Today we will discuss the propagation of electromagnetic
waves and energy flow this is a very interesting
topic, and will be useful throughout our discussion
energy flow and Poynting theorem .
So, we will be discussing under the following
heads that energy transport in electromagnetic
waves in a certain volume of the space through
which the electromagnetic waves are travelling
how the energy is entering and how much of
the energy is leaving out of that space, that
will clearly quantify will try to understand
how the loss is taking place within that volume.
Then in the process we will workout the Poynting
vector and the Poynting theorem look at each
of the terms of the Poynting theorem. Followed
by this we will take two example cases the
energy transport in a dielectric medium and
we will also considered the transport of energy
in a conducting medium.
So, Poynting theorem Poynting theorem basically
concerns the conservation of energy for a
given volume in space. As I have mentioned
that when the electromagnetic waves are travelling
through certain region of space within the
volume the energy, how much part of the energy
is lost and how much part of the energy is
flowing out we will correlate these two things
and this will be a will be described through
this Poynting theorem, which is actually a
consequence of the Maxwell's equation and
we will start with the Maxwell's. equation
to arrive at the conclusion
So, let us begin with the Maxwell's curl equations,
we have these two famous curls equations that
del cross E equal to minus del B del t and
del cross H equal to j plus del D del t where
B is the magnetic field and D is the displacement
current j is the induction current .
.
So, starting with these two equations we will
organize the Poynting vector let us recall
that the vector identity that is, del dot
E cross H. This left hand side can be written
in the form that H dot del cross E minus E
dot del cross H.
So, if we plug in the curl equations for this
del cross E and for del cross H; from here
that is del cross equal to we will substitute
minus del B del t and for del cross H will
substitute J plus del d del t. If you do that
this quantity H dot del cross E becomes H
dot del B del t, because del cross E is now
replaced by minus of del B del t and for del
cross H if I substitute this quantity J plus
del d del t, then it becomes minus del j dot
E minus E dot del d del t. So, this is the
equation which involves the E cross H divergence
of that .
So, let us look at this, del dot E cross H
minus B equal to minus j E minus E dot del
d del t minus H dot del t. We have just rearrange
this equation in a fashion that, we have taken
this quantity and this quantity together which
are a like and this quantity we have placed
separately. The mission the reason is very
clear that we want to look at these terms
these two terms put together and also this
term J dot E d v .
Now, let us consider a certain volume of space
through which this electromagnetic wave is
travelling and let us considered that the
area bound surface area bound by this volume
is S is a close surface. So, we can we can
write this equation if you take if integrate
over this volume, we can write this equation
del dot E cross H d V is equal to minus J
E d V minus E dot del D del t plus H dot del
B del t.
Now, look at these two terms we will use Gausss
theorem to see what is meant by this quantity
del dot E cross E E cross H. So, Gausss theorem
states that it is it relates the volume integral
to a surface integral, for a vector field
H the divergence of this vector field over
is a over a certain volume is equal to the
closed surface integral of the vector field
.
So, by replacing this F by E cross H that
is if we F if we substitute for F this E cross
H, then we can re write this equation del
dot E cross H d V equal to the right hand
side which you have seen as it is. Now if
I if I apply this Gausss theorem, I can write
this equation as as surface integral of E
cross H that is E cross H dot n unit vector
d A integral of that will be equal to the
right hand side .
Now, we will define the Poynting vector the
Poynting vector is defined as the cross product
of E field and the magnetic field. So, s is
equal to E cross H, you can see from this
definition that S is perpendicular to E also
S is perpendicular to H; that means, this
Poynting vector is a vector which is perpendicular
to the plane containing the electric field
and the magnetic field; that means, if this
Poynting vector has to represent certain quantity,
then the direction of that quantity will be
perpendicular to the the to the plane that
contains the electric field and magnetic field
and we know that electric and magnetic field
they are also perpendicular to each other,
as a result this S E and H they form a right
handed tired of vectors .
Now, if we replace this E cross H by this
quantity Poynting vector s then s dot n d
A surface integral of that will be equal to
the quantity, which we have described before
that is J dot E d V minus E dot del D del
t H dot del B del t volume integral of that.
So, it tells you that the Poynting vector
measures the rate of energy transport and
the energy transport it happens in a direction
which is perpendicular to both the magnetic
field and electric field and; that means,
it is along the direction of propagation direction.
So, the propagation direction is the direction
along which the energy transport takes place
.
Now, let us recall that using the constitutive
relations, the energy density that is associated
with the electric field is half E dot D, that
is equal to half epsilon E square and the
magnetic field energy density associated with
the magnetic field is in the same way half
H dot B, which is equal to half mu dot H mu
into a H square .
So, if we use definition of the of the electric
and magnetic field energy density, then we
can write that E dot del D del t which is
equal to half of epsilon epsilon is replacing
this D that is D equal to E into epsilon.
So, half epsilon into del E square del t which
can be rewritten in this form half del d del
del t of E dot D which is equal to del del
t of the energy density w E electrical energy
density.
So, we have used this electrical energy density
here in place of this you can take half inside.
So, it gives you this quantity and in the
same way if we proceed then for H dot del
B del t, we can write half mu into del t del
del t of H square which is equal to half of
del del t of H dot B and in turn this gives
you del del t of w B, which is the energy
density associated with the magnetic field.
Therefore, these two terms which are appearing
in the equation can be replaced by the sum
of these two terms; that means, we can represent
this sum of these two terms is equal to del
del t of the sum of the energy density is
associated with the electric and magnetic
fields .
So, if we rewrite this equation that S dot
n this left hand side the Poynting vector
surface [inter/integral] integral of that
will be equal to J dot E d V and this quantity.
Just now we have seen that del del t of del
B w B and w E d V is the translated form of
the individual variation of the electric and
magnetic fields. So, we can write this equation
if you take del del t outside the integral,
then we can write that it represents the total
energy, energy density associated with the
magnetic field and electric field over this
volume and del del t of that that is the rate
of change .
So, let us let us look at this term that that
that the energy density some of the energy
density is for the magnetic field and for
the electric field is equal to this just now
we have seen. But because this is the total
energy energy density with the electric field
and magnetic field so, this quantity actually
represents the total electromagnetic energy
stored in the volume V.
So, total we have to magnetic energy in a
volume V and if you take del del t of that
that will represent the change in the total
electromagnetic energy that is stored within
that volume. If we place a minus sin before
that it should definitely represent that this
amount of energy will be the loss .
And if you look at this term that is S dot
n d A which is equal to this, this actually
represents the instantaneous power flow of
instant instantaneous power instantaneous
power out of the volume V through the surface
S .
Now, again rewriting this equation in this
form, this left hand side that is the term
which contains the Poynting vector represent
the the total power flowing out of the volume
V which is equal to the the power in the volume
dissipated power in the volume and also the
total electromagnetic energy change in the
volume. This is just now I have mentioned
that this quantity del del this quantity w
B plus w E that is the energy densities of
the electric field and the magnetic field
integrated over the volume d V represents
the total volume within that space, total
energy stored within that space and if you
take the time derivative of that it represents
the change in that energy and since there
is a minus sign. So, it represents that the
loss of the energy from that space.
So, if you look at this equation, the energy
flow through that volume will be equal to
the loss of the energy from that volume and
the dissipated power within that volume. So,
that is very consistent and it really represents
the conservation of energy. Del dot S because
this quantity we have called that power dissipation
because del dot S del dot S and this del B
del t the continuity equation is equal to
minus J dot E .
So, we will consider two different cases one
is the dielectric and through which a plane
wave propagating, and how this Poyting vector
measures the flow of energy and followed by
that we will consider another situation when
the electromagnetic wave is propagating in
a conducting medium. So, in this case we can
represent the electric field E as x cap E
naught cosine omega t minus k z as if the
wave is propagating along the z direction,
then the magnetic field can be written as
this because electric field is x polarized
and the magnetic field is y polarized.
So, the energy density associated with the
electric field can be calculated as the integral
of this then half of E 0 square cosine square
omega t k z . So, this quantity if I take
the time average time average of this, that
is integral over a complete cycle then we
can write that the value of this will be because
for 0 to twice pi by omega time that is t
will give you the value half and you already
have half.
So, omega E the time average of the electrical
energy density associated with electric field
will be equal to one upon four E epsilon E
naught square where E naught is the amplitude
of the electric field.
In the same way if I consider the magnetic
field, energy density associated with the
magnetic field then we can calculate to be
equal to 1 upon 4 mu H 0 square where H 0
is the amplitude of the magnetic field. This
we could arrive at using this relation H 0
equal to omega by mu into k into E 0. So,
if you substitute this we can see that the
electrical energy density, the energy density
associated with the electric field and that
associated with the magnetic field the average
value of that over a complete cycle will be
the same that is 1 upon of 4 mu epsilon naught
square
So, I mean known this fact then we can write
the total energy density average value of
that which represents energy. So, will be
equal to twice the contribution is both from
the electric field and the magnetic field,
which will be equal to half E E 0 epsilon
naught E 0 square, which is again consistent
with the with the our previous information.
That the Poynting vector S there should be
a mod S is equal to mod of E cross H which
is equal to E 0 H 0 cosine square omega t
minus z, since if you take mod then this z
cap is not required. So, in that case the
average value of the Poynting vector S will
be equal to half of k by omega mu and E 0
square z that will give you this value. So,
we could so that the flow of energy which
is given by the Poynting vector is equal to
this. So, these two are the same equation.
Now, since the wave velocity of an electromagnetic
wave can be given by v equal to omega by k
and k can be replaced by mu epsilon naught
to the power of half. So, we can write that
the average value of the Poynting vector is
equal to the average energy density into the
velocity. That means, the average energy that
is associated with the both the electric and
magnetic field and over a distance which is
given by the velocity will be equal. So, a
S dot n d a that is the energy flow through
the area perpendicular to the area d a is
represented by this which will account for
the net energy flowing out of the closed surface
a .
Now, we will consider another interesting
case where we will see the Poynting vector
through a cylindrical conductor that carries
a steady current I now the electric field
inside this conductor is given by this E equal
to z cap I by sigma into A this is because
this Ohms law if you write J into sigma is
equal to E. So, sigma into E is equal to z
into J.
So, in place of J we have written I by A,
where A is a surface area and sigma is a conductivity
of the conductor and the magnetic field outside
this conductor can be represented by H equal
to phi I by twice pi rho. So, z and phi they
are mutually orthogonal z is along the axis
of the conductor phi is in the as azimuth.
So, if you take E cross H that should represent
a direction, which will be perpendicular to
z and phi and in the outward direction. So,
this will be evidently along the rho direction
that is the radius vector direction of the
cylindrical coordinate system representing
the cylindrical conductor
So, let us see that for the cylindrical conductor
that carries a current I, we can write the
Poynting vector in this way E cross H dot
n d a where we consider a certain length L
of the cylinder.
So, E cross H if we put together, this E cross
H and take the cross product of these two
quantities then we can represent that E cross
H equal to minus rho cap unit vector of rho
that is the radius vector direction by into
I square by twice pi rho sigma into A.
So, this is the value of the Poynting vector
and from the curved surface if you consider
the Poynting vector, then we can see that
we have to take the integration.
for phi which is about the complete azimuth
that is from 0 to 2 pi and for a certain length
L of the conductor which will be from z equal
to 0 to L then, we also write the area elementary
area in cylindrical coordinate systems that
is equal to rho d phi d z and that should
be equal, to this quantity is the constant
we have to take the integration of phi equal
to 0 to 2 pi and L and for z it will be z
equal to 0 to L. So, this is a mistake .
So, by doing that E cross H dot n d a a equal
to minus I square L by A sigma. That L by
A sigma is equal to the resistance, for resistance
we write rho L by A and in terms of conductivity
we write the resistance L by A sigma. So,
this quantity represents that minus I square
R. So, this is a known quantity which appears
in the textbook that this is the joules loss
heating loss .
So, in the case of a cylindrical conductor
which carries a current I we can see that
the Poynting vector if you calculate the Poynting
vector, but the curved surface then it really
gives you I square R with the negative sign,
it tells you that this amount of energy is
dissipated out and away from the conductor.
And J dot E d v will be equal to sigma E square
d V . So, this quantity is the loss this quantity
is the loss. So, that is equal to sigma integration
of E square d V if I substitute for E square,
then we again end up with the same value that
is I square of R; that means, the energy which
is flowing out is equal to the dissipation
of through at the conductor which is carrying
a current I.
So, by this discussion we have considered
the energy transport in electromagnetic waves
which is primarily represented by the Poynting
vector and through the poyntings theorem,
that the total amount of electromagnetic energy
that is flowing inside a certain volume of
space through which the electromagnetic wave
is propagating will be equal to the amount
of energy that is leaving that space plus
the amount of energy, which is lost within
that volume. So, energy transport in the case
of a dielectric medium as well as in the case
of a conducting medium we have also calculated
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