Good morning. Let us, start off by recapitulating
the things that we have learned in the last
class.
So, if you have an electron accelerating in
the direction a direction of this vector shown
over here. Then, in the last class we learned
that in addition to the electric field which
falls as 1 by r square this electron this
charged particle will also produce an electric
field component which falls as 1 by r. And
I had told you and that it is this part of
the electric field that falls as 1 by r which
is responsible for electromagnetic waves,
electromagnetic radiation.
So, if you ask the question what is this radiation
electric field, what is the electric field.
Due to this electron at the point over here
the electric field at this point will be.
So, the way to calculate the electric field
at this point is that you should take the
component of the acceleration perpendicular
to the line of site from the point where,
you wish to calculate the electric field to
the charge. So, this is the line of site from
the point where we wish to calculate the electric
field to the charge. We have to take the component
of the electric field perpendicular to this.
Which is what the dashed line over here shows.
So, the dashed line is the direction perpendicular
to the line of sight the component to the
electric acceleration in this direction, is
what causes the electric field over here.
And the electric field is given by the expression
over here. So, it is minus q where q is the
charge the value of the charge divided by
4 pi epsilon naught c square r r is the distance
from the point where we wish to calculate
the electric field to the charge.
This into the acceleration of the charge at
the retarded time, the retarded time is in
the past. So, if I wish to calculate the electric
field at a time t at this point. I have to
look at the acceleration at a time t minus
r by c. The term minus r by c takes into account
to fact that, the signal takes a finite time
to propagate from here to here. The signal
travels at the speed c the speed of light
and it takes the time r by c to propagate
from here to here.
T minus r by c is called the retarded time,
A is retarded acceleration. So, we have to
take, we have to use the acceleration of the
charged particle not at the time t, but at
a rated time t minus r by c. We have to take
the component of the acceleration, perpendicular
to the line of site which is which is what
give rise to this factor of sin theta. So,
the particle the charged particle over here
if it accelerates produce the electric field
given by this expression.
The key point is that the electric field falls
as 1 by r and it is proportional if the electric
field is in the direction perpendicular to
the line of site and it is proportional to
the component of the acceleration. In the
direction perpendicular to the line of site.
So, this was the first thing that I told you.
The second thing then, we moved on to a particular
situation where we can apply this. And we
were discussing the electric dipole oscillator.
This is the device of considerable technological
importance. So, we have 2 metal wires: A and
B which are aligned like this and these 2
metal wires are connected to a voltage generator,
which produces as an oscillating voltage at
these 2 ends of these wires.
Now, if have a positive voltage here and a
negative voltage here. So, the bottom of A
is given a positive voltage, the bottom of
B is given a negative voltage. Then, there
will be an excess of positive charge at this
tip of A and there will be an excess of negative
charge in the tip of B. These charges the
negative charge here and the positive charge
here will reverse. So, there will be a positive
charge here and negative charge here when,
the voltage is reversed.
So, if i have a positive voltage here, negative
voltage here and a positive voltage here these
charges will get reversed. So, as the voltage
oscillates the charges rush back from A to
B. So, we have a electrons rushing up and
down, from A to B and back and forth. So,
you can think of it as electrons moving up
and down this a single wire electrons accelerating
up and down. These electrons that accelerate
up and down, will produce electric fields
which fall of as 1 by r.
If I am sufficiently far away, you will see
an electric field component which falls of
as 1 by r this is dominate thing which, you
will see if you are far away. And so, we can
apply the expression for the acceleration,
which I just showed to you if you are sufficiently
far away. And if the time the it takes for
the electrons to go from here to here and
then, come back. If this time period is considerably
larger than the time period that light takes
to cross this you can think of it, you can
think of this whole device as an oscillating
electric dipole. So, this is called the electric
dipole oscillator.
So, when you are quite far away at a distance
which is much larger than the length of this.
So, when the oscillations here are quite slow
you can think of this as an oscillating electric
dipole. And I shall go into a little more
detail of this shortly later. Now, this kind
of an oscillating electric dipole has considerable
technological applications. So, much of the
radio receivers, radio transmitters, the antennas
used over there are electric dipole oscillators.
For example: your TV you might have seen the
TV antennas. They are a collection of electric
dipole oscillators. They look like electric
dipolar oscillator which is essentially, are
long metal rod which is cut in the middle
and has 2 leads connected to a voltage source
or possibly to a detector.
So, let me now consider the nature of this
dipole radiation. So, we have a dipole over
here we have this metal rod AB which I just
showed you over here. And we have a signal
voltage source connected to it. So, this is
the generator of the electromagnetic radiation,
the electrons accelerate up and down and these
accelerating electrons produce electric fields
elsewhere.
So, this is a common situation where I wish
to transmit some kind of an electromagnetic
radiation. I have a dipole oscillator and
I am, I have connected a voltage source; the
voltage source is the signal which I wish
to transmit. So, the voltage source is driving
my electrons up and down this metal, these
2 metal rods. And this gives rise electric
fields elsewhere.
Now, the question is then, I now wish to study
the electric field pattern So, let us consider
a situation where I have a detector located
elsewhere. So, this is what this picture shows
us. I have a source the generator over here
and I have a detector located a large distance
away from this. So, where is the detector
located over here? The detector also is a
dipole. So, we have a dipole here and a dipole
here.
The difference between the dipole here which
is the detector and the dipole which is the
generator is that the dipole over here. And
which, is the detector is connected to a voltage
detector could be an oscilloscope. So, if
I the electric field produced by the generator
when, it falls on a dipole located over here
D when, the electric field from the generator
falls on this.
If the electric field is aligned with the
dipole, it will produce a voltage across the
dipole and this will cause the current to
flow in this dipole. So, if you connect a
voltage detector then, it will measure a voltage
across the dipole. So, you could connect an
oscilloscope and you could measure the voltage
across the dipole. So, you can think of this
dipole as you can think of our radio antennas
or TV antennas as being some kind of typical
dipole.
So, I can use the dipole to measure the electric
field produced by this generator which is
also a dipole. So, let us consider the situation
where I have a dipole located. Let us, say
to start with over here. The dipole is aligned
in the same direction as the dipole which
is generating the signal. So, on the detector
dipole is aligned in the same direction as
the dipole which is generating the signal.
Now, the electron which is generating the
signal is running rushing up and down this
dipole the generator. So, let us ask the question
what kind of an electric field will be produced
over here? So, let us go back to our expression
for the electric field. The electric field
I told you at this point if the electron is
accelerating over here, the electric field
here will be parallel to the component of
the acceleration perpendicular to the line
of site.
So, this is the point where I wish to calculate
the electric field over here, this is the
point where I wish to calculate the electric
field. The line of site from here to here
is in this direction. So, the component of
the acceleration perpendicular to the line
of site is this direction. So, the electric
field here is basically parallel to this.
So, it is going to oscillate up and down,
the electric field is going to oscillate up
and down over here.
Then, it is going to produce a voltage in
this dipole which you can measure. Now, let
us consider another situation where I have
a moved the dipole on a circle. So, the think
of the dipole as being attached to this generator
through a rod or something like that. I move
it around, maintaining the same distance and
move it to a different position shown over
here. Now, if I keep my dipole over here and
I keep it aligned like this then, the electric
field then let us, ask the question. What
is the direction what is the electric field
here like?
So, you have to take the line of site from
this point to the generator, this line over
here shows us the line of sight from the point
where I wish to calculate the electric field
to the electron that is oscillating up and
down. You have to take the component of the
acceleration perpendicular to the line of
site. So, you have to take the component perpendicular
to this. So, the component perpendicular to
this line of site is going to go down by a
factor of sin theta.
So, if the electric field here is going to
be parallel to the component of the acceleration
which is projected perpendicular to the line
of site. So, it is going to be along the tangent
to the line of site which is the way the dipole
is oriented. Now, let us again consider as
another position for the detector. So, if
i move the detector all the way over here,
where it is directly overhead, directly overhead
to the dipole.
Let us, repeat the same exercise. The line
of site from the position where I wish to
calculate the electric field to the dipole
is this the particle the electron is accelerating
this way it has no component, the acceleration
has no component, perpendicular to the line
of site. So, the electric field produced over
here is 0. So, the electric field is maximum.
When, the at this point it falls as 1 sin
theta as I moved further, up as I move as
theta is varied is as theta. So, theta when
theta is 90 sin theta has the maximum value
as theta is decreased. The value of electric
field falls and it is 0 over here. The direction
of the electric field at this point is in
this direction it is the tangent to the radius.
So, it is the tangent to the line of site
which is in this case the radius. So, the
electric field is perpendicular to the radius
which is the tangent.
So, at this point the electric field is in
this direction at this point, the electric
field will be in this direction at this point
the electric field is 0. But this would be
the direction it could be normal to the line
of site, but it is has a magnitude 0. If I
move a slight distance away, it will be the
tangent. So, the electric field that is produced
is perpendicular to the line of site this
is the point which is important.
Another point, which is important is if I
put the dipole instead of putting the dipole
in this direction. If I were to put the dipole
in the direction which is perpendicular to
the electric field.
So, if I put the dipole like this and the
electric field oscillates this way. So, I
have a dipole in this direction and the electric
field oscillates like this. Then, this electric
field will not produce any voltage difference
across this dipole and this dipole this detector
will not be able to detect this electric field.
So, this is the point which we should bear
in mind.
If the detector dipole is perpendicular. So,
if the metal wire AB which is which I am using
to detect the electric field is perpendicular
to the direction of the electric field it
will not the experience any voltage difference
and there will be no signal picked up here.
Signal will be picked up only if the dipole
is aligned. So, if the dipole which I am using
to detect the electric field is aligned like
this. In the same direction it is only then,
that it will then there will be a voltage
difference produced across this and there
will be a signal picked up.
So, coming back to this picture over here
if I have got to put a dipole in the perpendicular
direction here I would get nothing. Right
the dipole here is aligned with the electric
field. So, I get a signal. So, this essentially
summarizes the direction dependence of the
signal of the electric field produced by the
dipole oscillator.
Now, we going to concentrate on a situation
where the signal that I am feeding in.
So, the signal that i am feeding in to the
dipole oscillator we are going to assume that
the signal is cosine is doing. So, the signal
which I am feeding into the dipole oscillator
the voltage generator is producing a sinusoidal
voltage pattern. So, if it if the voltage
generator produces a sinusoidal voltage pattern,
the electrons are also going to rush and rush
back and forth.
Then, the motion of the electron between those
2 wires: A and B across the 2 wires moves
up and down the 2 wires is going to be sinusoidal.
So, we are going to focus for the rest of
this we are going to focus on this of today’s
lecture. We are going to focus on this particular
situation. And in this situation, where the
voltage oscillates as a sinusoidal. We can
write down the displacement of the electron
up and down this dipole oscillator.
So, Y the dipole oscillator here is aligned
with the y axis and the motion of the electrons
as they move up and down this we can write
as y the displacement in this direction charged
particle moving up and down. The displacement
in this direction is yt equal to y naught
which is the amplitude of the displacement
into cos omega t. So, let me just remind you
of the situation again.
We have this kind of an electric dipole oscillator
the 2 metal rods: A and B or aligned with
the x with the y axis. And we had applied,
we have applied an oscillating voltage source
which is oscillating in a sinusoidal fashion
cos omega t. So, the electrons are also going
to go up and down in a sinusoidal fashion.
And we express the motion of the electrons
up and down the dipole oscillator as y naught
cos omega t. The electrons move back and forth
along the y axis. Now, we want to calculate
the electric field pattern at different points.
So, the electric field pattern at any arbitrary
position distance r away from the dipole can
be calculated in using this expression. So,
the electric field at a time t a distance
r away is minus q by 4 pi epsilon naught c
square r. Into the acceleration at a retarded
time t minus r by c into sin theta, where
sin theta is the angle between the point where
I wish to calculate the electric field and
then, direction of the dipole.
So, in this case it is the angle with respect
to the y axis. So, this is the expression
for the electric field in terms of the acceleration.
And in this case the particle, the displacement
of the particle is sinusoidal. So, it is y
naught cos omega t this is the displacement
I have to differentiate this twice to calculate
the acceleration. If I differentiate this
twice, I pickup up a minus sign and a factor
of omega square outside.
So, putting this into the expression for the
acceleration, we get this expression for the
electric field here. It is q into omega square
into y naught by 4 pi epsilon naught c square
r into cos omega t minus rc. So, is this is
the cosine at a retarded time t minus rc into
sin theta. So, this is the electric field
this gets us the electric field at the point
where we wish to calculate it and this can
be applied to any at a large distance r in
any arbitrary direction theta.
Now, it is convenient to think of this whole
thing. So, we have the whole thing is that
we have an electron, we have this these 2
things. So, you can affectively think of is
at 1 we have single we have some charge q
which is rushing back and forth up and down
this dipole. Now, it is convenient for certain
purposes to think of it as of 2 charges: 1
q by 2 and another minus q by 2.
And the charge q by 2 is doing this oscillation
yt which I just wrote down. And the charge
minus q by 2 is doing exactly the opposite
oscillation minus yt. So, a single charge
moving up and down. I can think of as 2 charges:
1 of magnitude q by 2, the other also of magnitude
the q by 2 with the opposite sign doing exactly
opposite oscillations. Now, the point is that
if I replace the a charge q with minus q and
replace the acceleration a with minus a such
a change does not change the electric field.
So, if I change its sign of the charge and
the sign of acceleration both, the electric
field is not changed. So, we are essentially
using this property. So, I am replacing 1
single charge q moving up and down with y
of t by 2 charges 1 q by 2 moving with yt,
another minus q by 2 moving exactly opposite
minus yt. So, the both of them produce exactly
the same electric field and sum of these 2
electric fields is exactly the same as the
electric field the radiation pattern of the
part of the electric field produced by this.
Now, if I have a charge q by 2. So, I have
a charge q by 2 and it is at a displacement
y. And I have another charge minus q by 2
at a displacement minus y this is a dipole
with a dipole moment d this d is the charge
into the distance. So, the charge is q by
2 the distance is 2y. So, this dipole moment
is q into y as a function of t. So, this signal
charge. So, as far as the radiation pattern
is concerned the 1 by r component of the electric
field is concerned a single charging move
up and down and the displacement being y of
t.
I can think of is 2 opposite charges of half
the magnitude moving up and down. And the
1 of them moving up and the other 1 moving
down by the same amount. And this can be thought
of. So, the signal charge moving up and down
can also be thought of as a dipole. Which
is doing the exactly the same kind of a oscillation.
So, the signal charge is moving as yt equal
to y0 cos omega t and we can think of this
as also being equivalent to a dipole dt which
is q y naught cos omega t.
Which I can write as d0 cos omega t. So, which
is what is given. So, what I have shown you
is that, a single charge moving up and down
this I can think of as a dipole which is oscillating,
as an oscillating dipole. So, the dipole moment
the magnitude of the dipole moment is d naught
which is equal to q into y naught into cos
the and the whole thing oscillates. So, it
is d naught into cos omega t.
And we can now, write the expression for the
electric field over here. Which we had the
expression for the electric field, it can
also be written in terms of the second derivative
of the dipole moment. So, when we have an
oscillating dipole it produces a radiation
electric field pattern which falls as 1 by
r. And the radiation electric field pattern,
is given by this particular expression over
here.
So, the accelerating charge moving up and
down you can also think of as a as an oscillating
dipole, as a as an oscillating electric dipole.
And the expression for the electric field
you can interpret in terms of an oscillating
dipole. So, what we see here is that if I
have an a dipole, an electric dipole positive
and negative charge which is oscillating.
It could be set into oscillations in in a
variety of ways.
So, I could have real physical oscillations
of a dipole or I could have a positive and
negative charge moving up and down. I could
have a dipole which is rotating, all of these
situations give you an oscillating dipole
and there are a large variety of other situations
also. All of these can be thought of as an
oscillating dipole and the electric field
pattern produced by this is given by the expression
over here.
Another convenient way and another convenient
way of expressing the electric field of the
same, electric field is in terms of the current.
So, this oscillating dipole let me is in terms
of the current. So, if when I have when, I
apply a positive voltage to 1 of these bits:
A and the negative voltage to B and then,
if the voltage is reversed there will be charges
rushing from A to B. And then, when the voltage
is reversed again the charges will rush back
and charges keep on moving back and forth.
Now, when charge is moved you have a current.
So, the current is the rate of change of the
charge, the rate the current is the rate at
which charge flows that is the rate of change
of the charge at the 2 tips. And the dipole
the rate of change of the dipole the second
derivative of the dipole over here in this
expression you have the second derivative
of the dipole.
So, the second derivative of the dipole moment
is the first derivative. So, you have the
dipole is the distance the separation into
the rate at which the charge changes is the
displacement of the charge. So, the second
derivative of the dipole moment you can write
in terms of this. So, the l into i dot the
rate of change of the charge, the rate of
the change of the current.
And if I have current which is minus I sine
omega t then, the rate of the change the second
derivative of the dipole moment is going to
be minus l into i into omega cos omega t.
So, we can also write the electric field in
terms of the current. The I here is the amplitude
of the current and this is the expression
for the electric field in terms of the current.
So, what we have done is we have written the
expression for the electric field in this
situation in terms of the dipole moment and
also in terms of the current. So, all of these
expressions have their utility depending on
the situation that you are analyzing.
A point which I should make here is, that
if you look at the expression for the electric
field in terms of the dipole moment or in
terms of the acceleration then, the electric
field depends on omega squared. So, if you
double the angular frequency the electric
field will go up 4 times. And this is true
if you maintain the amplitude of the displacement
fixed or if you maintain the amplitude of
the dipole moment fixed and double the angular
frequency.
But if you maintain the amplitude of the current
fixed and double than angular frequency. Then,
the omega dependence note that, the omega
dependence is different and the electric field
will only go up twice. So, this is the point
which needs to borne in mind that the omega
dependence depends on the variable in terms
of which you are expressing the electric field.
And it will have a omega or omega square dependence
depending on the variable in terms of which
you are writing the electric field.
Now, let us go back to this situation which
we were considering. So, we were considering
a situation where I have a dipole oscillating
along the y axis and we would like to calculate
the electric field pattern at difference positions.
Different positions means, different value
of theta and different distance r. And we
diagnosed a little we took a the little detour.
Where, we wrote down the electric field in
terms of different variables the rate of the
change of dipole moment, the rate of change
of the current. But let us, now come back
to a study of the electric field. So, this
is the dipole which is oscillating and let
us calculate the electric field at different
points along the x axis for this. So, to simplify
matters let us restrict our attention to points
along the x axis.
So, the expression for the electric field
at different points along the X axis is this.
So, we have the expression for the electric
field. And we have replaced the distance r
with x. And we also have this, simplification
that along the when, you wish to calculate
the electric field along the x axis. You have
to take the component of the acceleration
in the direction normal to the x axis.
In this case, the electrons, the charged particles
accelerate along the y axis the dipole is
aligned with the y axis. So, the charged particles
accelerate along the y axis. If I take the
component of this acceleration in the direction
perpendicular to the x axis it is still along
the y axis there is no change. The electric
field will be parallel to this component perpendicular
to the x axis.
So, the electric field will be along the y
direction you will only have a y component
of the electric field which is what I have
written here. So, you are going to have an
electric field along the x axis, you are going
to have an electric field only along the y
direction. And the magnitude of the electric
field is given by this. It is has only the
y component and at a distance x away you will
this the magnitude of the electric field.
So, the point to note is that it will depend
on the acceleration of the charges at a retarded
time, the retarded time is t minus x by C.
And there is a factor of omega which is there
for both of these. So, this takes into account
the retardation of the, retardation which
has to be put in. Now, we shall consider a
situation.
So, let me now explain to you as a the situation
which we shall consider. We shall be looking
at a situation where, the dipole is located
at a large distance from the point where I
wish to calculate the electric field. So,
we could take for example: the point where
I wish to calculate the electric field is
located at a large distance say 1 kilometer
away from the dipole.
Then, we now move from a point which is 1
kilometer away, we move to a point which is
1 meter away from this. So, this figure is
not to scale you should bear in mind. So,
I wish to calculate the variation in the electric
field. When, I move from a point which is
1 kilometer away, to a point which is 1 kilometer
and 1 meter away. So, the change in the x
is 1 meter and the distance was 1 kilometer
to start with.
So, you see that the fraction of the change
in the distance is very small it is 1 by 1000.
Which is equal to 10 to the power minus 3.
So, it is a 0 1 percent change in x. So, if
I move 1 meter from 1 kilometer to 1 kilometer
plus 1 meter, there is a 0 1 percent change
in x. Now, let us look at the expression for
the electric field and ask the question how
will the electric field vary? If I change
x by this small amount.
So, if I change x by this small amount the
term outside over here is going to change
by only 0 1 percent. And I can think of this
term as being roughly constant. The change
in x is only 0 1 percent. So, this term here
is also going to change by only 0 1 percent.
But look at this term the cosine term. If
x changes by point one percent it is not guaranteed
that this the change in this term is going
to be small there could be a very large change
in this term over here. And the magnitude
of this change is going to depend on the value
of omega by c. Now, if omega by c is very
large then, a small change in x will cause
a large change in the phase. So, it will cause
a large change in the argument of this cos
term and this change in the argument of the
cos term, may cause a significant oscillation.
See cosine is oscillating
So, a small change in x may cause a well a
considerable oscillation in this cosine term.
If omega by c is a large number. So, let us
consider the situation where we can ignore
the change in x in the first term, but we
have to take into account the change in x
for the term inside this cosine. So, I can
think of this as a constant, but I cannot
think of this x as a constant.
So, if I make this assumptions. So, if I work
in this regime we can write down the expression
for E the y component which has a y component.
So, the electric field which has only a y
component the expression for that, as i vary
x and as a function of time is a constant
into cos omega t minus kx where I have used
k to denote omega by c. So, I have used k
to denote this coefficient omega by c which
multiplies x.
So, the point to note is that the expression
for the electric field. If I vary x by a small
amount at a large distance from the dipole
oscillator is just the expression for the
sinusoidal plane wave which we had studied
earlier. So, this is the expression for the
sinusoidal plane wave which we had studied
earlier. So, what does it tells tell us it
tells us that if we give a sinusoidal voltage
to this dipole oscillator far away it produces
an electric field pattern.
So, at a fixed time this electric field pattern
is sinusoidal and as time varies the whole
sinusoidal electric field pattern will move
forward. The electric field is only along
the y axis and it has a sinusoidal pattern
like this as time increases, this whole thing
will move forward it will behave like a sinusoidal
plane wave. So, this oscillating dipole produces
a sinusoidal plane wave at a large distance
from the dipole.
So, what we see here is that if I have a sinusoidal,
if I have a dipole oscillating like this and
it is doing sinusoidal oscillations it will
produce a sinusoidal plane wave at a large
distance. So, if I go a large distance away
from this oscillator. The electric field pattern
produced by this will be a sinusoidal plane
wave. The electric field pattern at a fixed
time will look like the sinusoidal pattern
shown over here. And this whole pattern will
move forward in time as we have studied in
the lecture on sinusoidal plane waves.
So, coming back to our expression for the
electric field you can write it in this form
omega t minus kx where we have identified
the wave number k with omega by c. This is
a sinusoidal plane wave which, propagates
along the plus x direction. Now, let us ask
the question what is the wave number? The
wave number here is omega by c. Now, once
you know the wave number and the angular frequency,
you can calculate the phase velocity and the
phase velocity is omega divided by k. So,
omega divided by k gives us c.
So, you find that the electromagnetic wave
in this situation, the electromagnetic wave
has the phase velocity which is the speed
of light. So, the dipole oscillator over here
if that be the sinusoidal voltage produces
a sinusoidal plane wave at a large distance.
So, over here this the electric field pattern
produced by this is a sinusoidal plane wave
which keeps on propagating outwards.
Then, you could express this in the complex
notation. So, E tilde y as a function of x
and t is E tilde e to the power i k omega
t minus kx, this complex amplitude has both
the magnitude and the phase of the electric
field.
Then, Coming back to the our picture we have
the dipole oscillating like this. And if you
ask the question, what is the electric field
at some arbitrary point here far away. Then,
you have to take the component of this oscillation
that it is the component of the y axis in
the direction perpendicular to the line of
site. So, you have to take the component perpendicular
to this and this is the direction of the electric
field which is shown over here.
So, the electric field is going to oscillate
perpendicular to the direction of line of
site. So, it is going to oscillate in this
direction. The wave is going to propagate
along the line of site. So, which is the direction
k this is the wave vector it points from the
dipole to the point where I have calculate
the electric field. So, the wave is going
to propagate in this direction and the electric
field is going to oscillate in this direction.
So, it is magnitude is going to be smaller
than the magnitude of the electric field here.
And this is going to be smaller by a factor
of sin theta, but theta is the angle between
this direction and the dipole. And if I look
in this direction theta is going to go to
0. So, sin theta becomes 0, there will be
no electric field produced by the dipole oscillator
in this direction.
So, for any other direction other than these
this dipole oscillator far away is going to
produce a sinusoidal plane wave. The magnitude
of that, the magnitude of that sinusoidal
plane wave is going to fall as sin theta.
Now, next let us calculate the magnetic field.
The expression for the magnetic field is given
over here. You have to take the unit vector
along the line of site to the dipole. And
do a cross product with the electric field
divide by c and there is a minus sign here.
So, in the problem which we are dealing with
this is the point. Let us, again go back to
the x axis. The line of site is along the
minus x axis. So, the E cap vector is minus
i.
So, the E cap vector is minus i and the electric
field is along the y axis. So, it is along
j the magnetic field is going to be minus
E cap which is i cross E by c. So, the magnetic
field is E by c, the magnitude of the magnetic
field is E by c. It is in the same phase so,
it has cos omega t minus kx and it is in the
z direction i cross j is the unit vector k.
So, it is in the z direction.
So, going back to our picture. Along the x
axis the electric field is going to oscillate
in the same direction as a dipole which is
along the y axis. And the magnetic field is
going to be perpendicular to both, the direction
of the propagation of the wave and the electric
field the magnetic field is going to be along
the z axis. So, this is a typical feature
of electromagnetic waves.
In electromagnetic waves, the electric the
wave propagates in a particular direction.
The electric field is perpendicular to that
and the magnetic field is perpendicular to
both the direction of propagation and the
electric field. The oscillation of the electric
field and the magnetic field or both in exactly
the same phase. The magnitude of the, magnetic
field is a factor one by c smaller than the
magnitude of the electric field. So, this
is the feature which is typical of all electromagnetic
waves.
So, in this part in the previous lecture and
this part of the lecture we started of with
a loss which govern the electric field and
magnetic field produced by an by a charge.
And I showed you that, there is a term which
falls as 1 by r. So, this is term arises only
when there is an accelerating charge and if
I have a sinusoidally accelerating charge
or a dipole which is oscillating it then,
I showed you that such a thing produces a
sinusoidal plane wave and electromagnetic
wave at large distances.
Then, this electromagnetic wave has a direction
of propagation the electric field is perpendicular
to that the magnetic field is perpendicular
to both electric field and that direction
of propagation and it is in phase with the
electric field. So, this is very generic feature
of electromagnetic radiation. And I also showed
you, how such electromagnetic radiation can
be produced, how it can be generated using
dipole oscillators. Let us, now move ahead.
So, let us calculate the energy density in
this electromagnetic radiation. Now, all of
you must have already learnt that energy that
there is some energy in electric and magnetic
field configuration. And the energy in the
electric field configuration is half epsilon
naught e square. The energy in the magnetic
field configuration is half b square by mew
naught.
So, let us now calculate the energy density
that is energy density. So, let us now calculate
the energy density in this electromagnetic
wave produced by the oscillating dipole. So,
this is the expression for the energy density
as I just told you which I am sure is familiar
to all of us. And for the electromagnetic
wave we saw, that the electric and magnetic
field they are not independent they are both
produced by the same source.
The magnetic field the magnitude of the magnetic
field, is the magnitude of the electric field
divided by c. So, I can replace the term arising
from the magnetic field and write it in terms
of the electric field as 1 by 2c square mu
naught square into E square. We also know,
that the constants epsilon naught and mu naught
are related to c square they are not independent,
they are related to c square and c square
is 1 by epsilon naught into mu naught.
So, using this expression we can write the
expression for the energy density in terms
of the electric field as epsilon naught E
square The 2 terms exactly turn out to be
exactly equal. The electric field and the
magnetic field it turns out contribute the
same amount to the energy density and the
energy density is epsilon naught E square.
So, we let us just go back to the situation
we have an oscillating dipole or we have a
charge going up and down that an angular frequency
omega.
So, this produces an electric field far away
which also does oscillation that exactly is
the same angular frequency. So, the electric
field is also oscillating at an angular frequency
omega. Now, the instantaneous energy density
in the electric and magnetic field, the magnetic
field is also oscillating at the same frequency.
So, the instantaneous energy in the energy
density in this electromagnetic field we just
calculated that it is epsilon naught into
E square.
So, if the electric field oscillates with
the angular frequency omega, the energy density
depends on E square. So, we have already seen
right in the first lecture that the energy
density is going to also oscillate. The instantaneous
energy density is also going to oscillate
and it is going to oscillate at twice the
angular frequency of the electric field. So,
the instantaneous energy density that is what
we have calculated here is going to oscillate
at twice the angular frequency at which the
electric field is oscillating.
Now, the quantity of interest is not the instantaneous
in most situations the quantity of interest
is not the instantaneous energy density, but
the time average energy density. In most situations,
the electromagnetic field oscillates quite
fast and the quantity that we measure is not
the oscillating energy density, but the time
average energy density. For example: the bulb
which illuminates the room is the is emitting
radiation, this radiation is oscillating at
a frequency the value of that frequency we
shall discuss after 1 or 2 lectures.
But it is an oscillating as we have seen the
electromagnetic radiation is an oscillating
electric field. So, the value of the electric
field is oscillating. But we see a steady
illumination that is because, our eye and
most optical devices measure only the time
average energy density. They the record the
energy over a time period which is much faster
than the time at the, rate at which the electric
field is oscillating.
So, they only measure the time averaged energy
density. So, we have we wish to calculate
this quantity which is of practical interest.
So, to calculate this we express we electric
field in the complex notation as you can see
here. And then, the time averaged energy density
is the, time average of E square which in
the complex notation is E into E star by 2.
So, this gives us gives us the average energy
density it is half epsilon naught E square.
Where E is the magnitude of the oscillating
electric field. This the expression for the
average energy density in an electromagnetic
radiation.
Now, remember that the we have, we have energy
in the electromagnetic field, but the electromagnetic
field pattern is a sinusoidal plane wave.
The sinusoidal plane wave is propagating forward
in this case along the x axis we were discussing
waves along the X axis. So, the sinusoidal
plane wave is propagating forward along the
x axis and it is propagating forward at a
speed c.
So, the energy density in this electromagnetic
field does not remain fixed over here the
whole thing propagates forward at a speed
c the speed of light. So, let us now ask the
question we take a surface perpendicular to
the direction in which the wave is propagating.
So, in this case the wave is propagating along
the x axis. So, we take a surface perpendicular
to the x axis. And ask the question how much
energy crosses this surface, crosses a unit
area of this surface per second.
So, what is the energy density? By energy
density we mean, the energy the energy flux.
So, this is what energy flux: the energy flux
is the energy which crosses per unit area
of the surface in a unit time. And this is
the energy density into c because, the whole
thing is moving forward at a speed c. So,
if you take a unit area and ask the question
how much energy will cross it in a second.
The amount of energy that will cross it in
a second is the energy density into the speed
at which it is moving which is c. So, this
gives us the energy flux.
Let me, just elaborate a little of this. So,
energy flux is the energy per unit area per
time per unit second. And this is the quantity
which is of considerable practical importance
because, whenever we have a detector for example:
we have a detector of some sort which is measuring
radiation. The quantity that we detect is
the amount of radiation, the amount of radiation
energy per unit area.
So, we have the area of the detector if i
the double the area of the detector, I will
get twice the radiation of the energy. So,
the amount of the energy per unit area of
the detector per second. And this has a unit
of so, energy per unit time is power this
is also you can think of it as power which
crosses per unit area. And this has got units
of joules per meter square per second or you
can also say, that it has units of watt per
meter square.
The power which crosses per unit area, the
power of the radiation the power from the
radiation per unit area. And this is the energy
density into c.
So, calculating this energy density into C
we have half epsilon naught into c into E
square. Which gives us the energy flux. You
should also remember, that the energy flux
is a vector, it is a vector in the direction.
So, this energy is moving in the direction
in which the wave is propagating. So, it is
a vector and if I want to calculate it at
some other point the energy would be moving
in some other direction.
So, along the x axis the energy is flowing
along, the x axis it is the energy density
into c. If I want to calculate the energy
flux here it could be a vector along the this
direction along the direction of the wave.
And it would be the energy density here into
c. It should be in this direction. So, in
the next let me stop here for today. In the
next class, we shall calculate this expression
for the flux and go ahead for the further
from there.
