transmission lines and e.m. waves lec 19

8/20/2019 Transmission Lines and E.M. Waves Lec 19

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Transmission Lines and E. M. Waves

Prof. R. K. Shevgaonkar

Department of Electrical Engineering

Indian Institte of Technolog!" #om$a!

Lectre % &'Ma()ell*s E+ations

In the last lecture we got the mathematical formulation of the physical laws of

electromagnetics that were the Gauss’ law, Ampere’s circuit law and Faraday’s law of

electromagnetic induction. The compilation of these physical laws in mathematical form

is called the Maxwell’s equations. o Maxwell essentially compiled these laws, ga!e the

mathematical representation" howe!er, while doing this we found a difficulty. #e found

there was inconsistency in the Ampere’s circuit law. $et us understand what was the

difficulty which was encountered %y Maxwell while compiling these equations.

o let us consider a closed surface of certain !olume let us say & and let us say these are

charge distri%uted inside this surface then there is a possi%ility that the charges might

lea!e this surface and when the charges will lea!e there will %e a rate of change of charge

so there will %e current flow from this surface. o we will ha!e a current density

distri%uted on the surface and whate!er charges will %e lea!ing in the form of current that

should %e reflected in the reduction of the net charge inside this surface.

o let us say we ha!e the conduction current density on the surface which is gi!en %y ',

then as we saw from the concept of di!ergence that if you ta(e this !ector field ' the

di!ergence if ' essentially is a net flux coming out of this !olume. o if I )ust ta(e a

di!ergence of ' that essentially should represent li(e the outflow of this !ector field

which is the conduction current density. *efore getting into that let us say that I ha!e a

small infinitesimal area on this surface which if gi!en %y da. o, if I ta(e the dot product

of conduction current density and the area that gi!es me the outward current from this

infinitesimally small area and if I integrate that o!er the entire surface I get the net

outward current from this surface.


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o I ha!e here from this surface the net outward current which is nothing %ut integrated

o!er the closed surface let us say the surface is , the conduction current density o!er that

infinitesimal area. o this integral as we saw earlier also that it represents the total current

coming out of this surface. As we (now the current is nothing %ut the rate of change of

charge and since this current is coming outwards from the surface it should %e equal to

the rate of charge from this !olume and since the current is coming outwards the net

charge must %e decreasing inside this !olume.

+efer lide Time -/-01

o if I say that this !olume is ha!ing a charge density which is gi!en as rho and this is the

conduction current density then the net decrease or rate of decrease of charge inside this

!olume must %e same as the total current which is coming from this surface. o if I ha!e

the charge density rho then the total charge enclosed %y this surface will %e the integral of

rho o!er the !olume. o if I ta(e now then rate of decrease of charge rate of change of

charge in the !olume that quantity will %e equal to minus d %y dt for rate of change of the

total charge enclosed in this !olume so that is the integral rho into d.


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+efer lide Time/ 2/341

5hat we are now saying these that, simple %y definition of the current, that is the current

leads the rate of change of charge, this net flow of the current from this surface that

should %e equal to the rate of change of charge from this is !olume. o essentially from

this we get that the closed surface integral o!er , this is o!er !olume ' dot da that is

equal to minus d %y dt of integral o!er the !olume rho into d!.

If I say the !olume is not changing with time as we assumed earlier we can change the

charges are changing as a function of time, the !olume is fixed, we can interchange the

integral dt so that quantity that we can get as minus integral o!er the !olume d rho %y dt

into d!.

6ow as we ha!e done earlier we can con!ert this surface integral +efer lide Time/

7/801 into the !olume integral %y applying the di!ergence theorem. o this thing %y

applying di!ergence theorem we can write as... from the di!ergence theorem or using

di!ergence theorem we can get this integral as the !olume integral & del dot ' d! that is

equal to minus the triple integral o!er the !olume d rho %y dt o!er the !olume.


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6ow I can %ring this term on this side and write the integral as the triple integral o!er the

!olume del dot ' plus d rho %y dt o!er the !olume that should %e equal to -. And as we

did in other equations, since this relation should %e true for any ar%itrary !olume this

quantity must %e identically 9ero so the integrand must %e identically 9ero so from here

essentially we get that del dot ' that is equal to minus d rho %y dt" this equation is called

the continuity equation.

o if we ha!e the time !arying charges then this quantity is the finite quantity and then

the di!ergence of the conduction current density is not 9ero" if we ta(e a static case where

the charges are not changing then the di!ergence of the conduction current density is

identically 9ero.


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+efer lide Time -:/71

And that ma(es physically sense %ecause if you are saying the di!ergence of ' as we saw

earlier from the concept of di!ergence the di!ergence tells you the net quantity coming

out of this !olume and if the net current is coming out of this !olume that means there

must %e... the net charges must %e lea!ing this !olume so there must %e a change in the

total charge or the charge density inside that !olume. If the charges are not changing

inside the !olume then whate!er charges are entering in the !olume those charges must

%e lea!ing so the net charge inside the !olume remains same and in that case the

di!ergence of the conduction current density is equal to 9ero %ecause there is no net flow

from this !olume for the current.

o whene!er we are ha!ing in general the time !arying quantities the quantitati!e

equation must %e satisfied %y the conduction current density and the charges. ;recisely

this was the equation which led to difficulties for compiling the other equation %y

Maxwell. $et us see what the pro%lem is.

5e ha!e seen earlier that from the Ampere circuit law we get an equation which is del

cross h that is equal to ' +efer lide Time/


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without getting into physical aspects we can simply say I can ta(e the di!ergence of this

equation on %oth sides so I can say di!ergence of curl of # is equal to di!ergence of '. o

if I )ust ta(e di!ergence I will get from this relation del dot del cross # that is equal to

dell dot '. 5e (now in the !ector product we can interchange the sign dot and cross so

this thing can %e also written as/ del cross del dot # equal to del dot '.

+efer lide Time/


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+efer lide Time/


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+efer lide Time/


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+efer lide Time nce we say that then what we are saying is now the net current which we are tal(ing

a%out is not only the conduction current %ut it is the conduction current plus the

displacement current. And then Maxwell said/ if you incorporate this term appropriately

for defining the total current then the Ampere’s law can %e modified to say that the

magneto moti!e force around a closed loop is equal to the total current enclosed %y that

loop which includes the conduction current as well as the displacement current.

o earlier since this concept was not there the current was always meant %y the flow of

the charges. #owe!er, in this case there is no flow of charges that e!en if you... I do not

ha!e charges still I will get this quantity %ecause this quantity = is related to the electric

field. o, if I ha!e an electric field which is time !arying this term will %e equi!alent to a

current flow. o this quantity which is simply representing the rate of change of electric

field represents the quantity which is equi!alent to current.

o, if I ta(e the total current which is a com%ination of the conduction current which is

%ecause of the moment of the charges and %ecause of the time !arying electric field, this

total current is equal to the magneto moti!e force around the closed loop. And once you

do that then this equation of the Ampere’s circuit law is modified %ecause now this


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current is %eing changed to the conduction current density plus the displacement current

density.

o, once you get that then the Ampere’s law now has %een modified is modified to del

cross # that is equal to the conduction current density as we had earlier plus we are

ha!ing now an additional term which is d= %y dt. That essentially resol!es the difficulties

which were faced %y Maxwell and now this gi!es you the complete description of the

phenomena of electromagnetics.

o essentially now the four equations which you ha!e got/ the Gauss’ law equation, the

Gauss’ law equi!alent applied to the magnetic fields for magnetic charges, the Faraday’s

law of electromagnetic induction and the Ampere’s law appropriately modified to

accommodate what is called the displacement current density ma(es the complete set of

equations which represent the static and time !arying electric and magnetic fields. o

these equations are called the Maxwell’s equations.

o as we mentioned earlier the Maxwell’s equations can %e written in the differential

form or they can %e written in the integral form and depending upon the suita%ility the

equations can %e used either in differential form or they can %e used in the integral form.

o we finally write the four Maxwell’s equations. 5e can write this in two forms so we

can get these equations in differential and integral form.

o I write here the differential form or I can write this equation in the integral form. o if

I go to the the first equation which is the Gauss’ law that gi!es you del dot = that is equal

to rho. The same equation when written in the integral form it says that = dot da o!er the

surface that is equal to the rho into d!. The second equation when applied to the magnetic

charges that gi!es me del dot * is equal to - and that can %e written in the integral form

* dot da that is equal to -.


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The third equation which is Faraday’s law of electromagnetic induction that gi!es me del

cross ? that is equal to minus d* %y dt and the integral form of it can %e written as

integral o!er the contour ? dot d@ that is equal to minus d* %y dt dot da.

And finally the Ampere’s law which we ha!e got after introducing displacement current

density that is del cross # will %e equal to ' plus d= %y dt and the integral form the same

thing is o!er this contour # dot d@ that is equal to the total current enclosed so that is '

plus d= %y dt into da.


o now these are the set of equations which go!erns the total phenomena of

electromagnetics for static as well as time !arying fields. o once you ha!e these

generali9ed equations which are for time !arying fields we can reduce the equation for

the static fields that is %y putting all time deri!ati!es to 9ero" I will get the equations for

the static, electric and magnetic fields. o depending upon whether the permitti!ity or the

permea%ility of the medium is !arying as a function of space that means if the medium is

homogeneous I will use this equation, if the permitti!ity is constant in the function of

space then I can write the same equation as we saw earlier" I can ta(e epsilon now and I

can say del dot ? is equal to rho upon epsilon and so on.


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o you may ha!e !arious forms of these equations depending upon the condition applied

to the medium whether we are dealing with the electric fields, magnetic fields or we are

dealing with the displacement !ector and the magnetic flux densities and so on. And also

as we said if we are tal(ing a%out the static fields then the time deri!ati!e should %e put

identically to 9ero so this quantity will %e 9ero for the static field, this quantity will %e

9ero for the static fields.

o if we consider the static fields then this quantity will %e equal to 9ero so the curl of the

electric field will %e always equal to 9ero. o we will see depending upon application

later on that we may put this quantity equal to 9ero if you are dealing with the static

fields. #owe!er, in this particular course which is a course on electromagnetic wa!es we

are dealing with the quantities which are time !arying.

The static fields you would ha!e studied in the earlier course which is now

electromagnetic fields. o here we assume that all the fields and all the quantities which

we are dealing here they are time !arying in general. o the quantity rho, the quantity *,

the quantity =, ', # all these quantities are essentially time !arying quantities. o we will

loo( for the solution of these equations later for the time !arying fields.

6ow as we mentioned earlier that this equation which is in differential form cannot %e

applied in a situation where medium has discontinuity. That means if we tal( a%out the

media interfaces where the medium property suddenly change the permitti!ity may

change or permea%ility may change and there the deri!ati!es the space deri!ati!es cannot

%e defined %ecause the medium is discontinuous so the differential form of this

Maxwell’s equations is not useful in those situations.

#owe!er, as we had mentioned, the integral form is always useful and this can %e applied

in a situation. #owe!er, when we apply the integral form to the discrete media interfaces

we get a relationship %etween the quantities = * ? # in the two media )ust across the

interface and that relationship we call as the %oundary condition. o the same set of

equations when written in differential form they gi!e you what is called point relation so


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these relations are !alid at e!ery point in space. The equation in integral form when

applied to the discrete media interfaces they gi!e you what is called the %oundary

conditions.

+efer lide Time/ 37/


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equal to d. o if the thic(ness is gi!en then I can say if I see from the top that is the

charge I am going to see which is/ rho into = in this area which is < into


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+efer lide Time/ 0


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+efer lide Time 03/


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6ow again if we ma(e this d tending to - and if ' goes to infiniti!e then again this

product ' into = will %e a finite quantity so you will ha!e a current which is truly flowing

on the surface and that current we call as the surface current. o if I ta(e a limit, so if I

define what is called surface current density and that is gi!en %y ' s that is equal to the

limit when d tends to - ' into d times d is not a dot product it is )ust the ' multiplied %y

the area and direction of ' s is same as the direction of ' so this quantity will gi!e me a

current which is flowing on the surface.

And again as I mentioned, this will %e true pro!ided this quantity of the conduction

current density ' is infinite then and then only when d goes to - this product will %e finite

and we will ha!e the surface current density. This again will %e applica%le to the

%oundary which are conducting %oundaries so when the conducti!ity of the medium

%ecomes infinite that time the current which will %e sigma times ? for finite electric field

will %ecome infinite and then you will ha!e what is called the surface current.

ince the unit of the conduction current density is Ampere’s per meter square the unit for

this will %e the Ampere’s per meter square multiplied %y d which is the length so the unit

for ' s will %e Ampere’s per meter" that is the reason this quantity is also called linear


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surface current density %ecause this is now as the per unit length you are defining this

current.


o now we are ha!ing o!erall, the quantities li(e/ the charge density which means

!olume charge density, we ha!e a quantity li(e current density which means conduction

current density, we ha!e a quantity li(e displacement current density then we got surface

charge density and then we ha!e got the surface current density. o one may say these are

the sources which are related to the fields which are the electric and magnetic fields.

o in general then one can esta%lish relationship %etween these quantities which we can

call sources to the fields which are electric and magnetic fields and these relationships are

called the %oundary conditions.


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+efer lide Time 04/


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And in this case this surface is a spherical surface and if the charge is distri%uted in a

spherically symmetric manner the electric displacement will %e uniformly distri%uted on

the surface of this sphere. If I consider this sphere li(e that the electric displacement will

%e uniform all across the sphere.

o now we can find out the electric displacement on the surface of this sphere which is

nothing %ut the total charge enclosed which is this charge B which is pi upon


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+efer lide Time 83/8


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o now the electric field or the electric displacement density is gi!en in the !ectorial

form and you ha!e to find the charge density inside a cu%e of side 3 meters placed

centered at the origin with its side along the coordinate axes.

o now we can apply the Gauss’ law in the differential form to find out first the charge

density and once you get the charge density then we can find out the charge enclosed

inside the cu%e.

o firstly we can go... we (now from the Gauss’ law that the del dot = is equal to the

!olume charge density rho. ?xpanding this in the artesian coordinate system we get d

=x %y dx plus d =y %y dy plus d =9 %y d9 that is equal to the !olume charge density in

that region. u%stituting now for the = which is gi!en here where the x component of =

is x cu%e, y component of = is - and 9 component of = is x square y. If I su%stitute this

into the expression essentially we find that rho is equal to d %y dx of x cu%e, the y

component is - so this quantity is - plus d %y d9 of x square y. ?!en this quantity is -

+efer lide Time/ 8/


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o now inside the !olume of the cu%e the charge density essentially !aries only as a

function of x and the charge density is constant as a function of y and 9. o now if I ha!e

a coordinate system a cu%e essentially is placed something li(e this +efer lide Time/

8/-1 around the center and then you want to find out what is the total charge enclosed

inside this cu%ical !olume.

o you can get the total charge enclosed B that will %e integrating this charge density

o!er the cu%ical !olume that will %e equal to minus < to < minus < to < minus < to < rho

into dx to dy d9. u%stituting for rho in this that will %e minus < < minus < to < minus < to

< 0 x square dx dy d9 which we can write this quantity integration !ersus !ector y will

gi!e me 3 !ersus to 9 will gi!e 3 so essentially we ha!e here


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$et us consider now a pro%lem which is related to the magnetic fields. o let us say now

the pro%lem is in a conducting medium" the magnetic fields is gi!en as # is equal to y

square 9 x cap plus 3 into x plus < y9 y cap minus x plus < 9 square 9 cap. Find the

conduction current density at -.3- minus < meters. Also find the current enclosed %y a

square loop which is defined %y y equal to < x %etween - and < and 9 %etween - and


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o at location 3, -, minus < the conduction current density ' will %e equal to... if I

su%stitute now x equal to 3 y equal to - and y equal to minus < the conduction current

density will %e nce I (now the conduction current density then I can go and find out what is the total

current enclosed %y the loop which is defined %y y equal to < and x going from - to < and

9 going from - to


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%ecause we are su%stituting y equal to < in this dx d9. If I sol!e this integral I get the total

current enclosed inside this loop which will %e 8 upon 0 Amperes.

+efer lide Time 0/-81

o in this case we are using the Ampere’s law the differential form. The magnetic field

was gi!en in the differential form. then %y using the Ampere’s law in differential form we

find out from the curl the conduction current density and once you (now the conduction

current density then we can find out %y integrating o!er the area of the loop the total

current enclosed %y the loop.

o these are some of the !ery simple pro%lems which essentially gi!e me some feel for

how to apply in practical life the laws the physical laws li(e Gauss’ law or the Ampere

law either in differential form or in integral form.

transmission lines and e.m. waves lec 19

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