. So, in this section, we will discuss electromagnetic
wave propagation, In observing medium, we
have seen the propagation characteristics
of electromagnetic waves in dielectric medium
isotropic dielectric and now in an observing
medium mostly absorbing medium primarily where
conductors conducting medium.
So, today in this discussion, the following
topics will be discussed that the transient
decay of free charge in a conducting medium,
then we will address the wave equations for
a the for the conducting medium and the resulting
plane waves phase, we will look at the phase
and amplitude of the of the electric field
and the magnetic field of the plane waves,
then will categorize the good conductors and
poor conductors, will see that the electromagnetic
waves cannot propagate in conducting medium
as a result, there will be only a small skin
depth penetration of the electromagnetic waves
will characterize that and look at the quantity
expression, how it depends on the propagation
core and the frequency of the electromagnetic
waves, then we will see the average power
dissipation joules loss in conductor.
So, a conducting medium will look at the Maxwell's
equation; for example, conducting medium ah
salt, water, metals in such a medium the charge
flow cannot be controlled and the current
density due to free charge carriers is not
0 free charge, free current density J f is
given by the Ohm's law sigma into E a should
be a vector here
So, now will look at the Maxwell's equation,
in this case, del dot E equal to free charge
rho f by epsilon and the curl equation for
the electric field del cross E equal to minus
del B del t, then del dot B equal to 0 and
del cross B equal to J f the current density
due to free charges and mu into epsilon del
E del t. So, this J f is equal to sigma into
epsilon which will substitute in place of
J f and will reorganize this Maxwell's equation
to form the wave equation.
Also we have this equation of continuity;
the del dot J equal to del rho f del t is
the free charge rate of change of the free
charge ah there should be a minus sign and
then this J f is the charge density, the current
density free charge due to the free charges
now ohms law states that J f equal to sigma
into E Gauss's law gives you the del dot E
equal to sigma rho f by epsilon.
Now, if I combine these two equations, then
we can write that del dot J is equal to del
dot J is equal to sigma del dot E using this
equation in this and then if I use del dot
E equal to sigma rho f by epsilon, then I
can write del dot J is equal to sigma rho
f by epsilon . So, I have been able to write
an expression which relates this J f and rho
f through sigma and epsilon.
Now, because the left hand side of this del
dot J is equal to del rho f del t and the
right hand side this left hand side is equal
to this which is equal to this. So, I can
write del rho f del t is equal to sigma rho
f by epsilon naught. So, equating this to
this, we can write this the equation of continuity.
In this; from the del rho f del t is equal
to sigma rho f by epsilon and if you solve
this equation, this gives you the solution
rho f equal to rho f 0 that is a time t equal
to 0, the free charge times E to the power
of minus sigma by epsilon times t. So, it
shows that the free charge decays exponentially
and the decaying depends on this factor sigma
by epsilon naught that is the conductivity
and the sigma . So, free charge density dissipates
in the characteristic time tau sigma by ah
epsilon by sigma that is the time over which
the charge decays to 1 upon E times of the
original.
So, looking at this consequence, we can classify
the conductors, if sigma tends to infinity
and tau this characteristic time, the dissipation
time is equal to 0 that is instantaneous,
then it is a perfect conductor and if sigma
is large reasonably large and tau is small,
then this conductor is a good conductor , but
the reverse will happen, if sigma equal to
small and tau is very large that is the characteristic
time for charge dissipation is quite large
in that case the conductor will be classified
as a poor conductor.
So, this equation that rho f equal to rho
f 0 e to the power of minus sigma by t into
t sigma by epsilon into t, it reveals that
if some free charge is placed inside the conductor,
the charge flows out immediately to the edges
that is in no time, the charge dissipates
. So, we will not consider such transient
behavior while looking at the electromagnetic
wave propagation in a conductor therefore,
we will put this free charge rho f is equal
to 0 in the equation arising out of the Maxwell's
equation.
So, del dot E equal to 0 del dot B equal to
0 and del cross E equal to as it is and del
cross B will now be represented as mu sigma
E which is for J; that is J equal to sigma
into a and mu epsilon del E del t. So, this
is the radio this is the new form of the Maxwell's
equation when we have use this rho f equal
to 0.
Now, if we take the curl of the equation del
cross E equal to minus del B del t that is
the third Maxwell's equation taking the curl
on either side. So, the lef left hand side
gives you del cross del cross E which is equal
to del of del dot E and E of del dot del which
is a very well known identity. And this identity
in this case will result in only minus del
square E because del dot E is equal to 0 which
has come from the consequence of this equation.
So, as a result, we can write this left hand
side del cross del cross E is equal to minus
del square E and the right hand side that
is minus del B del t, you should take the
curl of this. So, you have to write minus
del del t of del cross B which is equal to
minus mu epsilon del square E del t square
minus mu epsilon del E del t, this has come
from this equation, I take curl of del B del
t which will be equal to mu epsilon del E
del t plus mu epsilon del to E del t square
exactly that is what is written here. So,
I have been able to calculate the del cross
del cross E and del cross del B del t.
So, equating these 2, we can write this equation
del square E equal to mu epsilon del square
E del t square plus mu epsilon mu sigma del
E del t, this is for the electric field and
if we proceed by taking the curl of the forth
equation that is del cross del cross B del
cross del cross B, then for del cross E, if
I write this and also for del cross E, if
I write this, then we can arrive at the this
equation del square B equal to mu epsilon
which is a similar equation except that E
is now replaced by the B vector these equations,
if we look at these 2 equations these equations
admit plane wave solution.
So, we will try we will look for the plane
wave solution and we will try and look and
on substitution of this plane wave solution
into these equation, we will see what is the
nature of the propagation constant what is
the amplitude of the electric field and the
magnetic field and is there any phase relation
between the electric field and the magnetic
field.
So, accordingly, we assume a plane wave solution
consider and x propagating electromagnetic
wave. So, in that case we can write this electric
field it is propagation along x as a function
of x, but it is polarized in the y direction.
So, we can write this equation y cap E naught
E to the power of i k x minus omega t substituting
this equation into the wave equation that
is del square E equal to mu epsilon del square
E del t square plus mu epsilon del E del t,
we get that for an operator del square E operating
on this; this equation plane wave equation
we get i k and i k twice.
So, that gives you minus k square and operating
twice that is taking the time derivative twice
will take out i minus i omega twice. So, that
gives you minus omega square multiplied by
mu epsilon and del E del t will give you i
into omega times mu into omega. So, this equation
when substituted with this solution, then
we get that minus k square equal to minus
mu epsilon omega square minus i mu sigma omega
all of them are having a minus sign.
Now, it tells you that the square of the propagation
constant is also imaginary; that means, the
complex k of the form you can consider that
k is k plus plus i k minus and if you take
the the the mod of this k square k square
will be equal to k plus square plus k minus
square minus twice i k plus k minus.
So, you have been able to write the imaginary
part and the real part for k square and now
we can compare these two equation k square
equal to this also k square equal to this
and on comparing we get that k square plus
plus k minus square is equal to mu sigma omega
square whereas, k plus k minus the product
of these 2 quantities; the real and imaginary
part of the propagation constant will give
you mu sigma omega by 2. So, we have been
able to see the complex propagation constant
whose real part is mu epsilon omega square
and the real and the imaginary part is this
of course, this is the real and imaginary
part of the k square quantity.
So, k square equal to we can write that mu
epsilon omega square plus i mu sigma omega
. So, this on solving because I have these
equations k square plus k square minus then
k plus k minus. So, I can use this we can
2 and then you minus into this, then we can
add, you can frame k plus minus k minus square.
Which will yield the solution of this k plus
minus the k plus and k minus put together
will give you the solution of this equation
is a quadratic equation whose solution will
be this. So, once it is plus 1; next for the
plus value of k and minus 1 for the minus
value of k.
So, the plane wave solution having known that
the k the propagation constant has 2 values
k plus and k minus and can be written as k
equal to k plus plus i k minus we can write
the plane wave solution we can rewrite.
In fact, the plane wave solution as E 0 E
to the power of minus k minus x into E to
the power of i k plus x minus omega t; in
fact, you can see that this part accounts
for the phase k plus is the real part of the
propagation constant k minus is the imaginary
part and this imaginary part has given you
the attenuation of the amplitude E 0 E to
the power of minus k minus x.
So, it tells you the the amplitude of the
electromagnetic wave which is traveling along
the x direction in a conductor decays as E
to the power of minus k minus x. So, this
k minus k plus and k minus these 2 quantities
are given by this which are the functions
of omega epsilon mu and sigma. So, all these
4 quantities are going to decide the k plus
value and the k minus value that is how what
will be the phase instantaneous phase and
what will be the what will be the loss in
the amplitude of the electromagnetic wave
in the conducting medium for the magnetic
field we know that B equal to this can be
connected to the electric field through this
equation.
Now, since k is complex B and E, they are
connected by a complex quantity; that means,
the phase related with this k will also be
attached to the phase of this electric field
and in general electric and magnetic field
will have a phase difference and writing this
B field in term using this electric field
value known; we can write in this equation.
So, k by omega which is a consequence of this
equation; so, I am trying to write this electric
if the magnetic field in terms of the amplitude
of the electric field the attenuation factor
of the electric field and the phase of the
electric field, but this is a complex amplitude.
So, let us look at this. So, we could write
the electric field as this and magnetic field
in the same way k by omega and the electric
field quantity, but you can see that they
are oriented in a mutually orthogonal direction
this is along the y direction this is ah this
electric field direction is y that is and
y polarized the electromagnetic wave and the
magnetic field is along the z direction.
Now, this k plus and k minus these 2 quantities
which constitute the complex propagation constant
can be written as k mod into E to the power
of i phi where phi is the phase of the phase
factor at associated with this propagation
constant and this phi the value of the phi
will be equal to tan inverse k minus by k
plus. So, this is the phase that is associated
with the propagation constant and we will
see how this phase is going to be related
to the phase of the electric field and the
magnetic field now we could write k equal
to k E to the k mod E to the power of I phi
with phi equal to this and this is the relation
which connects electric field and the magnetic
field k therefore, k decides the phase of
B with respect to E.
So, let us see that we can write this B and
E in this form as we have seen B and E their
no more in phase. So, we attach different
phases delta E for the electric field and
delta B for the magnetic field this and we
also remember recall that k has a phase of
E to the power of phi. So, you have three
quantities electric field has a phase of delta
E magnetic field has a phase of delta B and
the propagation constant has a phase of phi.
So, B and E are then related through this
equation because B equal to k E by omega.
So, because k and E they are at right angles.
So, k into E by omega if you look at the magnitude
therefore, B is equal to k, then E by omega
into E to the power of i phi is the phase
for this k and E to the power of I delta is
the phase due to the electric field. So, this
is the way the electric field and the magnetic
field they are related.
So, the amplitude and phase delay between
E and B fields, you have this real part of
the amplitude of the electric field and the
magnetic field they are related in this way
where the value of k and omega we have to
consider. So, this is the connection that
this gives you the value of k by omega and
the imaginary part that gives you the phase
difference between the magnetic field and
the electric field.
So, through this figure we tried to depict
how the electromagnetic wave is propagating
in a conducting medium you can see that the
field amplitude decays as it moves along the
x direction and also the electric field and
the magnetic field they have a phase difference
as it propagates along the magnetic ah along
the conducting medium.
So, these are the electric field directions
and these are the representation for the magnetic
field. So, it shows that the electric field
and the magnetic field the decay as it propagates
through a conducting medium this; that means,
that an electromagnetic wave cannot propagate
cannot penetrate a conductor, it only penetrates
wont up to a certain depth which is referred
to as the skin depth and will try to quantify
the skin depth of electromagnetic waves in
a conductor
So, in a conducting medium the skin depth
is the distance at which the amplitude reduces
to 1 upon E times the original amplitude.
So, within the before original ampli by saying
original amplitude, I mean that before it
heats the conducting medium the or within
a within a point at a point within the conducting
medium the field if it is E 0, then after
the skin depth the field will be E 0 by e.
So, the electric field in a conducting medium
the plane wave solution is E 0 complex amplitude
and this is the decay factor and this is the
phase factor and in the similar fashion the
the magnetic field has a complex amplitude
and a decay factor and a phase factor.
So, the skin depth is is to account for this
quantity that is by the distance over which
this E 0 the amplitude falls to one upon falls
to E 0 by E one upon E times E 0 that is I
have to put this value to be equal to one.
So, k x into d will be equal to one that gives
you d equal to 1 by minus k. So, this is the
skin depth this k plus and k minus they have
different significances k minus gives you
the skin depth whereas, k plus accounts for
the wavelength of the electromagnetic wave.
So, twice pi by lambda equal to k plus we
can write in this form also by putting twice
pi equal to 1.
Now, for a poor conductor for a poor conductor
the skin depth can be written as to account
for the skin depth for a poor conductor we
have sigma which is much much less than omega
epsilon.
So, k plus becomes approximately equal to
omega under root mu into epsilon k minus will
be approximately equal to sigma by two under
root mu by epsilon you can see that the skin
depth d which is one upon k minus is equal
to one upon k minus if you right this will
become the reciprocal of this quantity that
is two by sigma epsilon by mu this is this
does not involve the frequency of the electromagnetic
wave that is it is independent of the frequency
for a poor conductor the skin depth does not
depend on the frequency of the [freque/frequency]
frequency of the electromagnetic wave that
is trying to travel through the medium.
In the case of a good conductor, we have sigma
which is much much greater than omega epsilon
this is the way for define this good conductor
that is conductivity is very high.
Then this k plus will be approximately equal
to k minus because you can ignore that one
in the solution of k plus and k minus, then
both of them are approximately same. So, you
can write this k plus or k minus is equal
to this therefore, the skin depth d which
is the reciprocal of k plus will be equal
to lambda by twice pi and d equal to one sixth
of the wavelength. So, in a in a good conductor
the electromagnetic waves only travel through
one sixth of the of the wavelength distance,
the energy density of an electromagnetic wave
is given by this we have seen that.
So, if you substitute in the sin cosine solution
of the electric field and magnetic field,
we can rewrite this equation in this form
and then you can calculate the average energy
density associated with the electromagnetic
wave is given by this.
So, for a good conductor, you can calculate
the average energy density associated with
the electromagnetic wave is equal to this.
And the energy flow is characterized by the
pointing vector that is s equal to one upon
mu E cross B and if we evaluate this expression
where for a cycle the integration is equal
to half into cosine phi in this case, it is
not equal to half this is because there is
a phase difference between the electric and
magnetic field which is like the power factor.
So, it appears in the expression of k plus
as k into cosine phi. So, we can calculate
the pointing vector in this case as well.
So, the average power; we can calculate in
this form with the progression of the wave
the energy density decreases with a factor
of this and this lost energy heats of the
conductor which you call the conductor heating.
Next will consider the average power dissipation
in a conductor let us consider a slap in the
slab; this power we can write in this form
that E into J delta x into a J is equal to
sigma E square. So, we can rewrite this equation.
Now if you calculate the power; average power,
then we can write this equation.
Now, let us consider the average power that
is flowing in to the slab, we have calculate
the expression for average power whether it
is going in or going out will be decided by
this factor.
So, power flowing in will be given by this
into a whereas, the power flowing out from
the slab this k minus k into x will be replaced
by twice minus k into x plus delta x. So,
this factor will account for the thickness
of the slab.
So, the difference between the power that
is going in and going out will give you the
power that is lost within the conductor, if
we evaluate this equation, you can rewrite
this equation in this form using k plus k
minus equal to this which we have seen in
the solution for k. So, you can calculate
the power loss is equal to this.
So, this is how we have discussed the transient
decay of the free charge in conducting medium
wave equations and plane wave solutions in
a conducting medium phase and amplitude relations
for the electric field and the magnetic fields,
we have also classified the good conductor
and poor conductor and we looked at the skin
depth for good conductor and poor conductor
for a good conductor, it is only the one sixth
of the wavelength of the electromagnetic wave.
Whereas, for a poor conductor, it does not
depend on the frequency and we also have calculated
the average power dissipation through a conducting
medium which actually accounts for the joules
loss in the conductor.
Thank you.
