We have been talking about the propagation
of electromagnetic waves, mostly in vacuum
and also in linear dielectric media. What
we want to do today is to talk about propagation
in a conducting medium and also what happens,
when electromagnetic waves falls on the
interface between dielectric like for example
dielectric like vacuum or air and a
conductor. So, this is basically what we will
be talking about, but let us for the moment
think in general terms and see what happens
when an uniform plane wave is propagating
let us take it along the z direction, and
we will assume systems are linear, so that
we have
electric field, the magnetic field and the
direction of propagation are still mutually
perpendicular as we have talked about.
And for being specific, let us say that the
electric field is along the z direction, the
electric field is along the x direction, the
wave is propagating along the z direction
and
the electric, the magnetic field is along
the y direction.
.
.So, we we come back to the set of Maxwell’s
equations, we had the del cross of E which
is minus d b by d t, but since we have said
this is a linear medium, so we will write
it as
minus mu d H by d t and del cross H is well,
d b by d t which is epsilon d E by d t plus
J.
Now we will assume that our conductor is a
ohmic conductor, so that J is equal to sigma
E is what we will take. So, sigma is the conductivity,
so J is equal to sigma E is what we
have taken. So, since we have said that electric
field is along the x direction, the
magnetic field H is along the y direction
and the direction of propagation is along
the z
direction. Let us take the appropriate components
of this equation. So, the appropriate
component of this equation would be, since
H is along y direction, I will say that del
cross E is y component, I will bring this
term to the other side, mu times d H y by
d t that
is equal to 0 and del cross E is y component
is d by d Z of E x minus d by d x of E z.
But I do not have an E z, so this plus mu
times d H y by d t is equal to 0, this is
one of
the equation. The second equation is obtained
by, rewriting the, this equation because we
have said E is along the x direction, So,
I will take del cross H’s x component.
.
So, I will take del cross H’s x component
and that is equal to epsilon d E x by d t,
I had
sigma E, so which is sigma times E x. So,
del cross H’s x component is d by d y of
H z
and I do not have an H z minus d by d z of
H y and that is equal to whatever we have
written down d E x by d t plus sigma E x.
So, this is the pair of equations I have got
.connecting E x and H y and so let us write
them together. And I will write this as d
by d
z of H y plus epsilon d E x by d t plus sigma
E x equal to 0.
So, these are the two equations, which we
will be handling. This is and this equation.
Now, what we, what to do now is, that since
I know that I am looking for harmonic wave
solutions, so I will assume that E and H go
as E to the power i omega t.
.
So, if you take E x and H y their time dependence
goes as E to the power i omega t. d by
d t essentially means multiplying by i omega,
therefore these pair of equations, the time
dependence is removed I will be rewriting
these two equations like this.
..
So, I have got d E x by d z 
plus i omega mu H y equal to 0 and the other
equation is d by
d z of H y plus I have got two terms in E
there. So, I have got i omega epsilon and
plus
sigma E x is equal to 0. So, this is the pair
of equations that I need to solve. So, as
we
have done several times, the way to solve
these equations coupled equations, is to take
a
further differentiation of any one of these
equation. For example, if I took a
differentiation of the first equation, I will
get d square by d z square of E x plus i omega
mu d H y by d z and d H y by d z will give
me minus sign i omega epsilon plus sigma
times H y E x, that is equal to 0.
So, d square by d z square E x minus i omega
mu into i omega epsilon plus sigma of E x
is equal to 0. Parallelly we could get all
most an identical equation for H y which will
be
d by d z square of H y minus i omega mu into
i omega epsilon plus sigma H y equal to 0.
I need to solve, so these are of course, decoupled
now. But of course, we have to pay a
price these are the second order differential
equation and therefore, let us let us define
this quantity here, or this quantity as equal
to gamma square.
..
So, that gamma square is defined by this and
my equations become d square by d z
square of either E x or H y, I will not it
twice, minus gamma square H y E x is equal
to 0,
where gamma square is a complex quantity,
i omega mu i omega epsilon plus sigma. The
the solutions of this equation is of course
very well-known, we could because this is
d by
d z square is proportional to gamma square.
So, the solutions are in terms of hyperbolic
functions of cosine hyperbolic, of gamma z.
So, let us write down E x is equal to some
constant A cosine hyperbolic gamma z plus
let us say, B sine hyperbolic gamma z and
parallelly H y will become C cos gamma z plus
d sine h gamma z.
Now, obviously we need to determine the constants
A, B and C and to do that let us put
in a condition, that supposing at z equal
to 0, let us say E x is equal to E 0, so if
you say
at z equal to 0, E x is equal to E 0 this
immediately determines the these two constants.
So, A is equal to E 0 and C is equal to H
0 and I am saying H H y is equal to H 0. So,
A
is equal to E 0 and C is equal to H 0. I still
have two more constants to determine and for
that what I do is to go back to the first
order differential equation that we had, where
we
related the curl of E with d B by d t or things
like that.
So, for instance we had an equation which
said d E x by d z plus i omega mu H y was
equal to 0. Now, what we do is this, that
we plug in these solutions into this equation
so
if I do that, that will give me a relationship
between B and d. Let us look at what it gives
me? So, d E x by d z remember that cosine
hyperbolic and sine hyperbolic, they just
.when you differentiate them, cos hyperbolic
gives you sine hyperbolic and sine
hyperbolic gives you cos hyperbolic, there
are no no sine changes.
.
Therefore, I get d by d z of E x gives me
A gamma sine hyperbolic gamma z plus B
gamma cos gamma z plus i omega mu C cos gamma
z plus d sine gamma z. Out of that
you remember that, we have already determined
that A is equal to E 0 and C is equal to
H 0, so this relationship which has both sine
and cosine and can be valid for all z. If
I say
that A gamma plus i omega mu times d is equal
to 0 plus b gamma plus i omega mu C is
equal to 0, so that gives me that d, which
I needed to determine is minus gamma by i
omega mu times A. But then A we had seen to
be equal to mu 0.
So, this is the relationship between d and
A, but let us put the gamma value there.
Remember gamma square was shown to be equal
to i omega mu into i omega epsilon
plus sigma. So, this is square root of i omega
mu into i omega epsilon plus sigma divided
by i omega mu times E 0 which gives me square
root of i omega epsilon plus sigma
divided by i omega mu times E 0 that is my
d. And this quantity that we have got here,
I
will denote it by 1 over eta, so this will
be written as minus E 0 divided by eta. I
repeat
the same thing, what with respect to the second
equation and I will then relate the, I have
already determined C. I will relate B to H
0 and what we will find, if you do that is
B is
given by...
..
So, in this case I have defined eta to be
given by square root of i omega mu divided
by
sigma plus i omega epsilon. And so this constant
B works out to minus eta times H 0.
This completes our derivation and therefore,
let us combine them, I get E x equal to E
0
times cosh gamma z minus eta H 0 that is my
B sine hyperbolic of gamma z and
parallelly H of y is equal to H 0 cosh gamma
z minus E 0 by eta sine hyperbolic of
gamma z. Now, that is the pair of solutions
that I have got.
Now, suppose I am talking about the surface
of the conductor being at z equal to 0 and
the conductor the wave is propagating in the
downward direction, and therefore, what I
could do is for example, I could say, suppose
at z equal to minus l. Let us assume that
the, I know the electric field, but I can
write down, so if I, the reason for putting
minus l
is very trivial. I just do not want these
minus sign not that it matters. So, at z equal
to
minus l my E x will be given by E 0 cosh gamma
l plus eta H 0 sine hyperbolic of
gamma l, and a very similar expression for
H y namely H 0 cosh gamma l plus E 0 by eta
sine hyperbolic of gamma l.
Now, supposing the electromagnetic wave has
propagated a large distance into the
medium, so that my l is very large. Now, if
l is very large, then remember the cosh
gamma l is E to the power gamma l plus E to
the power minus gamma l by 2 and sine
hyperbolic of gamma l is E to the power gamma
l minus E to the power minus gamma l
by 2.
..
So, if l is very large, so l very large then
cosh and sine both of them have the same value
because E to the power minus gamma l, I can
neglect them and write this as E to the
power gamma l by 2, so then I will, I can
write my E x is equal to E 0 plus eta times
H 0
and H y is H 0 plus E 0 divided by eta. Now,
take the ratio of E x to H y, so you find
this
is equal to E 0 plus eta H 0 divided by H
0 plus E 0 by eta. And you notice this is
nothing
but just eta. Therefore, this quantity eta
which is called characteristic impedance,
the
reason for that name will become clear as
we go along. We will see that eta as we have
obtained as the dimension of resistance and
therefore, actually it is an impendence and
so
the ratio of E x to H y is the value eta.
So, let us let us look at what that eta actually
is?
..
So, this eta that we have defined earlier
was given by root of i omega mu divided by
sigma plus i omega epsilon. Now, suppose I
have a lossless medium, lossless medium
means my conductivity is actually 0. So, if
sigma is equal to 0, then what I have is eta
is
simply equal to root of mu by epsilon. And
supposing this is my vacuum, that is mu is
equal to mu 0 and epsilon is equal to epsilon
0. Then you can calculate this this is
remember that this is 4 pi 10 to minus 7 and
this is about 8.85 or let say approximately
9
into 10 to minus 12, so if you calculate this
works out to 377 ohms. So, this is the value
in vacuum.
And gamma which was written as equal to square
root of i omega mu into sigma plus i
omega epsilon, since sigma is equal to 0 and
therefore, the, I have two i’s under the
square roots, I will pull out an i there omega
and root of mu epsilon. So, these these are
the characteristic impendence and the propagation
vector. In case well, I have just taken
so far a pure dielectric. So, let us let us
talk about the metal, but once again I will
not, I
will write down the full three dimensional
version of that equation, which we talked
about.
..
So, the process is essentially the same, I
have got these two pair of curl equations.
del
cross H is equal to J plus epsilon d E by
d t, this is essentially partly repetition
because I
am simply doing it in three dimension and
J is equal to sigma E plus epsilon d E by
d t
and del cross E is minus mu d H by d t. So,
a while back, we have this specialized and
talked about that. Let us suppose that E is
along the x direction, H is along the y
direction, but let us do it in general. Supposing,
I do del cross del cross E, I get, we have
seen that this is del of del dot of E minus
del square E and we do not have any charges.
So, it is minus del square E that is equal
to minus mu d by d t of del cross H and for
this
del cross H, I substitute from here, and so
that this is equal to minus mu d by d t of
sigma
E plus epsilon d E by d t. So, this is a,
an equation which is minus del square E equal
to
minus mu sigma d E by d t, this is first ordering
time minus mu epsilon d square E by d t
square. So, this is what we have and we are
looking for the solution of the form. Let
us
say E is equal to E 0, E to the power i k
dot r minus omega t. So, if you do that, I
have a
del square, which is which will give me minus
k square and I have a both a d by d t and
d
square by d t square d. By d t as we have
seen, will give you minus i omega and d square
by d t square will give you minus omega square.
So, as a result the right hand side of this
equation will give me E.
..
So, we have written the harmonic solution
E is equal to E 0 E to the i k dot r minus
omega t, this is what we want. So, the correspondingly
this equation, will give then k
square, that is what comes out of the del
square operation, is equal to i omega mu sigma
plus mu epsilon omega square. So, this is
this is the relationship between the propagation
well is not really propagation, but full.
So, k is equal to i omega mu sigma plus mu
epsilon omega square. And this, let me pull
out the omega outside, that gives me if I
pull
out omega mu, I get omega epsilon plus i sigma
raise to the power half.
Now, what I will do is this that you notice
that this is the of course, real, but this
is a
complex quantity. So, the k has a both the
real part and imaginary part. The imagine
since it is appearing in the complex E to
the power i k dot r. The real part of k will
give
you the propagation and the imaginary part
of k will give you the attenuation of the
wave. So, let us look at this is fairly straight
forward algebra, so what you want to do is
to write down this quantity. This is done
standard, just write omega epsilon equal to
some A cos theta and sigma is equal to some
sine A theta.
So, this quantity becomes A cos theta plus
i times A sine theta. So, this whole thing
will
become A to the power half E to the power
i theta by 2 and you can determine each one
of them by this trivial exercise. So, if you
did all that this is a fairly straight forward,
so
this as to be separated into real and imaginary
parts. This will give you omega this the,
other complicated expression, but let us write
it down. mu epsilon by 2 you get 1 plus
.square root of 1 plus sigma square by omega
square epsilon square raise to the power
half there is a square root within a square
root plus i times square root 1 plus sigma
square by omega square epsilon square minus
1 raise to the power half.
So, you notice that I identify this quantity
here, which is the real part of k as my
propagation vector beta. So, beta is the propagation,
which is omega root mu by mu
epsilon by 2 into square root of 1 plus square
root of 1 plus sigma square by omega
square epsilon square raise to the power half
and alpha which is the attenuation factor,
because i alpha when it goes to E to the power
i k dot r, that will give you E to the power
minus alpha r. And this is a very similar
expression which is omega mu epsilon by 2
and
this is this square root 1 plus sigma square
by omega square epsilon square minus 1 raise
to the power half So, what you have done is
in general determine the propagation vector
beta and the attenuation factor.
Now, at this stage we need to talk about what
is meant by a metal? So, for we have had
both the dielectric and the metal property
together, so it is this ratio omega by sigma
by
omega epsilon, which determines whether something
is a good dielectric or a good
conductor. Now, if something is a good conductor,
then sigma is much greater than
omega epsilon.
.
So, for a good conductor 
sigma is much greater than omega epsilon.
As a result in this
factor here, since sigma is much greater than
this. You notice that I can write down all
.these ones can be then neglected. If the
one is neglected, then for instance what I
get for
beta? So, beta will be equal to alpha and
is approximately equal to omega into root
mu
epsilon by 2 and I have, first I take this
square root giving me sigma by omega epsilon
and then I take the second square root, which
gives me square root of sigma by omega
epsilon. So, if you look at that that becomes
square root of omega mu epsilon by 2.
So, the velocity in in the conductor, which
is given by omega by the propagation
constant, so that is simply equal to square
root of 2 omega divided by omega mu sigma
mu and this is much less as you can see it
because your sigma is a very large quantity.
So, as a result the propagation speed sort
of gets reduced. The other thing is that,
since
the alpha is this quantity and the electric
field attenuated as E to the power minus alpha
z, the distance for at which the strength
of the electric field becomes 1 over E th
of its
initial value is what is called as the skin
depth. Now, clearly since alpha is given by
this,
the value of the skin depth is 1 over alpha,
which is square root of 2 over omega mu
omega sigma mu.
So, firstly you notice that this tells me
that the skin depth of course, becomes smaller
and
smaller, as the conductivity raises that is
the electromagnetic field does not quite
penetrate for into the medium into a conducting
medium. But but let us sort of have an
idea about how much is this skin depth? For
example, let us take a good conductor like
copper.
.
.So, delta there is square root of 2 over
omega sigma mu. So, let me take, let us take
the
omega is the frequency corresponding to, let
us say 1 mega hertz. That is mu is equal to
1 mega hertz that is 10 to the power 6. Let
us take copper whose conductivity sigma is
approximately 6 into 10 to the power 7 ohm
inverse meter inverse. It is slightly less,
something like 5.58 also 6 into 10 to the
power 7. So, if you calculate this sigma you
get
2 divide by, now there is a omega there, so
I get 2 pi mu, 2 pi into 10 to the power 6
sigma is 6 into 10 to 7. And for mu I will
take the permeability of the vacuum, which
is 4
pi into 10 to minus 7. You can immediately
see 10 to 7, 10 to minus 7 etcetera.
Go away and this number is a rather small
number it works out to something like 0.67
milli meter. You could compare this with what
happens for instance in sea water, which
is not as good a conductor as copper. And
the, in sea water it is about 25 centimetre
remember because of celerity sea water as
certain amount of conductivity, but if you
take
fresh water it goes something like 7. So,
this is this is what skin depth about. So,
basically what we have said is, that when
electromagnetic wave propagates in vacuum,
it
attenuates, it moves with a much lower speed
and there is a thin area, thin surface layer
at the surface where the amplitude will quickly
divert.
So, in other words since I am talking about
a conductor, any current that arises because
of the electric field will have to be confined
within this small thickness. Let us look at
the reflection, we have already talked about
reflection and transmittance from a dielectric
medium. Now, one can very similarly, work
out the reflection and the transmission from
the conducting medium.
..
Now, the principle is exactly the same. We
have seen that at the surface at the boundary
between, let us assume that electromagnetic
wave is falling from vacuum on to this
surface here. So, this surface I will take
to be z equal to 0 and as we have done earlier
I
will assume that the electric field is you
know perpendicular to this propagation
direction. And of course, the and the magnetic
field will be this way. So, therefore if the
electric field is to be continuous at the
boundary, then I can write down E incident
plus E
reflected must be equal to e transmitted.
.
.And parallelly the magnetic field is also
tangential component of magnetic field is
also
continues. So, I will write h incident plus
h reflected is equal to H transmitted. Now,
little while back, we had worked out the relationship
between the electric field magnetic
field and we had seen that the ratio of the
electric field to the magnetic field happens
to
be the characteristic impedance. So, for as
the incident ray is concern your H I is E
I by
eta. So, let us call it eta 1 because that
is the medium number 1. Now, so for as the
reflected part is concerned, this is minus
E reflected by eta 1 minus because the direction
of propagation has changed and since I have
assumed the electric field direction to be
the
same corresponding the magnetic field direction
as to reverse.
I have taken eta 1 because both incidence
and reflected ray they, belong to medium
number 1. So, eta 1 is medium 1. The transmitted
ray however H T will be E T by eta 2.
So, eta 2 is my transmitted ray. Now, can
easily solve this set of equations this is
and you
can show that E reflected by E incident happens
to be equal to eta 2 minus eta 1 by eta 2
plus eta 1 and E transmitted by E incident
is 2 eta 2 by eta 2 plus eta 1. And you could
also solve for the corresponding magnetic
field thing and the only difference that you
find is that the H R by H I instead of eta
2 minus eta 1 by eta 2 plus eta 1 happens
to be
eta 1 minus eta 2 by that is the numerator
that 2 is replaced by E 1 and this is the
only
difference. That is, there is a small error
here, which should be H R by H I, this equation
should be H R by H I.
.
..
So, what we have now done to, but before we
proceed, let us look at the consequence of
some of these things. So, let me look at the
supposing my eta on1e the medium number
1.
.
Well let me write down eta 2 first because
that is a conductor. So, eta 2 is i omega
mu
divided by sigma plus i omega epsilon. Now,
if I assume that the, it is a good conductor
then sigma of course, is much larger than
the omega epsilon. Though it is a complex
quantity, but in a first order I can always
neglect this part because this is be a rather
small
.number. So this quantity is equal to square
root of i omega mu divided by sigma. You
know that square root of i can be written
as 1 plus i by 2 root 2. So, you could actually
calculate by putting numbers here.
For example, let us take the same copper which
we had calculated, so I get square root of
i which is 1 plus i by root 2. Then I have
got square root omega which I am taking as
2 pi
10 to the power 6 1 Mega Hertz mu as again
is 4 pi 10 to minus 7, copper I am taking
conductivity is to be 6 into 10 to 7. You
can see what is this number? See this 10,
10 to
the power 7 here and there is a 10 to the
power minus 1 there, so this number is a rather
small number and if you calculate everything
here, you get a 1 plus i into 2.57 into 10
to
minus 4 this is approximate because I have
done you know neglected this, but you could
put in other things there.
Now, so far as eta 1 is concerned, I know
it is the vacuum. Therefore, I take the standard
value characteristic impedance 377 ohms. Now,
if I am now calculating the ratio of the
electric field reflected component to the
incident component, E R by E I, which if you
remember is eta 2 minus eta 1 divided by eta
2 plus eta 1. Of course, eta 2 is a, has a
small complex now part, but the real part
is rather small. So, there would be, this
would
be in general complex, but on the other hand
the angle complex part will not be all that
much. That is because it is eta 1, which determines
this ratio primarily and this is
approximately equal to minus 1, that is the
ratio of E R to E I is minus 1, this is what
you
would expect for a perfect reflector.
If you want a better value you need to calculate
these things little better and and this will
be sort of may be not minus 1 may be minus
0.99 times a, an angle, so that is that is
what
you will get. Therefore, a good metal is also
a perfect reflector, now notice something
very interesting that comes out of it.
..
Suppose, you were to calculate E transmitted
by E i you could actually immediately see
what it will be. Now, E transmitted by E i
is 2 eta 2 by eta 2 plus eta 1. Now, we know
that the numerator denominator is approximately
eta 1, which is 377 and eta 2 is a rather
small number. So, you multiply this 2.5. So,
I get 5.14 into 10 to minus 4 1 plus i divided
by 377, I am not hiding the little complex
part there. So, this is approximately because
you see their numbers, this is 10 to minus
4, this is 10 to 2. Therefore it is approximately
a number which is of the order of 10 to the
power minus 6 times 1 plus i.
So, in other words the transmitted electric
intensity or the electric field amplitude
is
substantially reduced. It is 10 to the power
minus 6 times 1 plus i. Now, if you did the
same thing which your H T by H I, you would
realize this will become approximately
equal to 2 and the reason is not very far
to see, difficult to see while the electric
field is
reflected with a phase change the magnetic
field is not reflected with the phase. But
that
its direction of propagation as changed, so
as a result in order to maintain continuity,
since the magnetic field is reflected without
a reversal i expect H T by H I should be
twice the, should be a approximately equal
to, so this all stands to the last thing.
I would talk about is, we have just now seen
that the electric field does not penetrate
much into the medium. So, if since the electric
field does not penetrate much into the
medium, gets attenuated very quickly, I expect
the electric field whatever is it penetrates,
.depending upon the skin depth to be contained
within a small thickness of from the
surface.
.
So, what we do is, we define what is known
as a surface impendence. So, the surface
impendence, remember the impedance is defined
basically in terms of the ratio of the
electric field to the current.
.
Now, in this case, thus surface impendence
will be defined as the parallel component
of
the electric field, which is giving raise
to that current divided by the surface current
.density. That is the current density, which
is running or the current density, which is
on
the surface. Now, I do, I calculate this.
Now, I know the current is confined within
a
small thickness, now let us assume that there
is no reflection of the electric field from
that thickness. That is since the electric
field gets attenuated, I can assume that the
essentially it is the infinite medium and
therefore, the current profile the current
density
profile can be J 0 into the E to the power
minus gamma z.
And surface current density will then be,
I simply calculate, I simply integrate over
the
all space. Ideally it is to be integrated
over that thickness, but then the I assume
that the
skin depth is rather small. So, as a result
the electric field sort of does not penetrate
much, so if I integrate this from 0 to infinity
that is perfectly legitimate. And that is
equal
to simply J 0 divided by gamma and if you
recall J 0 is nothing but sigma E parallel
this
divided by gamma. So, this is my surface current
density, now what is my gamma now?
Gamma if you recall is i omega mu into sigma
plus i omega epsilon and since it is a good
conductor I replace it with i omega mu sigma
and as before since square root of i is 1
plus i by root 2, this is omega mu sigma by
2 into 1 plus i. So, therefore my z s which
is
e parallel by k s 
and that is E parallel by sigma E parallel
into gamma. So, this is equal
to, well gamma by sigma and I have already
calculated what is gamma and I divide it by
sigma.
.
.So therefore, my surface impendence is one
by sigma into root of omega mu sigma by 2
into 1 plus i and since you recall that the
delta the skin depth is square root of 2 by
omega mu sigma. So, this is nothing but 1
over sigma delta into 1 plus i, so as a result
what I get is that, the surface impedance
which has a resistance part and a reactance
part.
.
So, this is z s is equal to 1 over sigma delta
into 1 plus i and this is surface resistance
plus
i times surface reactance and this surface
resistance. Then is, simply equal to 1 over
sigma delta.
.
.So, what it means is that the current profile
is like this. The current is confined to a
small
thickness and the resistance that the surface
provides depends of course on the
conductivity of the medium. So, 1 over sigma
as expected and also it is inversely
proportional to the skin depth delta.
.
