnetic fields, recall ourdiscussion in Chapter 30 (on two kinds of fields) wherewe saw that any vector field can be separated into twoparts; a divergent part like the electric field of staticcharges, and a solenoidal part like the electric field ina betatron or inductor. To completely specify a vectorfield, we need two equations – one involving a surfaceintegral or its equivalent to define the divergent part ofthe field, and another involving a line integral or itsequivalent defining the solenoidal part.

CHAPTER 32 MAXWELL'S EQUATIONS

32-2 Maxwell's Equations

All Maxwell did was to add one term to the fourequations for E and B , and yet the entire set ofequations are named after him. The reason for this isthat with the correct set of equations, Maxwell was ableto obtain solutions of the four equations, predictions ofthese equations that could not be obtained untilAmpere’s law had been corrected. The most famous ofthese predictions was that a certain structure of electricand magnetic fields could travel through empty spaceat a speed v = 1/ µ0ε0 . Since Maxwell knew that

1/ µ0ε0 was close to the observed speed 3 × 108 m/sfor light, he proposed that this structure of electric andmagnetic fields was light itself.

In this chapter, we will first describe the missing Gauss'law for magnetic fields, then correct Ampere’s law toget the complete set of Maxwell’s four equations. Wewill then solve these equations for a structure of electricand magnetic fields that moves through empty space ata speed v = 1/ µ0ε0 . We will see that this structureexplains various properties of light waves, radio waves,and other components of the electromagnetic spec-trum. We will find, for example, that we can detectradio waves by using the same equipment and proce-dures we have used in earlier chapters to detect andmap electric and magnetic fields.

GAUSS’ LAW FORMAGNETIC FIELDSLet us review a calculation we have done several timesnow—the use of Gauss’ law to calculate the electricfield of a point particle. Our latest form of the law is

E ⋅ dAclosed

surface

=Qin

ε0(29-5)

where Qin is the total amount of electric charge insidethe surface.

In Figure (1), we have a point charge Q and haveconstructed a closed spherical surface of radius r cen-tered on the charge. For this surface, E is everywhereperpendicular to the surface or parallel to every surfaceelement dA, thus E ⋅ dA = EdA. Since E is ofconstant magnitude, we get

E ⋅ dA

closedsurface

= E dA

closedsurface

= E 4πr2

=Qin

ε0

(1)

where Qin = Q .

rQ

EFigure 1Field of point charge.

32-3

The solution of Equation (1) gives E = Q/4πε0r2 asthe strength of the electric field of a point charge. Asimilar calculation using a cylindrical surface gave usthe electric field of a charged rod. By being clever, orworking very hard, one can use Gauss’ law in the formof Equation (29-5) to solve for the electric field of anystatic distribution of electric charge.

But the simple example of the field of a point chargeillustrates the point we wish to make. Gauss’ lawdetermines the diverging kind of field we get from apoint source. Electric fields have point sources, namelyelectric charges, and it is these sources in the form ofQin that appear on the right hand side of Equation(29-5).

Figure (2) shows a magnetic field emerging from apoint source of magnetism. Such a point source ofmagnetism is given the name magnetic monopole andmagnetic monopoles are predicted to exist by variousrecent theories of elementary particles. These theoriesare designed to unify three of the four basic interactions– the electrical, the weak, and the nuclear interactions.(They are called “Grand Unified Theories” or “GUT”theories. Gravity raises problems that are not handledby GUT theories.) These theories also predict that theproton should decay with a half life of 1032 years.

In the last 20 years there has been an extensive searchfor evidence for the decay of protons or the existence ofmagnetic monopoles. So far we have found no evi-dence for either. (You do not have to wait 1032 yearsto see if protons decay; instead you can see if one outof 1032 protons decays in one year.)

The failure to find the magnetic monopole, the fact thatno one has yet seen a magnetic field with the shapeshown in Figure (2), can be stated mathematically bywriting a form of Gauss’ law for magnetic fields withthe magnetic charge Qin set to zero

B ⋅ dAclosedsurface

= 0 (2)

When reading Equation (2), interpret the zero on theright side of Equation (2) as a statement that thedivergent part of the magnetic field has no sourceterm. This is in contrast to Gauss’ law for electricfields, where Qin /ε0 is the source term.

r

B

Figure 2Magnetic field produced by a point source.


path on the wire so that B and sections d of the pathare everywhere parallel. Thus B ⋅ d = Bd , andsince B is constant along the path, we have

B ⋅ d = B d = B×2πr = µ0i

which gives our old formula B = µ0 i/2π r for themagnetic field of a wire.

To see the flaw with Ampere’s law, consider a circuitwhere a capacitor is being charged up by a current i asshown in Figure (4). When a capacitor becomescharged, one plate becomes positively charged and theother negatively charged as shown. We can think of thecapacitor being charged because a positive current isflowing into the left plate, making that plate positive,and a positive current is flowing out of the right plate,making that plate negative.

Figure (4) looks somewhat peculiar in that the currenti almost appears to be flowing through the capacitor.We have a current i on the left, which continues on theright, with a break between the capacitor plates. Toemphasize the peculiar nature of this discontinuity inthe current, imagine that the wires leading to thecapacitor are huge wires, and that the capacitor platesare just the ends of the wires as shown in Figure (5).

Now let us apply Ampere’s law to the situation shownin Figure (5). We have drawn three paths, Path (1)around the wire leading into the positive plate of thecapacitor, Path (2) around the wire leading out of thenegative plate, and Path (3) around the gap between theplates. Applying Ampere’s law we have

MAXWELL’S CORRECTIONTO AMPERE’S LAWAs we mentioned in the introduction, Maxwell de-tected a logical flaw in Ampere’s law which, whencorrected, gave him the complete set of equations forthe electric and magnetic fields. With the complete setof equations, Maxwell was able to obtain a theory oflight. No theory of light could be obtained without thecorrection.

Ampere’s law, Equation (29-18), uses the line integralto detect the solenoidal component of the magneticfield. We had

B ⋅ d = µ0ienclosed

Ampere'sLaw

(29-18)

where ienclosed is the total current encircled by theclosed path used to evaluate B ⋅ d . We can say that

µ0ienclosed is the source term for this equation, inanalogy to Qin /ε0 being the source term for Gauss’law.

Before we discuss Maxwell’s correction, let us reviewthe use of Equation (29-18) to calculate the magneticfield of a straight current i as shown in Figure (3). In(3a) we see the wire carrying the current, and in (3b) weshow the circular magnetic field produced by thecurrent. To apply Equation (29-18) we draw a closedcircular path of radius r around the wire, centering the

Figure 3Using Ampere's law.

Figure 4Charging up a capacitor.

ri

r

B

(a)

(b)

––––––

i i

++++++

32-5

B ⋅ d

path 1= µ0i

path 1 goesaround a current i

(3)

B ⋅ d

path 2= µ0i

path 2 goesaround a current i

(4)

B ⋅ d

path 3= 0

path 3 does not goaround any current

(5)

When we write out Ampere’s law this way, thediscontinuity in the current at the capacitor plateslooks a bit more disturbing.

For greater emphasis of the problem, imagine that thegap in Figure (5) is very narrow, like Figure (5a) onlyworse. Assume we have a 1 mm diameter wire and thegap is only 10 atomic diameters. Then according toAmpere’s law, B ⋅ d should still be zero if it iscorrectly centered on the gap. But can we possiblycenter a path on a gap that is only 10 atomic diameterswide? And even if we could, would B ⋅ d

be zero

for this path, and have the full value µ0 i for the path 10atomic diameters away? No, we simply cannot havesuch a discontinuity in the magnetic field and theremust be something wrong with Ampere’s law. Thiswas the problem recognized by Maxwell.

Maxwell’s solution was that even inside the gap at thecapacitor there was a source for B ⋅ d , and that thestrength of the source was still µ0 i. What actuallyexists inside the gap is the electric field E due to the +and – charge accumulating on the capacitor plates asshown in Figure (6). Perhaps this electric field cansomehow replace the missing current in the gap.

The capacitor plates or rod ends in Figure (6) have acharge density σ = Q/A where Q is the presentcharge on the capacitor and A is the area of the plates.In one of our early Gauss’ law calculations we saw thata charge density on a conducting surface produces anelectric field of strength E = σ/ε0, thus E betweenthe plates is related to the charge Q on them by

E =

σε0

=Q

ε0A ; Q = ε0EA (6)

The current flowing into the capacitor plates is relatedto the charge Q that has accumulated by

i = dQdt (7)

Using Equation (6) in Equation (7), we get

i = ε0

ddt

EA (8)

++++++

––––––

i

i

i

i

i

i

Path 1 Path 3 Path 2

Figure 5Current flows through Paths (1)& (2), but not through Path (3).

Figure 6An electric field E exists between the plates.

Figure 5aVery narrow gap

i

i

i

i

i

i

++++++

––––––

i

i

i

i

i

i

E

E

E

+Q –Q

capacitor plate of area A


Noting that the flux ΦE of electric field between theplates is

ΦE = E ⋅ dA = E A⊥ = E A (9)

and multiplying through by µ0, we can write Equation(8) in the form

µ0i = µ0ε0

dΦE

dt(10)

We get the somewhat surprising result that µ0ε0 timesthe rate of change of electric flux inside the capacitorhas the same magnitude as µ0 i , where i is the currentin the wire leading to the capacitor. Maxwell proposedthat µ0ε0dΦE dtdΦE dt played the same role, inside thecapacitor, as a source term for B ⋅ dl, that µ0 i didoutside in the wire.

As a result, Maxwell proposed that Ampere’s law becorrected to read

B ⋅ d = µ0i + µ0ε0

dΦE

dt

correctedAmpere'slaw

(11)

Applying Equation (11) to the three paths shown inFigure (7), we have

B ⋅ dpaths 1&2

= µ0i (ΦE = 0) (12)

B ⋅ d

paths 3= µ0ε0

dΦE

dt( i = 0 ) (13)

I.e., for Paths (1) and (2), there is no electric fluxthrough the path and the source of B ⋅ d

is the

current. For Path (3), no current flows through the pathand the source of B ⋅ d is the changing electricflux. But, because µ0 i in Equation (12) has the samemagnitude µ0ε0dΦE dtdΦE dt in Equation (13), the term

B ⋅ d has the same value for Path (3) as (1) and (2),

and there is no discontinuity in the magnetic field.

Example: Magnetic Fieldbetween the Capacitor PlatesAs an example of the use of the new term in thecorrected Ampere’s law, let us calculate the magneticfield in the region between the capacitor plates. To dothis we draw a centered circular path of radius r smallerthan the capacitor radius R as shown in Figure (8).There is no current through this path, but there is anelectric flux ΦE r = EA r = Eπ r2 through thepath. Thus we set i = 0 in Ampere’s corrected law, andreplace ΦE by the flux ΦE r through our path to get

B ⋅ d = µ0ε0

dΦE r

dt(14)

Equation (14) tells us that because we have an in-creasing electric field between the plates, and thus anincreasing electric flux through our path, there must bea magnetic field around the path.

Figure 8Calculating Bin the region between the plates.Figure 7

Path (2) surrounds a changing electric flux. Inside thegap, µµ00εε00(dΦΦ E/dt)replaces µµ00 i as the source of B.

i

i

i

i

i

i

E

E

E

1 3 2

i iEr

R

32-7

Due to the cylindrical symmetry of the problem, theonly possible shape for the magnetic field inside thecapacitor is circular, just like the field outside. Thiscircular field and our path are shown in the end view,Figure (9). Since B and d are parallel for all the stepsaround the circular path, we have B ⋅ d = Bd . Andsince B is constant in magnitude along the path, we get

B ⋅ d = B d = B× 2πr (15)

To evaluate the right hand side of Equation (14), notethat the flux through our path ΦE r

is equal to the total

flux ΦE total times the ratio of the area π r2 of our pathto the total area πR2 of the capacitor plates

ΦE r = ΦE total π r2

πR2 (16)

so that the right hand side becomes

µ0ε0dΦE r

dt = µ0ε0

dΦE totaldt

r2

R2

= µ0i r2

R2 (17)

where in the last step we used Equation (10) to replaceµ0ε0dΦE total /dt by a term of the same magnitude,namely µ0 i.

Finally using Equation (15) and (17) in (14) we get

B×2πr = µ0i

r2

R2

B =

µ0i

2πr

R2

magneticfield betweencapacitor plates

(18)

Figure (10) is a graph of the magnitude of B bothinside and outside the plates. They match up at r =R, and the field strength decreases linearly to zeroinside the plates.

Exercise 1Calculate the magnetic field inside the copper wiresthat lead to the capacitor plates of Figure (5). UseAmpere’s law and a circular path of radius r inside thecopper as shown in Figure (11). Assuming that there isa uniform current density in the wire, you should getEquation (18) as an answer. Thus the magnetic field iscontinuous as we go out from the copper to between thecapacitor plates.

Figure 9End view of capacitor plate.

B

B

��r

circularMagneticField

circularPath ofradius r

Figure 10Magnetic fields inside and outside the gap.

R

B(r)

r

Figure 11

ir

R


MAXWELL’S EQUATIONSNow that we have corrected Ampere’s law, we areready to write the four equations that completelygovern the behavior of classical electric and magneticfields. They are

(a) E ⋅ dAclosedsurface

=Qin

ε0

Gauss'Law

(b) B ⋅ dAclosedsurface

= 0NoMonopole

(c) B ⋅ d = µ0i + µ0ε0

dΦE

dtAmpere'sLaw

(d) E ⋅ d = –dΦB

dtFaraday'sLaw

(19)

The only other thing you need for the classical theoryof electromagnetism is the Lorentz force law andNewton’s second law to calculate the effect of electricand magnetic fields on charged particles.

F = qE + qv ×B

LorentzForceLaw

(20)

This is a complete formal summary of everything wehave learned in the past ten chapters.

Exercise 2This is one of the most important exercises in the text.The four Maxwell’s equations and the Lorentz force lawrepresent an elegant summary of many ideas. Butthese equations are nothing but hen scratchings on apiece of paper if you do not have a clear idea of howeach term is used.

The best way to give these equations meaning is toknow inside out at least one specific example thatillustrates the use of each term in the equations. ForGauss’ law, we have emphasized the calculation of theelectric field of a point and a line charge. We have thenonexistence of the divergent magnetic field in Figure(2) to illustrate Gauss’ law for magnetic fields. We haveused Ampere’s law to calculate the magnetic field of awire and a solenoid. The new term in Ampere’s law wasused to calculate the magnetic field inside a parallelplate capacitor that is being charged up.

Faraday’s law has numerous applications including theair cart speed meter, the betatron, the AC voltagegenerator, and the inductance of a solenoid. Perhapsthe most important concept with Faraday’s law is that

E⋅dl is the voltage rise created by solenoidal electricfields, which for circuits can be read directly by avoltmeter. This lead to the interpretation of a loop of wirewith a voltmeter attached as an E⋅dl meter. We usedan E⋅dl meter in the design of the air cart speeddetector and experiment where we mapped the mag-netic field of a Helmholtz coil.

Then there is the Lorentz force law with the formulas forthe electric and magnetic force on a charged particle.As an example of an electric force we calculated thetrajectory of an electron beam between charged plates,and for a magnetic force we studied the circular motionof electrons in a uniform magnetic field.

The assignment of this exercise is to write out Maxwell’sequations one by one, and with each equation writedown a fully worked out example of the use of each term.Do this neatly, and save it for later reference. This is whatturns the hen scratchings shown on the previous pageinto a meaningful theory. When you buy a T-shirt withMaxwell’s equations on it, you will be able to wear it withconfidence.

We have just crossed what you might call a conti-nental divide in our study of the theory of electricityand magnetism. We spent the last ten chaptersbuilding up to Maxwell’s equations. Now we de-scend into applications of the theory. We will focuson applications and discussions that would not havemade sense until we had the complete set of equa-tions—discussions on the symmetry of the equationsand applications like Maxwell’s theory of light.

32-9

Equations (22) immediately demonstrate the lack ofsymmetry caused by the absence of magnetic mono-poles, and so does the Lorentz force law of Equation(20). If the magnetic monopole is discovered, and weassign to it the “magnetic charge” QB, then for ex-ample Equation (22b) would become


= QB (22b')

If we have magnetic monopoles, a magnetic fieldshould exert a force FB = QBB and perhaps an electricfield should exert a force something like FE = QBv × E.

Aside from Equation (22b), the other glaring asymme-try is the presence of an electric current i in Ampere’slaw (22c) but no current term in Faraday’s law (22d).If, however, we have magnetic monopoles we can alsohave a current iB of magnetic monopoles, and thisasymmetry can be removed.

Exercise 3Assume that the magnetic monopole has been dis-covered, and that we now have magnetic charge QB

and a current iB of magnetic charge. Correct Maxwell’sEquations (22) and the Lorentz force law (20) to includethe magnetic monopole. For each new term you add tothese equations, provide a worked-out example of itsuse.

In this exercise, use symmetry to guess what termsshould be added. If you want to go beyond what we areasking for in this exercise, you can start with the formula

FB = QBB for the magnetic force on a magnetic charge,and with the kind of thought experiments we used in thechapter on magnetism, derive the formula for the elec-tric force on a magnetic charge QB . You will also endup with a derivation of the correction to Faraday’s lawcaused by a current of magnetic charge. (This is moreof a project than an exercise.)

SYMMETRY OFMAXWELL’S EQUATIONSMaxwell’s Equations (19 a, b, c, and d), display consid-erable symmetry, and a special lack of symmetry. Butthe symmetry or lack of it is clouded by our choice ofthe MKS units with its historical constants µ0 and ε0that appear, somewhat randomly, either in the numera-tor or denominator at various places.

For this section, let us use a special set of units wherethe constants µ0 and ε0 have the value 1

µ0 = 1 ; ε0 = 1

in a specialset of units

(21)

Because the speed of light c is related to µ0 and ε0 byc = 1/ µ0ε0, we are now using a set of units wherethe speed of light is 1. If we set µ0 = ε0 = 1 inEquations (19) we get

E ⋅ dAclosedsurface

= Qin (22a)


= 0 (22b)

B ⋅ d = i +

dΦE

dt(22c)

E ⋅ d = –

dΦB

dt(22d)

Stripping out µ0 and ε0 gives a clearer picture of whatMaxwell’s equations are trying to say. Equation (22a)tells us that electric charge is the source of divergentelectric fields. Equation (22b) says that we haven’tfound any source for divergent magnetic fields. Equa-tion (22c) tells us that an electric current or a changingelectric flux is a source for solenoidal magnetic fields,and (22d) tells us that a changing magnetic flux createsa solenoidal electric field.


Equations (24a, b) suggest a coupling between electricand magnetic fields. Let us first discuss this couplingin a qualitative, somewhat sloppy way, and then workout explicit examples to see precisely what is happen-ing.

Roughly speaking, Equation (24a) tells us that a chang-ing electric flux or field creates a magnetic field, and(24b) tells us that a changing magnetic field creates anelectric field. These fields interact, and in some sensesupport each other.

If we were experts in integral and differential equa-tions, we would look at Equations (24) and say, “Oh,yes, this is just one form of the standard wave equation.The solution is a wave of electric and magnetic fieldstraveling through space.” Maxwell was able to do this,and solve Equations (24) for both the structure and thespeed of the wave. The speed turns out to be c, and heguessed that the wave was light.

Because the reader is not expected to be an expert inintegral and differential equations, we will go slower,working out specific examples to see what kind ofstructures and behavior we do get from Equations (24).We are just beginning to touch upon the enormoussubject of electromagnetic radiation.

A Radiated Electromagnetic PulseWe will solve Equations (24) the same way we havebeen solving all equations involving derivatives orintegrals—by guessing and checking. The rules of thegame are as follows. Guess a solution, then applyEquations (24) to your guess in every possible way youcan think of. If you cannot find an inconsistency, yourguess may be correct.

In order to guess a solution, we want to pick an examplethat we know as much as possible about and use everyinsight we can to improve our chances of getting theright answer. Since we are already familiar with thefields associated with a current in a wire, we will focuson that situation. Explicitly, we will consider whathappens, what kind of fields we get, when we first turnon a current in a wire. We will see that a structure ofmagnetic and electric fields travels out from the wire,in what will be an example of a radiated electromag-netic pulse or wave.

MAXWELL’S EQUATIONSIN EMPTY SPACEIn the remainder of this chapter we will discuss thebehavior of electric and magnetic fields in empty spacewhere there are no charges or currents. A few chaptersago, there would not have been much point in such adiscussion, for electric fields were produced by charges,magnetic fields by currents, and without charges andcurrents, we had no fields. But with Faraday’s law, wesee that a changing magnetic flux dΦB dtdΦB dt acts as thesource of a solenoidal electric field. And with thecorrection to Ampere’s law, we see that a changingelectric flux is a source of solenoidal magnetic fields.Even without charges and currents we have sources forboth electric and magnetic fields.

First note that if we have no electric charge (or mag-netic monopoles), then we have no sources for either adivergent electric or divergent magnetic field. Inempty space diverging fields do not play an importantrole and we can focus our attention on the equations forthe solenoidal magnetic and solenoidal electric field,namely Ampere’s and Faraday’s laws.

Setting i = 0 in Equation (19c), the Equations (19c) and(19d) for the solenoidal fields in empty space become

B ⋅ d = µ0ε0

dΦE

dt(23a)

E ⋅ d = –

dΦB

dt(23b)

We can make these equations look better if we write µ0ε0 as 1/c2, where c = 3 × 108 m/s as determined in

our LC circuit experiment. Then Equations (23) be-come

B ⋅ d =

1

c2

dΦE

dt

E ⋅ d = –dΦB

dt

Maxwell'sequationsin emptyspace

(24a)

(24b)

32-11

A Thought ExperimentLet us picture a very long, straight, copper wire with nocurrent in it. At time t = 0 we start an upward directedcurrent i everywhere in the wire as shown in Figure(12). This is the tricky part of the experiment, havingthe current i start everywhere at the same time. If weclosed a switch, the motion of charge would begin at theswitch and advance down the wire. To avoid this,imagine that we have many observers with synchro-nized watches, and they all reach into the wire and startthe positive charge moving at t = 0. However you wantto picture it, just make sure that there is no current in thewire before t = 0, and that we have a uniform current iafterward.

In our previous discussions, we saw that a current i ina straight wire produced a circular magnetic field ofmagnitude B = µ0i/2πr everywhere outside thewire. This cannot be the solution we need because itimplies that as soon as the current is turned on, we havea magnetic field throughout all of space. The existenceof the magnetic field carries the information that wehave turned on the current. Thus the instantaneousspread of the field throughout space carries this infor-mation faster than the speed of light and violates theprinciple of causality. As we saw in Chapter 1, wecould get answers to questions that have not yet beenasked.

Using our knowledge of special relativity as a guide,we suspect that the solution B = µ0i/2πr every-where in space, instantaneously, is not a good guess. Amore reasonable guess is that the magnetic field growsat some speed v out from the wire. Inside the growingfront, the field may be somewhat like its final form

B = µ0i/2πr , but outside we will assume B = 0.

The pure, expanding magnetic field shown in Figure(13) seems like a good guess. But it is wrong, as we cansee if we apply Ampere’s law to Path (a) which has notyet been reached by the growing magnetic field. Forthis path that lies outside the magnetic field,

B ⋅ d = 0 , and the corrected Ampere’s law, Equa-tion (19c), gives

B ⋅ d = µ0i + µ0ε0

dΦE

dt= 0 (25)

In our picture of Figure (13) we have no electric field,therefore ΦE = 0 and Equation (25) implies that µ0i iszero, or the current i through Path (a) is zero. But thecurrent is not zero and we thus have an inconsistency.The growing magnetic field of Figure (13) is not asolution of Maxwell’s equations. (This is how we playthe game. Guess and try, and this time we failed.)

Equation (25) gives us a hint of what is wrong with ourguess. It says that

dΦE

dt= -

iε0

(25a)

thus if we have a current i and have the growingmagnetic field shown in Figure (13) we must also havea changing electric flux ΦE through Path (a). Some-where there must be an electric field E to produce thechanging flux ΦE , a field that points either up or down,passing through the circular path of Figure (13).

Figure 12A current i is startedall along the wire attime t = 0.

Figure 13As a guess, we will assume that the magneticfield expands at a speed v out from the wire,when the current is turned on.

B = 0B

i up at t = 0v

v

v

v

v

v

path (a)

i


In our earlier discussion of inductance and inducedvoltage, we saw that a changing current creates anelectric field that opposed the change. This is whatgives an effective inertia to the current in an inductor.Thus when we suddenly turn on the upward directedcurrent as shown in Figure (12), we expect that weshould have a downward directed electric field asindicated in Figure (14), opposing our trying to start thecurrent.

Initially the downward directed electric field should beinside the wire where it can act on the current carryingcharges. But our growing circular magnetic fieldshown in Figure (13) must also have started inside thewire. Since a growing magnetic field alone is not asolution of Maxwell’s equations and since there mustbe an associated electric field, let us propose that boththe circular magnetic field of Figure (13), and thedownward electric field of Figure (14) grow together asshown in Figure (15).

In Figure (15), we have sketched a field structureconsisting of a circular magnetic field and a downwardelectric field that started out at the wire and is expand-ing radially outward at a speed v as shown. Thisstructure has not yet expanded out to our Path (1), sothat the line integral B ⋅ d is still zero and Ampere’slaw still requires that

0 = µ0i + µ0ε0

dΦE

dtPath 1 ofFigure15

dΦE

dt= –

iε0

(26)

which is the same as Equation (25).

Looking at Figure (15), we see that the downwardelectric field gives us a negative flux ΦE through ourpath. (We chose the direction of the path so that by theright hand convention, the current i is positive.) And asthe field structure expands, we have more negative fluxthrough the path. This increasing negative flux is justwhat is required by Equation (26).

Figure 14When a current startsup, it is opposed by anelectric field.

Bpath (1)

vv

EE

Figure 15As a second guess,we will assumethat there is adownward directedelectric fieldassociated with theexpandingmagnetic field.Again, Path (1) isout where thefields have not yetarrived.

EE

iup, increasing

i

32-13

What happens when the field structure gets to andpasses our path? The situation suddenly changes. Nowwe have a magnetic field at the path, so that B ⋅ dis no longer zero. And now the expanding front isoutside our path so that the expansion no longer con-tributes to –dΦE dt–dΦE dt. The sudden appearance of

B ⋅ d is precisely compensated by the sudden lossof the dΦE/dt due to expansion of the field structure.

The alert student, who calculates E ⋅ d for somepaths inside the field structure of Figure (15) willdiscover that we have not yet found a completelysatisfactory solution to Maxwell’s equations. Theelectric fields in close to the wire eventually die away,and only when they have gone do we get a staticmagnetic field given by B ⋅ d = µ0i + 0.

The problems associated with the electric field dyingaway can be avoided if we turn on the current at timet = 0, and then shut it off a very short time later. In thatcase we should expect to see an expanding cylindricalshell of electric and magnetic fields as shown in Figure(16). The front of the shell started out when the currentwas turned on, and the back should start out when thecurrent is shut off. We will guess that the front and backshould both travel radially outward at a speed v asshown.

EE

B

vvvv

Figure 16Electromagneticpulse producedby turning thecurrent on andthen quickly off.We will see thatthis structureagrees withMaxwell'sequations.


Speed of an Electromagnetic PulseLet us use Figure (16), redrawn as Figure (17a), as ourbest guess for the structure of an electromagnetic pulse.The first step is to check that this field structure obeysMaxwell’s equations. If it does, then we will see if wecan solve for the speed v of the wave front.

In Figure (17a), where we have shut the current off,there is no net charge or current and all we need toconsider is the expanding shell of electric and magneticfields moving through space. We have no divergentfields, no current, and the equations for E and Bbecome

B ⋅ d = µ0ε0

dΦE

dt(23a)

E ⋅ d = –

dΦB

dt (23b)

which we wrote down earlier as Maxwell’s equationfor empty space.

In order to apply Maxwell’s equations to the fields inFigure (17a), we will focus our attention on a smallpiece of the shell on the right side that is moving to theright at a speed v. For this analysis, we will use the twopaths labeled Path (1) and Path (2). Path (1) has a sideparallel to the electric field, and will be used forEquation 23b. Path (2) has a side parallel to the mag-netic field, and will be used for Equation 23a.

Analysis of Path 1In Figure (17b), we have a close up view of Path (1).The path was chosen so that only the left edge of lengthh was in the electric field, so that

E ⋅ d = Eh (27)

In order to make E ⋅ d positive on this left edge, wewent around Path (1) in a counterclockwise direction.By the right hand convention, any vector up throughthis path is positive, therefore the downward directedmagnetic field is going through Path (1) in a negativedirection. (We will be very careful about signs in thisdiscussion.)

EE

B

vvvv

path (1)

path (2)

E

v

v

h

path 1

B into paper

Figure 17bSide view showing path (1). An increasing (negative)magnetic flux flows down through Path (1).

Figure 17aIn order to analyze the electromagnetic pulseproduced by turning the current on and off, weintroduce the two paths shown above. Path (1)has one side parallel to the electric field, whilePath (2) has a side parallel to the magnetic field.

32-15

In Figure (18), we are looking at Path (1), first at a timet (18a) where the expanding front has reached aposition x as shown, then at a time t + ∆t where the fronthas reached x + ∆x. Since the front is moving at anassumed speed v , we have

v = ∆x∆t

At time t + ∆t , there is additional magnetic flux throughPath (1). The amount of additional magnetic flux ∆ΦBis equal to the strength B of the field times the additionalarea ( h∆x ). Since B points down through Path (1), ina negative direction, the additional flux is negative andwe have

∆ΦB = –B(∆A⊥) = –B(h∆x) (28)

Dividing Equation (28) through by ∆t, and taking thelimit that ∆t goes to dt, gives

∆ΦB

∆t= –Bh

∆x∆t

dΦB

dt= –Bh dx

dt= –Bhv (29)

We now have a formula for E ⋅ d (Equation 27) andfor dΦE/dt (Equation 29) which we can substitute intoFaraday’s law (23b) to get

E ⋅ d = –

dΦB

dt

Eh = – ( –Bvh) = +Bhv

The factor of h cancels and we are left with

E = Bv

fromFaraday's law

(30)

which is a surprisingly simple relationship between thestrengths of the electric and magnetic fields.

v path 1

at time t+∆t

b)

v

at time t

a)

x

∆xx

path 1

B(down)

additional area = h∆x

h

Figure 18As the front expands, there is moremagnetic flux down through Path (1).


In Figure (19), we show the expanding front at time t(19a) and at time t + ∆t (19b). The increase in electricflux ∆ΦE is (E) times the increased area ( h∆x)

∆ΦE = E h∆x

Dividing through by ∆t, and taking the limit that ∆t goesto dt, gives

∆ΦE

∆t= Eh

∆x∆t

dΦE

dt= Eh

dxdt

= Ehv (33)

Using Equation 33 in 32, and then cancelling h, gives

Bh = µ0ε0 Ehv

B = µ0ε0 EvFromAmpere's law

(34)

which is another simple relationship between E and B.

Analysis of Path 2Path (2), shown in Figure (17c), is chosen to have oneside in and parallel to the magnetic field. We have gonearound clockwise so that B and d point in the samedirection. Integrating B around the path gives

B ⋅d = Bh (31)

Combining Equation 31 with Ampere's law

B ⋅d = µ0ε0

dΦE

dt

gives

Bh = µ0ε0

dΦE

dt(32)

To evaluate dΦE dtdΦE dt, we first note that for a clockwisepath, the positive direction is down into the paper inFigure (17c). This is the same direction as the electricfield, thus we have a positive electric flux through path(2).

Figure 17cAn increasing (negative) electric fluxflows down through Path (2).

v

h

path 2

B

vE into

paper

EE

B

vvvv

path (1)

path (2)

Figure 17a (repeated)We will now turn our attention to path 2 whichhas one side parallel to the magnetic field.

32-17

Exercise 4Construct paths like (1) and (2) of Figure (17), but whichinclude the back side, rather than the front side, of theelectromagnetic pulse. Repeat the kind of steps usedto derive Equation (35) to show that the back of the pulsealso travels outward at a speed v = 1/ µ0ε0 . As aresult the pulse maintains its thickness as it expands outthrough space.

Exercise 5

After a class in which we discussed the electromagneticpulse shown in Figure (20a), a student said she thoughtthat the electric field would get ahead of the magneticfield as shown in Figure (20b). Use Maxwell's equationsto show that this does not happen.

If we divide Equation (30) Bv = E , by Equation (34) B = µ0ε0 Ev, both E and B cancel giving

BvB

=E

µ0ε0Ev

v2 =1

µ0ε0

v =1

µ0ε0

speed oflight!!!

Thus the electromagnetic pulse of Figures (16) and (17)expands outward at the speed 1/ µ0ε0 which we haveseen is 3 × 108 meters per second. Maxwell recog-nized that this was the speed of light and recognizedthat the electromagnetic pulse must be closely relatedto light itself.

Using v = 1/ µ0ε0 = c in Equation (34) we get

B = E

c (36)

as the relative strength of the electric and magneticfields in an electromagnetic pulse, or as we shall see,any light wave. If we had used a reasonable set of unitswhere c = 1 (like feet and nanoseconds), then E and Bwould have equal strengths in a light wave.

v path 2

at time t+∆t

b)

v

at time t

a)

x

∆xx

path 2

E(down)

additional area = h∆x

h

Figure 19As the front expands, there is moreelectric flux down through Path (2).

E

B

v

E

B

v

Path 1

Figure 20aThe radiated electromagnetic pulse wesaw in Figures (16) and (17).

Figure 20bThe student guessed that the electric fields wouldget out ahead of the magnetic field. Use Path (1)to show that this does not happen.

(35)


ELECTROMAGNETIC WAVESThe single electromagnetic pulse shown in Figure (17)is an example of an electromagnetic wave. We usuallythink of a wave as some kind of oscillating sinusoidalthing, but as we saw in our discussion of waves on aSlinky in Chapter 1, the simplest form of a wave is asingle pulse like that shown in Figure (21). The basicfeature of the Slinky wave pulse was that it maintainedits shape while it moved down the Slinky at the wavespeed v . Now we see that the electromagnetic pulsemaintains its structure of E and B fields while it movesat a speed v = c through space.

We made a more or less sinusoidal wave on theSlinky by shaking one end up and down to producea series of alternate up and down pulses that traveledtogether down the Slinky. Similarly, if we use analternating current in the wire of Figure (17), we willget a series of electromagnetic pulses that travel outfrom the wire. This series of pulses will moreclosely resemble what we usually think of as anelectromagnetic wave.

Figure (22a) is a graph of a rather jerky alternatingcurrent where we turn on an upward directed current ofmagnitude i0, then shut off the current for a while, thenturn on a downward directed current i0, etc. This seriesof current pulse produces the series of electromagneticpulses shown in Figure (22b). Far out from the wirewhere we can neglect the curvature of the magneticfield, we see a series of pulses shown in the close-upview, Figure (23a). This series of flat or non-curvedpulses is called a plane wave of electromagnetic radia-tion.

If we used a sinusoidally oscillating current in the wireof Figure (22), then the series of electromagnetic pulseswould blend together to form the sinusoidally varyingelectric and magnetic fields structure shown in Figure(23b). This is the wave structure one usually associateswith an electromagnetic wave.

When you think of an electromagnetic wave, picturethe fields shown in Figure (23), moving more or less asa rigid object past you at a speed c. The distance λbetween crests is called the wavelength of the wave.The time T it takes one wavelength or cycle to pass youis

T second

cycle=

λ metercycle

c metersecond

= λ secondc cycle (37)

T is called the period of the wave. The frequency ofthe wave, the number of wavelengths or full cycles ofthe wave that pass you per second is

f

cyclesecond

=c meter

secondλ meter

cycle

=λ cyclec second (38)

In Equations (37) and (38) we gave λ the dimensionsmeters/cycle, T of seconds/cycle and f of cycles/sec-ond so that we can use the dimensions to remember the

Figure 22Fields produced by a series of current pulses.

B

E

0

down

upi

i

0 –i–t

b) Resulting electric and magnetic fields

a) Graph of current pulses in wire

cc

E

Figure 21Slinky wave pulse.

32-19

You can remember where the 2π goes by giving it thedimensions 2π radians/cycle. (Think of a full circle orfull cycle as having 2π radians.) We will indiscrimi-nately use the word frequency to describe eitherf cycles/second or ω radians/second, whichever ismore appropriate. If, however, we say that somethinghas a frequency of so many hertz, as in 60 Hz, we willalways mean cycles/second.

λOne wavelength l = the distance between similar crests

Electricfield

Magneticfield

Fields move as a fixed unitat the speed of light.

a) Electric and magnetic fields produced by abruptly switching the antenna current.

b) Electric and magnetic fields produced by smoothly switching the antenna current.

E

B

c c

cc

Figure 23Structure of electric and magnetic fields in light and radio waves.

formulas T = λ /c, f = c/λ . (It is now common to use“hertz” or “Hz” for the dimensions of frequency. Thisis a classic example of ruining simple dimensionalanalysis by using people’s names.) Finally, the angularfrequency ω radians per second is defined as

ω

radianssec ond

= 2πradianscycle

× fcyclessecond

= 2πfradianssecond

(39)


ELECTROMAGNETIC SPECTRUMWe have seen by direct calculation that the electromag-netic pulse of Figure (17), and the series of pulses inFigure (22) are a solution of Maxwell’s equations. It isnot much of an extension of our work to show that thesinusoidal wave structure of Figure (23b) is also asolution. The fact that all of these structures move at aspeed c = 1/ µ0ε0 = 3 × 108 m/s is what suggestedto Maxwell that these electromagnetic waves werelight, that he had discovered the theory of light.

But there is nothing in Maxwell’s equations that re-stricts our sinusoidal solution in Figure (23b) to certainvalues or ranges of frequency or wavelength. Onehundred years before Maxwell it was known frominterference experiments (which we will discuss in thenext chapter) that light had a wave nature and that thewavelengths of light ranged from about 6 × 10–5cm inthe red part of the spectrum down to 4 × 10–5cm in theblue part. With the discovery of Maxwell’s theory oflight, it became clear that there must be a completespectrum of electromagnetic radiation, from very longdown to very short wavelengths, and that visible lightwas just a tiny piece of this spectrum.

More importantly, Maxwell’s theory provided the clueas to how you might be able to create electromagneticwaves at other frequencies. We have seen that anoscillating current in a wire produces an electromag-netic wave whose frequency is the same as that of thecurrent. If, for example, the frequency of the current is1030 kc (1030 kilocycles) = 1.03 × 106 cycles/sec,then the electromagnetic wave produced should have awavelength

λ meters

cycle=

c meterssecond

fcycles

sec ond

= 3 × 108m/s103 × 106c/s

= 297 meters

Such waves were discovered within 10 years ofMaxwell’s theory, and were called radio waves. Thefrequency 1030 kc is the frequency of radio stationWBZ in Boston, Mass.

Components of theElectromagnetic SpectrumFigure (24) shows the complete electromagnetic spec-trum as we know it today. We have labeled variouscomponents that may be familiar to the reader. Thesecomponents, and the corresponding range of wave-lengths are as follows:

Radio Waves 106 m to .05 mm

AM Band 500 m to 190 m

Short Wave 60 m to 15 m

TV VHF Band 10 m to 1 m

TV UHF Band 1 m to 10 cm

Microwaves 10 cm to .05 mm

Infrared Light .05 mm to 6×10– 5 cm

Visible Light 6×10– 5 to 4×10– 5cm

Ultraviolet Light 4×10– 5 cm to 10– 6cm

X Rays 10 –6 cm to 10 –9 cm

γ Rays 10 –9 cm and shorter

infrared rays

10110 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -126 5 4 3 210

X-rays

wavelength, cm

lightvisible ultraviolet

raysradio, television, radar gamma rays

Figure 24The electromagnetic spectrum extends from long wavelength radio waves down to short wavelengthX rays and gamma rays. The visible part of the spectrum is indicated by the small box.

32-21

In each of these ranges, the most efficient way to emitor detect the radiation is to use antennas whose size iscomparable to the wavelength of the radiation. Forradio waves the antennas are generally some kind of astructure made from wire. In the infrared and thevisible region, radiation is generally emitted by mol-ecules and atoms. The short wavelength x rays and γrays generally come from atomic nuclei or subatomicparticles.

The longest wavelength radio waves that have beenstudied are the so-called “whistlers”, radio waves withan audio frequency, that are produced by lighteningbolts and reflected back and forth around the earth bycharged particles trapped in the earth’s magnetic field.On a shorter scale of distance are the long wavelengthradio waves which penetrate the ocean and are used forcommunications with submarines. The radio station inCutler, Maine, shown in Figure (25), has twenty-sixtowers over 1000 feet tall to support the antenna toproduce such waves. This station, operated by theUnited States Navy, is the world’s most powerful.

As we go to shorter wavelengths and smaller antennas,we get to the broadcast band, short wave radio, then tothe VHF and UHF television frequencies. (FM radiois tucked into the VHF band next to Channel 6). Thewavelengths for VHF television are of the order of

meters, while those for UHF are of the order of a foot.Those with separate VHF and UHF television antennaswill be familiar with the fact that the UHF antenna,which detects the shorter wavelengths, is smaller insize.

Adjusting the rabbit ears antenna on a television setprovides practical experience with the problems ofdetecting an electromagnetic wave. As the TV signalstrikes the antenna, the electric field in the wave acts onthe electrons in the TV antenna wire. If the wire isparallel to the electric field, the electrons are pushedalong in the wire producing a voltage that is detected bythe television set. If the wire is perpendicular, theelectrons will not be pushed up and down and novoltage will be produced.

The length of the wire is also important. If the antennawere one half wavelength, then the electric field at oneend would be pushing in the opposite direction from thefield at the other end, the integral E ⋅ dl down theantenna would be zero, and you would get no netvoltage or signal. You want the antenna long enoughto get a big voltage, but not so long that the electric fieldin one part of the antenna works against the field inanother part. One quarter wavelength is generally theoptimum antenna length.

Figure 25The worlds largest radio station at Cutler, Maine. This structure, with 75 miles of antenna wire and 26 towersover 1000 ft high, generates long wavelength low frequency, radio waves for communications with submarines.


The microwave region, now familiar from microwavecommunications and particularly microwave ovens,lies between the television frequencies and infraredradiation. The fact that you heat food in a microwaveoven emphasizes the fact that electromagnetic radia-tion carries energy. One can derive that the energydensity in an electromagnetic wave is given by theformula

energy density in anelectromagneticwave

=ε0E2

2+

B2

2µ0(37)

We have already seen the first term ε0E2/2, when wecalculated the energy stored in a capacitor (seeEquation 27-36 on page 27-19). If we had calculatedthe energy to start a current in an inductor, we wouldhave gotten the formula B2 /2µ0 for the energydensity in that device. Equation (37) tells us theamount of energy is associated with electric andmagnetic fields whenever we find them.

Blackbody RadiationAtoms and molecules emit radiation in the infrared,visible and ultraviolet part of the spectrum. One of themain sources of radiation in this part of the spectrum isthe so-called blackbody radiation emitted by objectsdue to the thermal motion of their atoms and molecules.

If you heat an iron poker in a fire, the poker first getswarm, then begins to glow a dull red, then a bright redor even, orange. At higher temperatures the pokerbecomes white, like the filaments in an electric lightbulb. At still higher temperatures, if the poker did notmelt, it would become bluish. The name blackbodyradiation is related to the fact that an initially cold,black object emits these colors of light when heated.

There is a well studied relationship between the tem-perature of an object and the predominant frequency ofthe blackbody radiation it emits. Basically, the higherthe temperature, the higher the frequency. Astrono-mers use this relationship to determine the temperatureof stars from their color. The infrared stars are quitecool, our yellow sun has about the same temperature asthe yellow filament in an incandescent lamp, and theblue stars are the hottest.

All objects emit blackbody radiation. You, yourself,are like a small star emitting infrared radiation at awavelength corresponding to a temperature of 300K.In an infrared photograph taken at night, you wouldshow up distinctly due to this radiation. Infraredphotographs are now taken of houses at night to showup hot spots and heat leaks in the house.

Perhaps the most famous example of blackbody radia-tion is the 3K cosmic background radiation which is theremnant of the big bang which created the universe.We will say much more about this radiation in Chapter34.

UV, X Rays, and Gamma RaysWhen we get to wavelengths shorter than the visiblespectrum, and even in the visible spectrum, we begin torun into problems with Maxwell’s theory of light.These problems were first clearly displayed by MaxPlanck who in 1900 developed a theory that explainedthe blackbody spectrum of radiation. The problemwith Planck’s theory of blackbody radiation is that itcould not be derived from Maxwell’s theory of lightand Newtonian mechanics. His theory involved arbi-trary assumptions that would not be understood foranother 23 years, until after the development of quan-tum mechanics.

Despite the failure of Newton’s and Maxwell’s theo-ries to explain all the details, the electromagneticspectrum continues right on up into the shorter wave-lengths of ultraviolet (UV) light, then to x rays andfinally to γ rays. Ultraviolet light is most familiar fromthe effect it has on us, causing tanning, sunburns, andskin cancer depending on the intensity and duration ofthe dose. The ozone layer in the upper atmosphere, aslong as it lasts, is important because it filters out muchof the ultraviolet light emitted by the sun.

X rays are famous for their ability to penetrate flesh andproduce photographs of bones. These rays are usuallyemitted by the tightly bound electrons on the inside oflarge atoms, and also by nuclear reactions. The highestfrequency radiation, γ rays, are emitted by the smallestobjects—nuclei and elementary particles.

32-23

POLARIZATIONOne of the immediate tests of our picture of a light orradio wave, shown in Figure (23), is the phenomena ofpolarization. We mentioned that the reason that youhad to adjust the angle of the wires on a rabbit earsantenna was that the electric field of the televisionsignal had to have a significant component parallel tothe wires in order to push the electrons up and down thewire. Or, in the terminology of the last few chapters, weneeded the parallel component of E so that the voltageV = E ⋅ dl would be large enough to be detected bythe television circuitry. (In this case, the line integralE ⋅ dl is along the antenna wire.) Polarization is a

phenomena that results from the fact that the electricfield E in an electromagnetic wave can have variousorientations as the wave moves through space.

Although we have derived the structure of an electro-magnetic wave for the specific case of a wave producedby an alternating current in a long, straight wire, someof the general features of electromagnetic waves areclearly present in our solution. The general featuresthat are present in all electromagnetic waves are:

1) All electromagnetic waves are a structure consist-ing of an electric field E and a magnetic field B .

2) E and B are at right angles to each other asshown in Figure (23).

3) The wave travels in a direction perpendicular tothe plane of E and B .

4) The speed of the wave is c = 3××108m/s.

5) The relative strengths of E and B are given byEquation (36) as B = E/c.

Even with these restrictions, and even if we consideronly flat or plane electromagnetic waves, there are stillvarious possible orientations of the electric field asshown in Figure (26). In Figure (26a) we see a planewave with a vertical electrical field. This would becalled a vertically polarized wave. In Figure (26b),where the electric field is horizontal, we have a hori-zontally polarized wave. By convention we say that thedirection of polarization is the direction of the electricfield in an electromagnetic wave.

Because E must lie in the plane perpendicular to thedirection of motion of an electromagnetic wave, E hasonly two independent components, which we cancall the vertical and horizontal polarizations, or the xand y polarizations as shown in Figures (27a) and (27b)respectively. If we happen to encounter an electromag-netic wave where E is neither vertical or horizontal,but at some angle θ, we can decompose E into its x andy components as shown in (27c). Thus we can considera wave polarized at an arbitrary angleθ as a mixture ofthe two independent polarizations.

Figure 26Two possible polarizations ofan electromagnetic wave.

Ex

yE

E

E

E

a) Vertical Polarization

b) Horizontal Polarization

c) Mixture

θ

Figure 27We define the directionof polarization of anelectromagnetic wave asthe direction of theelectric field.

Magneticfield

a) Vertically polarized electromagnetic wave.

b) Horizontally polarized electromagnetic wave.

Electricfield

Magneticfield

Electricfield

E

E

B

B

c c

c c


PolarizersA polarizer is a device that lets only one of the twopossible polarizations of an electromagnetic wave passthrough. If we are working with microwaves whosewavelength is of the order of a few centimeters, a framestrung with parallel copper wires, as seen in Figure(28), makes an excellent polarizer. If a verticallypolarized wave strikes this vertical array of wires, theelectric field E in the wave will be parallel to the wires.This parallel E field will cause electrons to move upand down in the wires, taking energy out of the incidentwave. As a result the vertically polarized wave cannotget through. (One can observe that the wave is actuallyreflected by the parallel wires.)

If you then rotate the wires 90º, so that theEfield in thewave is perpendicular to the wires, the electric field canno longer move electrons along the wires and the wireshave no effect. The wave passes through withoutattenuation.

If you do not happen to know the direction of po-larization of the microwave, put the polarizer in thebeam and rotate it. For one orientation the microwavebeam will be completely blocked. Rotate the polarizerby 90° and you will get a maximum transmission.

Figure 28Microwave polarizer, made from an array of copper wires. The microwavetransmitter is seen on the other side of the wires, the detector is on this side.When the wires are parallel to the transmitted electric field, no signal isdetected. Rotate the wires 90 degrees, and the full signal is detected.

32-25

Light PolarizersWe can picture light from the sun as a mixture of lightwaves with randomly oriented polarizations. (The Efields are, of course, always in the plane perpendicularto the direction of motion of the light wave. Only theangle in that plane is random.) A polarizer made of anarray of copper wires like that shown in Figure (28),will not work for light because the wavelength of lightis so short λ ≈ 5 × 10-5 cm that the light passes rightbetween the wires. For such a polarizer to be effective,the spacing between the wires would have to be of theorder of a wavelength of light or less.

A polarizer for light can be constructed by imbeddinglong-chain molecules in a flexible plastic sheet, andthen stretching the sheet so that the molecules arealigned parallel to each other. The molecules act likethe wires in our copper wire array, but have a spacingof the order of the wavelength of light. As a result themolecules block light waves whose electric field isparallel to them, while allowing waves with a perpen-dicular electric field to pass. (The commercial name forsuch a sheet of plastic is Polaroid.)

Since light from the sun or from standard electric lightbulbs consists of many randomly polarized waves, asingle sheet of Polaroid removes half of the waves nomatter how we orient the Polaroid (as long as the sheetof Polaroid is perpendicular to the direction of motionof the light beam). But once the light has gone throughone sheet of Polaroid, all the surviving light waves havethe same polarization. If we place a second sheet ofPolaroid over the first, all the light will be absorbed ifthe long molecules in the second sheet are perpendicu-lar to the long molecules in the first sheet. If the longmolecules in the second sheet are parallel to those in thefirst, most of the waves that make it through the first,make it through the second also. This effect is seenclearly in Figure (29).

Figure 29Light polarizers. Two sheets of polaroid are placed ontop of a drawing. On the left, the axes of the sheets areparallel, so that nearly half the light passes through. Onthe right, the axes are perpendicular, so that no lightpasses through. (Photo from Halliday & Resnick)


Magnetic Field DetectorSo far, our discussion of electromagnetic radiation hasfocused primarily on detecting the electric field in thewave. The rabbit ear antenna wire had to be partiallyparallel to the electric field so that E ⋅d and there-fore the voltage on the antenna would not be zero. Inour discussion of polarization, we aligned the parallelarray of wires or molecules parallel to the electric fieldwhen we wanted the radiation to be reflected or ab-sorbed.

It is also fairly easy to detect the magnetic field in aradio wave by using one of our E ⋅d meters to detecta changing magnetic flux (an application of Faraday’slaw). This is the principle behind the radio directionfinders featured in a few World War II spy pictures.

loop antenna

Figure 30bCar driving toward radio transmitter.

Figure 30aCar with radio direction finderloop antenna mounted on top.

vB

circular magnetic fieldradiated by the antenna

detector loopon carvertical antenna

In a typical scene we see a car with a metal loopmounted on top as shown in Figure (30a). It is chasinganother car with a hidden transmitter, or looking for aclandestine enemy transmitter.

If the transmitter is a radio antenna with a verticaltransmitting wire as shown in Figure (30b), the mag-netic field of the radiated wave will be concentriccircles as shown. Objects on the ground, the grounditself, and nearby buildings and hills can distort thispicture, but for now we will neglect the distortions.

Figure 31In a January 1998 National Geographicarticle on Amelia Earhardt, thereappeared a picture of a vintage Electraairplane similar to the one flown byEarhardt on her last trip in 1938. On thetop of the plane, you can see the kind ofradio direction finder we have beendiscussing. (The plane is being flown byLinda Finch.)

32-27

c

E

BMagneticfield

Electricfield

E

B

cc

cc

a) Loop oriented so that no magnetic flux goes through it

b) Loop oriented so that magnetic flux goes through it

cc

Figure 32Electromagnetic field impinging upon a loop antenna.In (a), the magnetic field is parallel to the plane of theloop, and therefore no magnetic flux goes through theloop. In (b), the magnetic flux goes through the loop.As the wave passes by, the amount of flux changes,inducing a voltage in the loop antenna.

Vvoltmeter

metal loop

Figure 33We can think of a wire loop connectedto a voltmeter as an E⋅d meter. Anychanging magnetic flux through theloop induces a voltage around the loop.This voltage is read by the voltmeter.

The most sensitive way to use this radio direction finderis to get a zero or “null” reading on the voltmeter. Onlywhen the loop is oriented as in Figure (32a), with itsplane perpendicular to the direction of motion of theradio wave, will we get a null reading. At any otherorientation some magnetic flux will pass through theloop and we get some voltage.

Spy pictures, set in more modern times, do not showantenna loops like that in Figure (30) because modernradio direction finders use so-called “ferrite” antennasthat detect the electric field in the radio wave. We geta voltage on a ferrite antenna when the electric field inthe radio wave has a component along the ferrite rod,just as it needed a component along the wires of a rabbitears antenna. Again these direction finders are mostaccurate when detecting a null or zero voltage. Thisoccurs only when the rod is parallel to the direction ofmotion of the radio wave, i.e. points toward the station.(This effect is very obvious in a small portable radio.You will notice that the reception disappears and youget a null detection, for some orientations of the radio.)

In Figures (32a) and (32b), we show the magnetic fieldof the radio wave as it passes the detector loop mountedon the car. A voltmeter is attached to the loop as shownin Figure (33). In (32a), the plane of the loop is parallelto B, the magnetic flux ΦB through the loop is zero, andFaraday’s law gives

V = E ⋅d =

dΦB

dt= 0

In this orientation there is no voltage reading on thevoltmeter attached to the loop.

In the orientation of Figure (32b), the magnetic fieldpasses through the loop and we get a maximum amountof magnetic flux ΦB. As the radio wave passes by theloop, this flux alternates signs at the frequency of thewave, therefore the rate of change of flux dΦB/dt is ata maximum. In this orientation we get a maximumvoltmeter reading.

E ⋅d meter


RADIATED ELECTRIC FIELDSOne of the best computer simulations of physicalphenomena is the series of short films about the electricfields produced by moving and accelerated charges.We will describe a few of the frames from these films,but nothing replaces watching them.

Two basic ideas underlie these films. One is Gauss’law which requires that electric field lines not break,do not end, in empty space. The other is that distur-bances on an electric field line travel outward at thespeed of light. No disturbance, no change in the electricfield structure, can travel faster than the speed of lightwithout violating causality. (You could get answers toquestions that have not yet been asked.)

As an introduction to the computer simulations ofradiation, let us see how a simple application of thesetwo basic ideas leads to the picture of the electromag-netic pulse shown back in Figure (16). In Figure (34a)we show the electric field of a stationary, positivelycharged rod. The electric field lines go radially out-ward to infinity. (It’s a long rod, and it has been at restfor a long time.)

At time t = 0 we start moving the entire rod upward ata speed v. By Gauss’ law the electric field lines muststay attached to the charges Q in the rod, so that the endsof the electric field lines have to start moving up withthe rod.

No information about our moving the rod can traveloutward from the rod faster than the speed of light. Ifthe time is now t > 0, then beyond a distance ct, theelectric field lines must still be radially outward as inFigure (34b). To keep the field lines radial beyondr = ct, and keep them attached to the charges +Q in therod, there must be some kind of expanding kink in thelines as indicated.

At time t = t1, we stop moving the positively chargedrod. The information that the charged rod has stoppedmoving cannot travel faster than the speed of light, thusthe displaced radial field next to the rod cannot be anyfarther out than a distance c t- t1 as shown in Figure(34c). The effect of starting, then stopping the positiverod is an outward traveling kink in the electric fieldlines. It is as if we had ropes attached to the positive rod,and jerking the rod produced an outward traveling kinkor wave on the ropes.

In Figure (34d), we have added in a stationary nega-tively charged rod and the inward directed electric fieldproduced by that rod. The charge density on thenegative rod is opposite that of the positive rod, so thatthere is no net charge on the two rods. When wecombine these rods, all we have left is a positiveupward directed current during the time interval t = 0 to

t = t1. We have a short current pulse, and the electricfield produced by the current pulse must be the vectorsum of the electric fields of the two rods.

In Figure (34e), we add up the two electric fields. In theregion r > ct beyond the kink, the positive and negativefields must cancel exactly. In the region r < t - t1 weshould also have nearly complete cancellation. Thusall we are left with are the fields E+ and E– inside thekink as shown in Figure (34f). Since electric field linescannot end in empty space, E+

and E– must add up toproduce the downward directed Enet shown in Figure(34g). Note that this downward directed electric fieldpulse was produced by an upward directed currentpulse. As we have seen before, this induced electricfield opposes the change in current.

In Figure (34h) we added the expanding magnetic fieldpulse that should be associated with the current pulse.What we see is an expanding electromagnetic pulsethat has the structure shown in Figure (16). Simplearguments based on Gauss’ law and causality gave usmost of the results we worked so hard to get earlier.What we did get earlier, however, when we appliedAmpere’s and Faraday’s law to this field structure, wasthe explicit prediction that the pulse expands at thespeed 1/ µ0ε0 = 3 × 108 m/s.

32-29

++++++++++

t < 0

t > 0

++++++++++

ct

++++++++++

t > t1

ct

c(t–t1)

++++++++++

–

–––––––

––

v b) At time t = 0, westart moving the entirerod upward at a speedv. The ends of the fieldlines must stayattached to the chargesin the rod.

c) We stop moving therod at time t = t1 . Noinformation about ourhaving moved the rodcan travel out fasterthan the speed of light.

e) When we add up theelectric fields of thepositive and negativerods, the fields canceleverywhere except atthe outward goingpulse.

g) Thus a shortupward directedcurrent pulseproduces a downwarddirected electric fieldthat travels outwardfrom the wire at aspeed c.

h) Add in the magneticfield of the currentpulse, and we have theelectromagnetic wavestructure seen inFigure (16).

a) Electric field of astationary, positivelycharged rod.

f) At the pulse, thevector sum of E+ and

E– is a downwarddirected field Enet asshown.

d) We have added inthe electric field of aline of stationarynegative charges. As aresult, the net chargeon the rod is zero andwe have only a currentpulse that lasted fromt = 0 to t = t1 .

Figure 34Using the fact that electric field lines cannot break in empty space (Gauss' Law), and the idea that kinksin the field lines travel at the speed of light, we can guess the structure of an electromagnetic pulse.

++–

E+

E–

Enet

E+

E–

Enet

B

Enet

c

E+

E–

c


Figure 35cField of a charge that stopped. Assume that thecharge stopped t seconds ago. Inside a circle ofradius ct, we have the field of a stationary charge.Outside, where there is no information that thecharge has stopped, we still have the field of amoving charge. The kink that connects the twofields is the electromagnetic radiation.

Figure 35aElectric field of a stationary charge.

Figure 35bElectric field of a moving charge. If the charge hasbeen moving at constant speed for a long time, the fieldis radial, but squeezed up at the top and bottom.

Field of a Point ChargeThe computer simulations show the electric field of apoint charge under varying situations. In the first, wesee the electric field of a point charge at rest, as shownin Figure (35a). Then we see a charge moving atconstant velocity v. As the speed of the charge ap-proaches c, the electric field scrunches up as shown inFigure (35b).

The next film segment shows what happens when wehave a moving charge that stops. If the charge stoppedat time t = 0, then at a distance r = ct or greater, we musthave the electric field of a moving charge, because noinformation that the charge has stopped can reachbeyond this distance. In close we have the electric fieldof a static charge. The expanding kink that connects thetwo regions is the electromagnetic wave. The result isshown in Figure (35c). The final film segment showsthe electric field of an oscillating charge. Figure (36)shows one frame of the film. This still picture does aserious injustice to the animated film. There is nosubstitute, or words to explain, what you see and feelwhen you watch this film.

v

32-31

Figure 36Electric field of an oscillating charge.

Figure 35c (enlarged)Electric field of a charge that stopped. Thedotted lines show the field structure wewould have seen had the charge not stopped.


Figure 37aElectric field emerging from ping pong ball.

E

Ball

B

Tube(end view)

E

Ball

Figure 37cElectric field emerging above ping pong ball.

Figure 37bMagnetic field emerging from ping pong ball.

Exercise 6Assume that we have a supply of ping pong balls andcardboard tubes shown in Figures (37). By looking atthe fields outside these objects decide what could beinside producing the fields. Explicitly do the followingfor each case.

i) Write down the Maxwell equation which you used todecide what is inside the ball or tube, and explain howyou used the equation.

ii) If more than one kind of source could produce thefield shown, describe both (or all) sources and show theappropriate Maxwell equations.

iii) If the field is impossible, explain why, using aMaxwell equation to back up your explanation.

In each case, we have indicated whether the source isin a ball or tube. Magnetic fields are dashed lines,electric fields are solid lines, and the balls and tubes aresurrounded by empty space.

32-33

B

Tube(end view)

x x

x x

x xx x

xx

xxx

xxxx

x

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xxxx

xx

xx x

x xx x

x

xx

xx

xx

xx

xx

xx

xx

xx

x x

x x

x x

x = electric field into paper

Figure 37fFor this example, explain what is happening to thefields, what is in the tube, and what happened inside.

E

B

Tube(end view)

Figure 37gThere is only ONE object inside this tube.What is it? What is it doing?

E

Tube(end view)

ETube(end view)

Figure 37dElectric field around tube.

Figure 37eElectric field passing through tube.


IndexAAmpere's law

Maxwell’s correction to 32- 4Analysis

Of path 1 for electromagnetic pulse 32- 14Of path 2 for electromagnetic pulse 32- 16

BBlackbody radiation

Electromagnetic spectrum 32- 22

CCapacitor

Magnetic field in 32- 6Colors

Blackbody radiation 32- 22Electromagnetic Spectrum 32- 20

DDetector for radiated magnetic field 32- 26

EElectric field

And light 32- 20In electromagnetic waves 32- 18Radiation by line charge 32- 28Radiation by point charge 32- 30

Electromagnetic radiationPulse 32- 10

Analysis of path 1 32- 14Analysis of path 2 32- 16Calculation of speed 32- 14

Electromagnetic spectrum 32- 20Electromagnetic waves 32- 18

FField

ElectricRadiation by line charge 32- 28Radiation by point charge 32- 30

Electromagnetic field 32- 18Magnetic field

Between capacitor plates 32- 6Detector, radio waves 32- 26Gauss's law for (magnetic monopole) 32- 2Surface integral 32- 2Thought experiment on radiated field 32- 11

Flux, magneticMagnetic field detector 32- 26Maxwell's equations 32- 8

Force

Lorentz force law 32- 8

GGamma rays 32- 20, 32- 22

Wavelength of 32- 20Gauss' law

For magnetic fields 32- 2

IInfrared light

Wavelength of 32- 20Integral, line

In Maxwell's equations 32- 8Integral, surface

For magnetic fields 32- 2In Maxwell's equations 32- 8

LLight

Electromagnetic spectrum 32- 20Electromagnetic waves 32- 18Polarization of light 32- 23Polarizers 32- 25Radiated electric fields 32- 28Speed of light

Electromagnetic pulse 32- 14Structure of electromagnetic wave 32- 19

Lorentz force lawAnd Maxwell's equations 32- 8

MMagnetic field

Between capacitor plates 32- 6Detector 32- 26Gauss' law for 32- 2In electromagnetic waves 32- 18Radiated, a thought experiment 32- 11Surface integral of 32- 2

Magnetic fluxMagnetic field detector 32- 26Maxwell's equations 32- 8

Maxwell’s correction to Ampere’s law 32- 4Maxwell's equations 32- 8

Chapter on 32- 1In empty space 32- 10Symmetry of 32- 9

Maxwell’s theory of light 32- 2Maxwell's wave equation 32- 18Microwaves

Electromagnetic spectrum 32- 20Microwave polarizer 32- 24

PPolarization of light waves 32- 23Polarizer

Light 32- 25Microwave 32- 24

32-35

Polaroid, light polarizer 32- 25Pulse

Of electromagnetic radiation 32- 10

RRadar waves

Wavelength of 32- 20Radiated electromagnetic pulse 32- 10Radiated magnetic field thought experiment 32- 11Radiation

Blackbody 32- 22Electromagnetic field

Analysis of path 1 32- 14Analysis of path 2 32- 16Calculation of speed 32- 14Spectrum of 32- 20

Radiated electric fields 32- 28Radiated field of point charge 32- 30UV, X Rays, and Gamma Rays 32- 22

Radio wavesIn the electromagnetic spectrum 32- 20Wavelength of 32- 20

Relativistic physicsElectromagnetic radiation, structure of 32- 19Maxwell’s equations 32- 8Radiated electric fields 32- 28Speed of light wave 32- 17Thought experiment on expanding magnetic

field 32- 11

SSpectrum

ElectromagneticWavelengths of 32- 20

Speed of an electromagnetic pulse 32- 14Speed of light

Calculation of speed of lightAnalysis of path 1 32- 14Analysis of path 2 32- 16Using Maxwell's Equations 32- 14

Surface integralFor magnetic fields 32- 3In Maxwell's equations 32- 8

Symmetry of Maxwell’s equations 32- 9

TTelevision waves

Wavelength of 32- 20

UUltraviolet light

Electromagnetic spectrum 32- 20, 32- 22Wavelength of 32- 20

VVisible light 32- 20

Wavelength of 32- 20

WWave

Electromagnetic waves 32- 18Light waves

Polarization of 32- 23

XX-Ch32

Exercise 1 32- 7Exercise 2 32- 8Exercise 3 32- 9Exercise 4 32- 17Exercise 5 32- 17Exercise 6 32- 32

X-raysElectromagnetic spectrum 32- 20, 32- 22Wavelength of 32- 20

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