Good Morning. The last two lectures I have
been talking about the magnetic field and
I hope I have sufficiently justified it is
existence. So, today I am going to do some
mathematics to make it possible to manipulate
this magnetic field when we make use of it.
So, let me go back to the starting equation.
The equation we had was that the magnetic
field was defined as mew not over 4 pi, I
still have not really discussed why mew not
over 4 pi. The reason I have not discussed
it is it pretty arbitrary integral, the current
at any other position. So, this is B at a
position r, it is an integral over other points.
Whereas current j of r prime-cross R 1 2 divided
by R 1 2 cubed dV prime. So, we are talking
about various current elements and we are
talking about a point r it is all coordinate
system.
So, this is r. This is r prime and this vector
is what I called R 1 2. So, you take the current
take the cross product where the R 1 2 divide
by the cube of the magnitude of R 1 2 integrate.
The result over all the current elements multiplied
by normalizing constant and that is a BC.
It is a relatively complicated expression,
but it is not that complicated because you
can compare it with the electric field which
is one over 4 pi epsilon naught again integral
over all other points the charge density of
that point times R 1 2 divided by R 1 2 cubed.
So, you see the same R 1 2 over R 1 2 cubed
appears both in the electric field and in
the magnetic field. Now what I would like
to do is to go from this equation is quite
useful as it is do something that is more
easier to this. For that first let us look
at this equation and look at the associated
equation which is the potential pie. We already
worked this out the potential was integral
row of r prime dV prime divided by R 1 2.
Now if you look at these two equations and
you look at the equation that electric field
is equal to minus of the gradient of pie.
You can write this this equation out. Let
us do that.
What it says is 1 over 4 pi epsilon naught
integral over other points row of r prime
R 1 2 vector divided by R 1 2 cubed dV prime
is equal to minus of the gradient of integral
1 over 4 pi epsilon naught outside row of
r prime over R 1 2. This is the same equation
as E is equal to minus grad pie I have just
written out what E is written out what pie
is. Now, this gradient if you look at what
it is it is x hat del del x plus y hat del
del y plus z hat del del z. That is to say
if you go back to this picture the gradient
represents shifting this point r because this
r is x x hat plus y y hat plus z z hat. This
point r prime if you root out its coordinates
it would be x prime x hat plus y prime y hat
plus z prime z hat.
So if I am doing a derivative with respect
to x it means I keep all the charges where
they are and I am move the point where I got
my measuring instrument where I am measuring
potential or I am measuring field. So, the
gradient operator there means move this points
slightly there see the change in potential
divide by delta L do it in all direction and
construct the gradient operator. But if you
look at this integral the integral is over
where the charge is, it does its not over
where the instrument is. So this integral
is not even involving x, y or z it involves
x prime, y prime, z prime.
So, I can take this derivative inside its
like if I did an integral dt of sin x plus
t and then I had del del x. Now this derivative
with respect to x does not is not the same
thing as the t. So, I can take this del del
x inside. Now, I can write this as an integral
dt del del x of sin x plus t and then I can
make this x act on here and do whatever you
want to do. So, we are going to do the same
thing here. The gradient is acting on a coordinate
that is not what we are integrating over.
I am going to pull that gradient inside and
make it act on all the parts of the integrand
that depend on x y z rather than x prime,
y prime, z prime, what do I get?
This electric field becomes equal to minus
one over 4 pi epsilon naught integral. Now
row of r prime the location whether the charge
is not a function of x, y, z; x, y, z, is
where I got mine meter by meter is not where
the charges. So, these are two separate piece
of information. So, I can feed the row of
r prime out dV prime before that but the gradient
does attack this piece. We go back to the
figure.
If I change the location here to another point
there r plus delta r then R 1 2 will change
R 1 2 will become R 1 2 plus delta R 1 2.
Therefore the gradient does affect the value
of R 1 2 it does not affect the value of charge.
But it affects the distance of the charge
from the measuring instrument.
So, I can put gradient acting on one over
R 1 2. Now look at this I have got I can take
both of these together and I get that 1 over
4 pi epsilon naught integral row of r prime
R 1 2 divided by R 1 2 cubed minus gradient
of 1 over R 1 2. I think from plus because
I have got a minus sign here and I took it
to other side dV prime is equal to 0. What
I have done is as said the electric field
is equal to minus grad pie I have rewritten
minus grad pie in this form.
So since both of these involve one over 4
pi epsilon integral over row of r prime dV
prime I took that common. What is remaining
R 1 2 over R 1 2 cubed and with this minus
sign when I take it on this side it will become
plus gradient of R 1 R 1 2 1 over R 1 2 this
is equal to 0 and this is equal to 0 for all
r. Now when this kind of integral happens
that, this is equal to 0 for all r, you might
in some very special cases find some non-zero
answers.
But invariably what it means is gradient of
1 over R 1 2 is equal to minus R 1 2 over
R 1 2 cubed. So it is a very important result.
This is in fact what allows us to have potential
electrostatic potential derived from coulomb’s
law. This is coulomb’s law 1 over r squared.
This is electrostatic potential 1 over R and
the reason why coulomb’s law and electrostatic
potential are related is because gradient
1 over R is this vector operator. So, this
is all electrostatics you already had it.
But now look back at the magnetic field equation.
Now, this magnetic field equation of course
involves current now rather than charge that
involves mew not rather than epsilon naught.
But it still involves R 1 2 over R 1 2 cubed.
The very same operator that electric field
uses. So, we can use this relationship between
R 1 2 over R 1 2 cubed and gradient of R 1
2 one over R 1 2 in here. So, what do you
get you get the magnetic field is equal to
mew not over 4 pi volume integral j of r prime
cross R 1 2 over R 1 2 cubed dV prime becomes
equal to minus mew not over 4 pi volume integral
j of r prime cross gradient of one over R
1 2 dV prime.
Have you used the same affects? This is R
1 2 over R 1 2 cubed. So, it must be equal
to minus gradient of R 1 2. But, this gradient
acts on x, y, z. This integral is over x prime;
y prime, z prime, in the current is also over
x prime, y prime, z prime. So, it means when
I vary the position of the instrument the
currents do not change therefore I can pull
this gradient right out. So, I get its equal
to well let me before I do that I have to
prove to you what that becomes.
Look at the integrand here it is j of r prime
cross gradient of 1 over R 1 2. Now in Cartesian
coordinates we write this as unit vector along
x, unit vector along y, unit vector along
z, jx, jy, jz, del del x of 1 over R 1 2,
del del y of 1 over R 1 2, del del z of 1
over R 1 2 and determinant of. Now this del
del x, del del y, del del z, are actually
not acting on jx, jy, jz. So, if I root out
the x component of this of the determinant,
the x component part would give me jx of r
prime del del z 1 over R 1 2 minus. Sorry,
jy jz of r prime del del y of 1 over R 1 2.
That is, when you take the determinant you
choose a row first entry times this, the product
of these two minus the product of these two.
Second entry times product of these 2 minus
product of these two third times this. So
the first entry the component multiply x hat
would be jy del del z 1 over R 1 2 inverse
minus jz del del y 1 over R 1 2. But, z and
y do not attack R 1 2 prime. They only act
on x and y and z. They do not act on x prime
y prime z prime. So, I can pull this out.
So, I get this is equal to del del z of jy
over R 1 2 and jy is a function of r prime
minus del del y of jz of r prime divided by
R 1 2. Now, this is still looking like a curl
like a cross product. But what kind of cross
product is it? Well, let us write it out.
We have to write it out.
I would say this is looking like the determinant
of I still have my x hat. But now I want del
del z of jy R 1 2 okay. So I would say that
I will put a minus sign so that this becomes
minus and this becomes plus. So del del y
of jz over R 1 2 minus del del z of jy over
R 1 2. So, you look at the y component the
same thing will happen will give you um jz
times del del x of 1 over R 1 2 minus jy sorry
minus jx del del z of 1 over R 1 2 once again
put a minus sign there.
So, you will get y hat times del del z of
jx over R 1 2 minus jz del del x acting on
jz over R 1 2. And similarly z hat times del
del x of jy over R 1 2 minus del del y jx
over R 1 2. So it looks like a cross product
again. Let us started with the cross product
this it was the cross product of j and gradient
of 1 over R. What I have ended up with is
a cross product that is strange kind of cross
product. This if I have to write what it is
minus the gradient operator itself. That is,
this cross product the function j divided
by R 1 2.
The first element is not really a quantity.
It is an operator because you can see just
del del x del del y del del z. It is a very
strange kind of cross product in fact it is
not a true cross product at all. It is a differential
operator and this gradient cross is called
the curl. And it is the most important operator,
once you have learned gradient and divergence.
Along with these other two operators, this
completes the set which will require for vector
analysis. So, let me repeat what I did.
I started with this magnetic field equation.
This equation was derived two lectures ago
and I repeated it last lectures. Then I said
that from electrostatics have already seen
that R 1 2 over R 1 2 cubed is nothing but
minus gradient of one over R 1 over R 1 2.
So, I can replace this piece by minus gradient
of one over R 1 2 then I looked at what is
in here it is a cross product. So, this cross
product can be written out as the determinant
of a 3 by 3 matrix that you must have done
quite a few times.
So you write down unit vectors on x, y, z,
the components of the first vector the components
of the second vector. So, perfectly reasonable
cross product but the operators here del del
x del del y del del z and you write it out
they can be pulled out because jy jx jz do
not depend on these coordinates. So, you can
write it this way and when you write it this
way, you can re-cause this determinant as
another determinant.
But the determinant that still uses unit vectors
x y z. But instead of jx jy jz I am putting
derivatives del del x del del y del del z
and instead of del del x one over r components
and I am putting j over R components. It is
still a cross product like operator except
that this first vector field is, appear operator
it is just gradient. So, if you have to write
it out you would say gradient cross field
this gradient cross is a very important operator
and that is what we called the curl alright.
Given that you have got the curl what do we
do with it?
So, I will write out the equation from B again
B is equal to minus mew not over 4 pi volume
integral j of r prime cross gradient of 1
over R 1 2 dV prime. Now I am going to take
this gradient out and rewrite this as minus
curl of j of r prime over R 1 2. That is what
we just proved. We proved that this cross
product is minus of the curl. But this curl
is acting on x, y, z. The integral is on x
prime, y prime, z prime. That is the curl
is acting on where the instrument is where
the measuring instrument is this integral
is over where the current elements are.
So, I will pull this right out so I get the
magnetic field this minus sign and this minus
sign cancel, it is equal to curl of mew not
over 4 pi volume integral j of r prime divided
by R 1 2 dV prime. This whole thing it depends
on x prime, y prime, z prime through j and
through R 1 2. But, that dependent on x prime,
y prime, z prime is integrated over. So, is
a as I had a sum n equals one to infinity
of various terms okay t sub n once you have
summed over all the n’s this sum no longer
depends on n.
Similarly, this integral no longer depends
on x prime y prime z prime. What it depends
on through this R 1 2 it depends on x y z.
Because R 1 2 is the vector R 1 2 is nothing,
but x minus x prime along the x direction
plus y minus y prime along the y direction
plus z minus z prime along the z direction.
So R 1 2 depends on 6 different coordinates
it depends on where the current elements are
it depends on where the measuring instrument
is depend on both so depends on six elements
six coordinates three of which get integrated
over its summed over. The remaining three
coordinates are what I have left and they
are what defined the value of B.
So, this square bracket is now a function
only of x y z and I can call its some function
I call it A vector it’s a vector because
j is a vector. So it has direction some vector
A which is the function of r. So I can finally
write this result out which is the magnetic
field B at any positions r is equal to the
curl of some other field A of r and what other
field is it. That field is mew 0 over 4 pi
integral over all the currents at any other
position divided by distance from the current
to the inter point where we want to calculate
the field.
So this is what A is and we take calculate
A the curl of A will give you B and as before
let me remind you once you have B the force
is nothing but row v cross B or j cross B
I will call it dF. So, the force per unit
volume is nothing but current cross B. So,
that is why we want B otherwise you would
not care about what B is. Once you have the
magnetic field you can calculate magnetic
forces. Now, there is a very important equation
probably the most important equation you can
calculate about magnetic field. But, an immediate
conclusion you can draw.
Supposing, I calculate divergence of B. Now
divergence of B would be divergence of curl
of A because we just worked out B is curl
of A. Let me write this out. It is equal to
divergence of determinant of x hat, y hat,
z hat, del del x del del y del del z Ax Ay
Az. That is what it is divergence of curl
of A. Now what this means is 
x hat del del x plus y hat del del y plau
z hat del del z this vector dot the determinant.
Now, if you look at what this is we can write
out x y and z components separately. So, the
x component of this vector will multiply the
x component of this vector.
So, it will become del del x of x component
of curl of A. The x component of curl of A
is the minus corresponding to 2 2 2 3 3 2
3 3 which is del Az del y minus del Ay del
z. The y component will be del del y of now
I have to take the y component of the curl
which is del Ax del z minus del Az del x.
Then, the z component del del z acting on
this which is del Ay del x minus del Ax del
y. So, there are six terms. Now, if you look
at this there is a del del del squared Az
del x del y. It is also del squared Az del
y del x. So I am going to collect terms out
now.
What I will get I will number this terms 1
2 3 4 5 and 6.
So taking terms 1 and 4 what I will get is
del squared del x del y of Az minus del squared
Az del y del x. If I take terms 2 and 5 then
I get del squared Ay del x del z minus del
squared Ay. Sorry, this is minus this, is
plus del z del x and then if I look at terms
I have taken care of the Az I have taken care
of A Ay. So, I will take care of Ax terms
3 and 6 I get del squared Ax del y del z minus
del squared Ax del z del y. So, this is what
divergence of curl of A is equal to all I
did was I took I wrote out what divergence
was I wrote out what curl was and I combined
all terms.
Now you look at each of this, if you look
at this one, you are taking the second partial
derivative of Az with first with respect to
y then with respect to x or you are taking
the second partial derivative first with respect
to x second with respect to y now you know
from your theory in partial derivatives. The
second partial derivatives commute. What that
means is that, I can interchange del y and
del x so I will do that one set of these.
What does that give me? Gives me del squared
Az del x del y minus del squared Az del x
del y. So, this is 0 gives me minus del squared
Ay del x del z plus del squared Ay del x del
z 0 del squared Ax del y del z minus del squared
Ax del y del z.
So, this whole thing is zero which means because
curl of A is be divergence of B is equal to
0. This is always true at it comes out of
simple the fact that curl of A is equal to
B. If you can find a function whose curl is
B then its divergence the divergence of B
is automatically 0. We proven it and there
is no assumption I put in to it. It is a very
general statement if any field is the curl
of another field its divergence is 0. Now,
why do we care we care for a very important
reason. If you look at electrostatics, the
divergence of the electric field is extremely
important.
The divergence of E is equal to row over epsilon
naught. The divergence of the electric field
that the source of the electric field because
row is what creates the electric field and
you take the divergence of the electric field
you get back to the source. The divergence
of B is equal to zero which means there is
no charge like source for the magnetic field.
There is no such thing as magnetic charge.
So it is very important statement and it purely
comes out of the fact we wrote B is equal
to mew not over 4 pi integral j cross gradient
one over R dV prime. That is we said B is
derived from currents and not from charges.
Moment you say that all the rest follow and
this is been verified by observation that
the magnetic field is derived from currents
and its also understandable because magnetic
field is due to relativity which means is
due to moving charges. Therefore it is due
to currents. But there is something surprising
about this statement.
Because the magnetic field was first discovered
because of permanent magnets and in permanent
magnets we always said north pole south pole.
And if you look at what we mean by a north
pole and what happens when a north pole is
placed in a magnetic field. A north pole is
like magnetic charge. So, I am saying there
is no magnetic charge it is intrusively true
yet the most basic magnetisable material has
what looks like magnetic charge. The answer
is a little quickly answer is if I cut this
magnet in two I am going to find I cannot
just get the north-pole in one piece and the
south-pole in the other piece.
Rather what will happen is I will get a south
pole at the bottom of the first piece and
a north pole at the top of the second piece.
No matter of how much I cut them I will keep
getting north and south and this is what it
means. It does not mean that we do not have
north like south behavior like behavior it
means you cannot extra just one charge out.
Anytime you try to save up a little bit of
this magnet and say I will take only the south-pole,
you will find that you look a little bit a
north pole as well so much. So, it is still
a magnet with the pair of poles attached to
it. That is what is meant by divergence B
is 0. We will never find a pure north pole.
Now, in terms of drawing lines of force, what
that means is if you drew for electric field
we drew a charge Q and then we drew arrows.
And so we had Gauss’s law and we could say
surface integral E dot dS was equal to Q enclosed
over epsilon naught. Now divergence B is 0
which means you can never have this situation
the kind of situation you will have from magnetic
field would be 
something like this you have a magnet. Now,
if you surround this by a surface, there are
many lines which never touch the surface there
are few lines where the magnetic field is
going out but corresponding each of this there
is a line where the magnetic field is going
in.
So, when you integrate over the whole surface
you get surface integral B dot dS which is
volume integral divergence of B is equal to
0. Because we have divergence B is equal to
zero. So, no matter of how you surround a
piece of magnetism you will always find that
the total amount of flex leaving that volume
is zero its no such thing as magnetic charge.
This is almost the most important thing that
you have to know about the magnetic field
and if you know one more thing, you know everything
there is to know about it.
So, we have B is equal to curl of A and divergence
of B is equal to 0. Similarly, we have from
electrostatics E is equal to minus gradient
of pie curl of E is equal to 0. Now let me
show this have actually up to now only shown
you loop integral E dot dl is equal to zero.
But now I introduce a new vector and I want
to talk about. I have E is equal to minus
grad pie. So, it means electric field is equal
to minus x unit vector along x del pie del
x minus unit vector along y del pie del y
minus unit vector along z del pie del z.
Now, I know what it means to take grad cross
it means take the determinant of x hat y hat
z hat del del x del del y del del z and then
a mistake Ex Ey Ez which is minus del pie
del x minus del pie del y minus del pie del
z. So, this is what it means we talk about
curl of E. Let me work out what one of the
components is.
Curl of E x component is equal to del del
y of minus del pie del z minus del del z of
minus del pie del y. So it is again equal
to del squared pie del z del y minus del squared
pie del y del z. Again we know from our mathematics
that I can exchange these two derivatives.
So, these two terms are identical one has
a minus sign. So, it is equal to 0. Similarly,
the curl of Ey will be 0. Similarly curl of
Ez will be 0. So, that is why the moment you
have anything call grad pie its curl becomes
zero anytime you have something, that is curl
of a vector field its divergences. Now I am
going to introduce a vector operation a vector
law called Stokes theorem.
It is a very easy theorem provided you approach
it in the right attitude. But because it involves
this thing called a curl most people get very
scared of it. If, you look at what curl is,
if you look at the x component it is del Ez
del y minus del Ey del z. Let us draw this
out this is the y z plane and I have electric
field with components in x and components
in y. What I am saying is if I keep z component
z fixed go along y if I keep y fixed go along
z there is a relationship. Let us look at
what it is trying to say.
Supposing I have electric field in some general
direction, so I have a certain amount of Ey
and I have Ey of z plus delta z. Similarly,
I have Ez at y Ez at y plus delta y. Supposing,
I do an integration round this square. I will
assume that this length is delta y and this
length is delta z. So, my loop integral E
dot dl we shown by surprised that we are talked
about E dot dl. Because, earlier I had written
down loop integral E dot dl equals 0 before
I wrote down curl of E equals 0. They are
saying the same thing.
Loop integral E dot dl becomes the bottom
loop is Ey at z times delta y then it becomes
plus Ez at y plus delta y delta z minus Ey
at z plus delta z delta y minus Ez at y delta
z. So, when I go in this direction increasing
y which plus sign increasing z is plus sign
decreasing y minus sign decreasing z minus
sign, so on. This first leg Ey at z, second
leg Ez ay y plus delta y, third leg Ey at
z plus delta z, fourth leg Ez at y. Now, I
will combine the Eys and I will combine the
Ezs. So, what do you get?
You get that that term is equal to Ey at z
minus Ey at z plus delta z times delta y plus
Ez at y plus delta y minus Ez at y delta z.
So I have combined this term and this term
and written out the first square bracket.
I have combined this term and this term and
written out this the second square bracket.
Now each of this can be written out as a derivative.
So, this can be written out as minus del Ey
del z delta y delta z plus del Ez del y delta
y delta z. So I have loop integral E dot dl
around this tiny loop is this.
But I can identify delta y delta z as nothing
but the area of this loop. So, it is equal
to dA the area multiplied by del Ez del y
minus del Ey del z and if you remember back
to what the definition of curl was this is
nothing but curl of E along x. Now, this area
is the area in the y z plane. So, you can
actually write this is dAx. So if you do this
you take the next other parts also this is
plus four other parts other terms, you will
get plus dAy curl of Ey plus dAz curl of Ez.
So, we add it all up you get a simple answer
you get that for the tiny loop very small
loop.
Loop integral E dot dl and this is any E I
call it E but I have not used the properties
of E here any E its equal to surface integral
over that small area curl of E dot dS. So
it is true for very tiny loop rectangular
loop that is still tiny in general direction.
The x component will give you dA projected
along x times curl of E in the x direction.
The y component will give you this. The z
component will give you this. So, in general
it can be written out is curl of E dot dS.
Now, if you look at the general surface you
can generalize this idea.
Supposing I have some closed loop and I want
to calculate E dot dl on this loop, what I
will do is, I will take this loop and connect
it through some surface I deliberately draw
on the surface so that it is bulging out.
This surface does not have to be flat it can
be any surface at all. Then on this arbitrary
closed surface I can draw little rectangles
on those tiny rectangles my theorem will hold
namely loop integral E dot dl is equal to
surface integral curl of E dot dS. Then I
can take the next rectangle and do it again.
But now let us look at this region bigger
I have a loop in integral then I have a loops
around it. I am going to draw nine of them.
Now, I am going to do my loop integral this
way I do my loop integral the same way in
all the nine rectangles. Now, if you look
at this particular leg when I am doing it
when I am doing E dot dl from rectangle one
I am moving in this direction. When I am doing
E dot dl from rectangle two I am going this
direction.
So, the E dot dl calculation if I add them
up will cancel out. Similarly the E dot dl
calculation in two will cancel E dot dl calculation
in 3. The E dot dl calculation in to at the
bottom will cancel the E dot dl in 4 and so
on and so forth. So, every interior edge will
cancel out. The only thing that one cancel
out it is the outside edge because there is
no, there is no rectangle outside to cancel
it. So, when I add up all this little rectangles
by some them all up I take this entire surface
tile it with lots and lots of little rectangles
and sum them all are. What I will get is,
loop integral over the our full curve E dot
dl because all the internal little integrals
canceled out.
Only thing that is left is the outside integral.
But what is on the other side? On the other
side is sum on all the little rectangles curl
of E dot dS. But this is nothing but a summation
and integration is nothing but summation anyway.
So this is surface integral over the whole
surface curl of E dot dS. So, this is what
it calls Stokes theorem. It says for any arbitrary
closed loop C I am just going to call it C.
The loop integral on c of any vector I am
going to call it E dot dl is equal to the
surface that connects c. So, usually I call
it S sub c to mean that it is the surface
whose edges correspond to c curl of E dot
dS.
It is an extremely important theorem and if
you did advance mathematics, you will learn
this theorem is a special case of divergence
theorem itself. However, in our use of electromagnetics
this theorem is exceptionally important; we
will use it in paradise law.
We will use it in Ampere’s law. We will
use it all over place. So, it is a theorem
you should become very familiar. I will continue
on next time and go for use this theorem and
gets some interesting results.
