In the previous lecture, we had started discussing
the time dependent phenomena, and I
had enunciated what is known as Faraday’s
Law, what faraday had proposed based on
certain experiments is that, whenever there
is a relative motion between a magnet and
a
circuit, or if the magnetic fields trying
this change in with time that is if any of
these
effects, results in the flux through a circuit
changing. Then it leads to an emf in the
circuit and as you had seen emf, we defined
emf as line integral of electric field. And
so
what we had seen is that the time dependent
phenomena, lead to a non conservative type
of electric field, because del cross of E
did not becomes 0, but had a value.
.
So, this what we did that the emf, which is
the line integral of the electric field is
rate of
change of flux.
..
So, emf is given by minus d by d t of B dot
d S, and emf by definition is the line integral
of the electric field, which by Stokes law
we know that we can write it as, a surface
integral of the curl of the electric field.
And by comparing this expression with this
expression, since this is valid for any surface,
we can write down a relationship which is
the differential form of faraday’s law that
is del cross of E is equal to minus d B by
d t.
So, notice that for the static phenomena,
my del cross of E was equal to 0, but now
I
write del cross of E as time derivative of
the magnetic field. So, I repeat again that
whatever is the cause for a flux through a
circuit changing, the we talked about motional
emf which arises when there is a relative
motion between the magnetic field and the
circuit. For instance either a magnetic is
moved towards a circuit, or if a circuit moves
towards or away from a magnetic field, which
is of course we know by relative motion it
is the same effect.
The other problem is that while this motional
emf could be explained by invoking the
Lawrence force, that is as the circuit moves,
the charge on the circuit also experiences
a
force due to the magnetic field. But, what
was surprising is that even if there is no
relative motion, if the magnetic fields strength
changes with time, even then there is an
emf, this is born out by the experiment. So,
therefore, we postulated that, whenever there
is a changing magnetic field, that is a change
flux changes through a circuit, for whatever
.reason; then there is an induced electric
field produced, and that is what is contained
in
this law that I have written down.
.
With this let me introduce the concept of
an inductance, so notice that supposing I
have
two circuits, I have called this loop 1 and
this I have called as loop 2. Now, let us
suppose that there is a magnetic there is
a current which is flowing through loop 1,
and
this current is changing with time, as a result
the magnetic field that it produces changes
with time. And the loop 2 is a loop which
intersects these magnetic field lines, so
therefore, as the magnetic field strength
changes, magnetic field strength produced
due to
the 1st loop, changes it results in a change
of flux in the 2nd loop.
..
Let us now look at, what it actually means,
now if you recall you are Biot-Savart Law
it
tells me B 1, because I have a loop 1, B 1
is the magnetic field due to the loop 1 that
is
lower loop in that picture; that is given
by mu 0 by 4 pi. Then the integral over the
loop
of d l 1 my index 1 subscript 1 is suppose
to be for the loop 1 cross r divided by r
cube.
So, this is the magnetic field produced by
the current in the 1st loop, so therefore,
what I
get is this mu 0 I have must put it there,
so the flux through the second circuit, this
circuit
is given by of course, integral B 1 dot d
S, if you like I will write it as d S 2.
So, B 1 is the magnetic field produced by
1st loop and d S 2 is because, I am integrating
over the surface of the 2nd loop, so you notice
since B 1 is proportional to the current,
let
me write it as a I 1 so phi 2 which is the
flux through, the second circuit is proportional
to the current in the first circuit. So, this
tells me that the flux I can write, the flux
to the
second circuit, I can write as equal to some
constant, let me write it as M 2 1 multiplied
by I 1.
So, therefore, flux in the second circuit
is proportional to the current in the first
circuit,
and that is what it is, and now the emf in
the second circuit, which is minus d phi 2
by d t
that is simply equal to minus M 2 1 times
d I 1 by d t. So, notice that this proportionality
constant, which comes when we express the
emf in the second circuit, with the rate of
change of current in the first circuit, this
is what I call as a mutual inductance. The
index,
the indices have been subscripts have been
written so that, this M 2 1 stands for the
.mutual inductance, when I am talking about
the emf in the second circuit due to the
cause is in the first circuit, the effect
is in the first index.
.
Now, we can do something interesting those
formula that I will derive is of no great
use,
but on the other hand, it establishes two
things. Firstly, it will establish that the
mutual
inductance is a property, which depends upon
the geometry of the loops, relative position
of that loops, and also I will show that M
2 1 is equal to M 1 2 that is whether, I am
changing in the changing current in the plus
circuit, and looking at the effect in the
second circuit, that is major the flux in
the second circuit. This gives me the same
proportionality constant, if I do the reverse
namely, if I change the current in the second
circuit, and look for the flux in the second
circuit.
..
So, therefore, let us look at what we did,
so for example, we said that phi 2 is integral
B
1 dot d S 2, B 1 because the magnetic field
is produced by the first loop. Now, let me
express this, in terms of the vector potential
produced by the current in the first circuit.
So, I will write this as del cross A, I will
write as A 1 again all the quantity dot d
S 2, and
my Stoke’s law this would become line integral
of A 1 dot d l 2 notice that, the vector
potential is produced due to the first circuit,
the loop integral is taken over the second
circuit.
But, I know that A 1 that is the vector potential,
produced by the current I 1 in the first
circuit is given by mu 0 I 1 over 4 pi, line
integral of d l 1 over r, this is something
which
we had discussed, while we defined the vector
potential. Now, if I plug this into this
expression, so this tells me that phi 2 is
given by mu 0 I 1 by 4 pi integral d l 1 dot
d l 2
over r; so this tells me that M 2 1 which
is what we wrote down, which is proportional,
the proportionality constant of phi 2 with
I 1 is given by mu 0 by 4 pi double loop
integral of d l 1 dot d l 2 divided by r.
Now, you can see that this expression is symmetric
in one, and two because, this r is
actually r 1 2, so therefore, this is identical
to what you would have, if you reverse the
procedure; namely M 1 2 is equal to M 2 1,
thus this is what is known as Neumann's
formula. Neumann's formula establishes that,
the number one the structure tells us, it
.depends only on geometry and the relative
position, the fact that M 2 1 is equal to
M 1 2
tells us that the mutual inductance is symmetric
in the indices of the loops.
Now, now I come to a rather curious facts,
if you change the current in loop number 1,
the flux through loop number 2, it cuts the
flux through loop number 2 if it change, and
establishes a emf. But, let us now ask the
following question supposing I do not have
two
loops, I just had one loop, now this loop
which I have when I change the current through
this loop itself, the magnetic field produced
by this current changes, and as a result flux
through this circuit also changes. So, in
some sense it is a self effect we are talking
about, and since the when there is a change
in magnetic flux through a loop, the loop
does not care about what caused this changing
flux, as long as there is changing in flux
it
will generate an emf.
So, what will happen that you change the current
through a loop, it leads to changing
magnetic field produced by that loop, and
which results in changing magnetic flux
through the loop itself. So, this would mean
that you have what is called a self effect,
that is the magnetic field when it changes
and leads to changing magnetic flux, there
will
be an emf generated in the same circuit.
.
So, I will write that, this emf this emf is
minus by definition minus d phi by d t, and
let
me write this as minus delta phi by delta
I and of course, delta I by d, d I by d, now
this
proportionality constant that you have got
here, rate of change of flux through a circuit
.with respect to the changing current in that
circuit itself is known as a self inductance,
and it is denoted by letter L. So, therefore,
the definition of self inductance is d phi
by d
t, where phi stands for changing flux in a
given circuit, so we have talked about what
is
self inductance and what is mutual inductance,
but let me first calculate just as
illustrative example, let me talk about calculation
of some simple cases of self and
mutual inductance.
.
So, here in this picture, you have find that
there is a current through a solenoid, and
this
current is going to be changed. Now, I know
f there is a current in a solenoid current
I,
then this leads to a magnetic field B given
by mu 0 n I.
..
So, magnetic field B in a solenoid is mu 0
n I, and actually it is direction I should
specify
that direction is along the axis, let me take
it along the z axis. Now so basically, what
is
happening is this, so I have this solenoid,
now when the current changes through the
circuit of course, the magnetic field changes,
but then there is a concept of how much of
flux is linked with each loop. So, we say
it is flux linked with each loop, with each
turn if
you like So, this is equal to simply the pi
times R square, which is the area of that
loop,
multiplied by the magnetic field which is
mu 0 n I. So, if you take a length L of the
loop,
let me write it as small l, so that I do not
confuse it with the notation for the self
inductance itself, now I know that these many,
these length since the number of terms
per unit length is small n, then number of
turns is equal to n times l. So, total flux
linked,
this is equal to pi R square mu 0 n I have
a another n from there, so I have a n square
l I,
and the now of course, the self inductance
is simply d phi by d I and this is linear
n I, so
this works out to pi R square mu 0 n square
times l.
So, this is the expression and you can see,
it depends upon only the dimensionalities
the
geometry of the radius, there are number of
terms, the length of that loop. So, this is
an
example of calculation of self inductance
of a solenoid, now let us do the following,
let
us calculate the mutual inductance of the
solenoid.
..
So, here what I done is this, simple I made
it somewhat simple, I have put one loop over
the other, you can see this in this picture,
there is blue colored loop, blue or green
which
I was talking about earlier; and now I have
superposed on it rather tightly. The another
solenoid, which let us say we will assume
that the first solenoid has let us say n 1
number
of turns, the second solenoid will have let
say n 2 number of turns, and they are tightly
bound. So, that we assume all that magnetic
field that is produced by one of the
solenoids passes, or a will be linked with
the flux will be totally linked with the other
one, So, let us look at how much flux will
be linked.
.
.So, I am talking about now, phi 2 I am changing,
I am passing a current I through the
first loop, and which we had seen gives me
a magnetic field mu 0 n I, I is the current
through the first loop, and the total area
is the area of each turn is pi R square, then
total
number of turns that I have in the second
loop is n 2 into l. So, since n 1 2 or M 2
1 is
nothing but, phi 2 divided by I that is simply
given by mu 0 this is n 1, mu 0 n 1 n 2 pi
R
square l; now this is a very interesting relationship
that can come out of here, remember
that we showed that the self inductance of
a solenoid is given by pi R square mu 0 n
square l.
So, therefore, self inductance of first solenoid
is given by pi R square mu 0 n 1 square
into l, self inductance of the second solenoid
will be pi R square mu 0 n 2 square into l,
remember I made an assumption that the two
solenoids are tightly bound to each other.
So, therefore, whenever flux changes, because
of change in current the entire flux is
linked with the turns. So, if you compare
this with this, you find M 1 2 can be written
as,
M 1 2 which we had seen as same as M 2 1,
can be written as square root of L 1 into
L 2.
This is not really a general expression, because
we have assumed that the two solenoids
are tightly bound to each other, but what
have what we find is that the relationship
between M 1 2. And the self inductances of
the two loops can be generally written as
some constant kappa times root of L 1 into
L 2, and this coefficient kappa will depend
upon relative orientation of the loop, and
this I can call it as a coefficient of coupling.
.
.Let us do another exercise, I am talking
about, I am taking two circular loops, and
again I
made the problem somewhat simple, there is
two co-planner are co-planner and
concentric circular loop, that is their centers
are the same and they are in the same plane.
And I have to assumed that the current is
changing through the second circle the bigger
circle.
.
Now, if I assume that the current is changing
through the bigger circle, and there is a
smaller circle inside I have, I am magnified
it but, let us assume that R 1, R 1 is much
smaller than R 2. Now, when that happens,
I know how to find out the magnetic field
due to a circular loop at it center, and that
expression the magnetic field due to the bigger
circle is given by mu 0 I 2, because I 2 is
the current in that divided by 2 R 2. Once
again
I could give it a direction, which is let
us say z axis, because I could change the
make the
current flow that way, so that if I, I use
the right handed rule and so therefore, thus
that
could be this direction. Now, it is reasonable
to assume, that if R 1 is much less than R
2,
then the magnetic field strength at the center
of the bigger circle is roughly the magnetic
field strength over the smaller circle. So,
the so once I make this statement that then
the
flux through the first circle, will be simply
B 2 into pi R 1 square, and B 2 we have seen
is mu 0 I 2 by 2 R 2 into pi R 1 square, so
therefore, M 1 2 is divided by phi 1 divided
by
I 2 is given by pi R 1 square mu 0 divided
by 2 R 2; once again it shows that mutual
inductance is a property of the geometry,
and mostly the dimensionality of the fix.
.So, with these examples, let me now go over
to a slightly different idea, let us talk
about
how to find the energy of a current distribution,
recall that when we did electrostatics, we
discussed how to find the energy of a charge
distribution. And what we did at that time
was, assume first initially all the charges
are away at infinity, and that I took as the
situation of 0 energy, then I brought one
charge put it in its place, no work is done,
because there is no field.
But, once the first charge comes, there is
another electric field than of course, I as
I bring
in the charges, I would require it would be
necessary for me to do work against the
electric field that have been established,
and we calculated how much is the energy.
Now, I have a problem now, I cannot remove
the current distribution to infinity so I
cannot do that but, so this technique of starting
with a 0 potential energy, and then
establishing the charges will not work here.
.
So, let us let us see what is the way in which
we will do it, now so these are the basic
principle, when we when we establish a current
in a circuit, we have just now seen an
induced emf is different, now I we must do
work to overcome this induced emf, this
induced emf we have seen by (()) opposes the
cause. So, therefore, if you are moving, for
example if you are moving a magnet towards
a circuit, the circuit will generate an
induced emf, circuit will have an induced
emf, and if that happens to be a conducting
circuit then of course, there will be a induced
current.
.So, therefore, if you want to move for example,
this magnet with a uniform velocity, you
will have to do work on it; now unlike the
case of resistance, this work is completely
recoverable, because the you know you get
if you are moving something, you need to do
some work, but then you can always get it
back in the reverse situation, that is situation
number one. And even if there is one circuit,
there will be an emf in that circuit and we
will have to overcome, do work to overcome
it, if there are more than one sub
conductors, then we had seen there are mutual
effects that is supposing, I concentrate on
a particular loop, let us call it loop number
1. Now, loop number 1 will have an emf
generated, because of changing current in
loop number 1 itself, secondly when there
are
changing currents in other conductors loop
number 2, 3, 4, etcetera, then also there
will
be an emf generated in loop number 1, and
we must do work for that question. And so
in
order to calculate the total energy, we have
to take into account both of these situations,
so let us see how it works.
.
So, let me I am going to calculate, how much
is the flux through I th circuit, supposing
I
have got very large number of loops, what
is the flux changed through the I th circuit
..
I th loop let us say, so we did that just
now, so let me call it phi i. Firstly, we
notice that,
there is a there is a flux generated, if there
is a current I is in the circle loop, I th
loop
itself, and we had seen that the proportionality
constant was the self inductance, since it
is the I th loop I will call it L i times
I i. The second part is that, we have seen
that, when
there are currents in the other circuits j
not equal to I, then I have a flux generated
in the I
th circuit which is given by M i j i j, where
M i j is the mutual inductance between i and
j; and I am assuming that M i j is equal to
M j i.
So, the emf through the first circuit, I th
circuits which is minus d phi i by d t that
is
equal to L i d I i by d t plus sum over j
not equal to i M i j d i j by d t. So, this
is the emf,
and this is the emf established in I th circuit,
so the rate at which a work must be done to
establish this emf or to overcome this emf.
So, rate at which 
work is done that is
obviously, given by E i times I i emf times
iso this is simply multiplying these with
I i all
over let us do that, L i I i d I i by d t
plus sum over j not equal to i M i j I i d
i j by d t.
Now, I will write it in a little more symmetric
fashion, firstly notice I i d I by d t can
be
written as a half of L i of course, is a constant
d by d t of I square, I i square. So, that
d
by d t of I i square is 2 I i d I i by d t
2 and half cancels out, the second term requires
a
little bit of explanation, but it turns out
I can write it like let me, first write it
down and
then we will see why half sum over j not equal
to I M i j d by d t of I I i j.
.Now, this is really nothing great, because
what we have done is to say this is d I i
by d t
multiplied by i j plus i i times d i j by
d t; now because, M i j is equal to M j I,I
could
interchange like this, that is write it as
your d I by d t I times d j by d t I j times
d I by d t.
So, you notice, so that these are written
very symmetrically, these are written very
symmetrically, so with this, we can now calculate
how much is the total work done, so
this is the rate at which the work is being
done, so that you need to integrate it over
d t.
.
So, that this will give me the work done,
which will convert to the potential energy,
which is integral sum over i E i I i strictly
speaking, this last line that I did it is
valid only
if I sum over , so that you can sum over this,
and there is a sum over I here, and then I
use M i j equal to M j i. So, then this once
I integrate over time, I find I get half sum
over
i L i I i square and plus half sum over I
sum over j, I not equal to j of M i j I i
I j.
Now, if you bring back the expression for
the flux, this is the expression for the flux,
you
notice this expression is nothing but, sum
over, this is nothing but, sum over I of I
i phi i
multiply I i with I i with I i square, and
multiply here you get I i I j and sum. So,
this is
equal to half sum over i I i times phi I,
now we will write it in a slightly different
fashion
now, this is half sum over i I i this is flux,
so flux by definition B dot d S, so I write
it as
B dot d S , so this is the flux through the
I th circuit. And I will write this again,
write B
as del cross of A, so del cross A dot d S
I, and then I use Stokes’s law, so that
this
becomes loop integral of A dot d l I, this
is for discrete. Now, how will it change if
I
.have it continues distribution, now see if
I have continues distribution then of course,
all
that I need is this that, this I i with a
surface will give me the current, and so therefore,
this will go to half of loop integral of A
dot J current, this is current density and
d cube r,
the current density because, this is per area
I mean we are current by area, so here since
there is just a length, so therefore, I have
got it d cube r there, because I have divided
by
this thing.
So, this is the expression for the work done
by did by establishing, while establishing
currents in various circuits, and this is
corresponding continues version, with this
much
of introduction of Faraday’s law, let me
go over to a slightly different story. Now,
let us
recall where are we with respect to the electromagnetism
at as of now.
.
So, we had del dot of E equal to rho by epsilon
0, del dot of B equal to 0 this actually
will not change generally, then we have just
now seen that del cross of E is minus d v
by
d t, we have not yet talked about, what happens
to this del cross of B equal to mu 0 j or
the corresponding expression del cross H equal
to just with J free. Similarly, here if you
want to talk in terms of the displacement
field than this is simply equal to rho free,
so
this is where we stand with respect to Maxwell’s
equation.
But, you notice that we understand the a symmetry
between del dot of E and del dot of
B, because this tells us there are charges,
this tells us that there are no magnetic
monopolies. Whereas this seems to be some
difference between the way these appear
.also, once again I understand why the term
corresponding to J free is not there, because
there are no magnetic monopoles which are
moving around, but what is not clear is why
is in there a time dependence here. So, this
was actually Maxwell’s contribution to the
electro dynamics, and that is why these set
of equation as modified later are known as
Maxwell’s equation, so, let me go through
that.
Firstly, let us look at a simple example of
charging of a capacitor, now remember that
when I pass a current through a circuit which
contains a capacitor, the there is no current
which passes through the gap of the capacitor.
But, however, there is an electric field
established in the capacitor, we will take
the simple example of parallel plate capacitor;
and we know that l I am charging a capacitor,
the electric field through this changes.
Because, electric field will depend upon what
is the instantaneous charge on the plates
of
the capacitor, and as you connect a battery
to a circuit, even though it is a small time
during that time the charge goes from 0 to
whatever its final values is. So, during that
time there is a change in electric field through
a capacitor.
.
But, there is a problem and the problem comes
up, let us look at the charging of a
capacitor, there is a capacitor with a current
I flowing in, and this current I is let us
say
momentarily changing, because of my having
connected a battery to the circuit, and the
electric field is of course, given like this.
Now, consider a loop which I have called as
C
.like this, now I know that I could calculate
I could calculate the magnetic field by using
the standard Ampere's Law, which says integral
B dot d l equal to mu 0 I.
Now, during the charging of the capacitor,
the charge accumulating in the plates of the
capacitor is changing, as a result there is
a current in the external circuit. Now, if
there is
a current in the external circuit, and I look
at this loop I can calculate the loop integral,
and say it is equal to mu 0 times I,I is the
instantaneous current, that instant at which
I
am calculating the magnetic field. Now, let
us look at what is the problem, now if I have
this loop, this is the artificial loop, mathematical
loop, now let us suppose that I put S
surface on the plane of the loop.
.
So, let me do that, this is what we said just
now integral B dot d l is equal to mu 0 I,
now
let me put a surface now on that loop, which
I have called as S 1. So, through that
surface a current is flowing, and the magnetic
flux line are intersecting that surface, so
so
far as this surface is concerned, I understand
that del cross of B equal to mu 0 J, because
B dot d l which I had written down integral
B dot d l equal to mu 0 I that B dot d l could
be converted to curl B dot d S; and over the
surface which I have called as S 1, the I
can
calculate the magnetic field at every point.
.And so therefore, del cross B dot d S is
my mu 0 I, and you since this will be true
for any
surface of this type, I expect del cross B
equal to mu 0 this is something which you
done
in the earlier, but let us pause for a moment.
We have said several times, if I have a loop
which I showed earlier, these result del cross
B equal to mu 0 J is independent of what
surface I take, as long as this is the boundary
of the surface that has been defined. So,
suppose instead of the surface S 1 through
which the current passes, and as a result
the
magnetic field lines also passes, let us look
at slightly different situation.
.
Again I had the same situation, I have however,
a slightly different surface you notice
this is a pot like surface, I have this same
loop C, but this time what is happened is
instead of choosing the surface on the plane
of the loop, I had taken at like this. So,
therefore, all this surface the through this
surface there is no current, now if there
is no
current through that surface, I expect del
cross the I dot d l to be equal to 0, as a
result del
cross B should be equal to 0, but there is
a dichotomy we had said earlier the result
should not depend upon which surface you choose.
Because, both of them represent the current
passing through the loop C 1, so how can I
get one result for this type of a service,
and another result for the surface S 1 which
was
on the plane of the loop. Now, what is the
problem, the problem is that there is no
magnetic field here, so that B dot d S should
be is equal to 0, but is there something
inside that, now you notice there is an electric
field inside that, and that electric field
is
.changing. So, as a result, this is this is
a dichotomy which Maxwell had realized, that
I I
get two different answers for the same problem
depending upon surface I choose.
So, what he did is to conjecture, they just
as by Faraday’s Law, whenever there is a
change in magnetic flux, where ever there
is a changing it change in magnetic flux
through a circuit, there is an emf generated
he wanted alternatively a changing magnetic
field, a time dependent magnetic field gives
rise to an induced electric field. So, Maxwell
now postulate, it going by parallelism that
if I time dependent magnetic field gives rise
to
an induced electric field, it is reasonable
to assume that a time dependent electric field
of
the type that I have here, the as the charging
is taking place, the electric field through
the
capacitor is changing; so a time dependent
electric field is equivalent to a magnetic
field,
the effects must be symmetric,
.
And this indeed was sort of proved or verified
by experiment, So, how does one fix this
dilemma, let us look at that, so just to understand
let us assume that the space between
the dielectric is filled with a space between
the capacitor plates is filled with dielectric.
Now, if there are dielectric, there are charges
and when you change the current through
it, the electric field changes that of course,
pushes these charges. So, what we say is this,
that this is the space in which the electric
field is changing with time.
..
And let us calculate how much is the what
is the effect, so the electric flux in this
region
is given by surface integral, I have this
time I am taking surface S 2 D dot d S, I
have
taken the flux of the D field, because I have
said it is filled with directive. Now, so
therefore, d phi by d t, well I should write
full, so this is equal to d by d t of integral
of D
dot d S, and this I will write this in a using
the divergence theorem, as d by d t of del
dot
of D d cube r volume integral, but del dot
of D is nothing but, rho, rho free d cube
r
which is nothing but, the total charge Q.
And so therefore, this is d Q by d t, notice
d Q by d t that is the rate of change of charge
on the capacitor plate is same as that of
the current that is flowing in, so therefore,
this is
this as the dimension of current. So, what
we have said is i D the name came from
displacement current, this was Maxwell gave
it a name displacement current, that is the
rate of change of the electric flux with that
time, so how did Maxwell handle it?
..
So, Maxwell said that let us still modify,
the last law that is the Ampere’s Law which
was del cross H equal to J by adding a term
d B by d t notice that del dot del cross H
is
the identically 0, because it is divergence
of a curl. So, therefore, I get from here
del dot
of J plus D by d t of del dot of D this equal
to 0, but del dot of D is rho, so that give
we
del dot of J plus D rho by d t is equal to
0, you identify recall this is nothing but,
our
continuity relationship, thus this is the
equation which should replace del cross of
H
instead of J, should now be added with a term
d B by d t. That completes our
establishment of the four Maxwell’s equation,
and we will discuss them all together in a
more coherent way from the next lecture.
.
