Field due to current loop - Electrical Engineering Video Lecture - Electrical

ha ha ha ha they just mean Allah was
beloved shall I
Oh God
you
you
good morning the entire previous lecture
was spent in explaining what exactly
curl meant and I hope you've got a good
understanding of what that operator is
it is the most complicated of the three
operators we've studied we have looked
at the gradient operator we have looked
at the divergence operator and we have
looked at now
the curl operator and we know that the
curl operator is basically the average
of the loop integral of any vector field
divided by the area of the loop and if
you take the area to be small that is
you take the limit as this loop becomes
tinier and tinier you get nothing but
the curl just as for divergence we had
that is equal to the surface integral of
e dot d s divided by volume enclosed by
the surface as you take the volume to 0
so both of these look very similar and
in fact in a more sophisticated
mathematical treatment you can show that
both of them are saying the same thing
in a certain sense this is the
divergence theorem in a different
dimension ok at the end of last lecture
I talked about what curl was in non
Cartesian systems and I wrote down the
answer I did not derive it it is derived
in the appendix of your textbook
II won e 2 e 3 our unit vectors
u 1 u 2 u 3 are the coordinates for
example R Theta Phi or R theta Z or X Y
Z and
so this is the expression for curl it
it's general in the sense that is true
for any orthogonal coordinate system and
last time we worked out what it looked
like in cylindrical coordinates now
today I am going to do a problem just
one problem is going to take the lecture
which will use curl in spherical
coordinates the problem is to find out
the magnetic field far away from a loop
of current this is an extremely
important problem because anyone in
electrical engineering has to do with
magnetic materials and the entire theory
of magnetic materials has to do with the
magnetic field due to loops of current
so all of material properties of
magnetics comes from this one problem so
we need to calculate with some intensity
and that is what we will do in this
lecture so for spherical coordinates
I've already explained where we get h1
h2 h3 the values are R R sine theta and
1 the coordinates this is the z axis is
any point the angle made with the z axis
is theta the distance from the origin is
R and if you draw a great circle through
that point and see where it intersects
the equator and find its angles with the
x axis this angle is Phi so the
coordinate system is R Theta Phi and
what is what this is saying is that if
you move a distance unity in are the
length is R itself dr itself so there is
no scaling factor if you move a unit
distance in theta the distance you move
this are d theta so the scaling factor
is R and if you move a distance you need
distance in Phi the scaling factor is
this length D Phi and this length is
nothing but R sine theta so that is
where these three coefficients come now
let us substitute this into the
expression for curl
so curl of e becomes 1 over H 1 H 2 H 3
which is 1 over R square sine theta
determinant of R hat R theta hat R sine
theta pi hat del del R del del theta del
del Phi e r r e theta R sine theta e Phi
I have just substituted for H 1 H 2 and
H 3 now we take the determinant by
taking the top row and expanding the
determine determinant of the minor so I
get first r hat times del del theta of
this minus del Del Phi of this you can
see that an R is there which does not
get acted upon by theta and Phi so 1 R
cancels out so R sine theta of
del del theta of E Phi sine theta minus
del Del Phi of Z theta then you have the
theta hat component there you have to
keep both the ARS but you do not have
any derivative with respect to theta
therefore the sine theta will cancel out
when you take the fighter R square sine
theta of
dele are Delphi
-
del Del R of R sine theta e v
I am going to write the last term at the
top itself so that I do not leave this
board plus PI hat again divided by R
square and theta del Del R of R a theta
minus del e R Delta let me just check
that I haven't made any mistakes I have
forgotten these these two terms so this
will come as R here and R sine theta so
this is the result our sine theta will
cancel out here and R will cancel out
here so if you so this is what you get
so it is a fairly complicated looking
expression but it is not as bad as it
looks
now what we are going to do is we are
going to take a loop and I am going to
put the loop in the origin so Center the
loop is at the origin the loop has a
radius a it has a current flowing of I
this is the x axis is the y axis and
this is the z axis now I am interested
in a point R and because I the symmetry
present I can always rotate it so that
this point points along X so this point
R is what I call Rho Rho hat plus Z Zed
hat in cylindrical coordinates that is
it's got a distance Rho and a height is
of Z
now different current elements on this
loop are going to cause magnetic field
here and we know the expression so
expression is V is equal to MU naught
over 4 pi integral over the whole volume
J of R prime cross R 1 2 divided by r 1
2 cubed what that means is I take each
of these pieces connect the line joining
them to R then take the cross product of
that line and this J whether J and this
line and integrate over it now you can
see there is going to be a very
complicated operation because the solid
angle between J and this line is going
to continuously change and I have to
work out that angle work out the
resultant direction and then integrate
over them luckily we have a better way
we know that D is equal to curl of a and
a is equal to MU naught over 4 pi volume
integral J of our prime divided by r1 to
DV Prime now at least we do not have to
do the vector cross product that vector
cross product is done later when we take
the curl to calculate a we only sum over
J's but we still have to divide by this
distance r12 now if you look at this
case consider that I take
to current elements which are at the
same angle with respect to X 1 is at
minus Phi and the other is at plus 5 now
if you look at it from above this
direction is X this direction is y I
have a current flowing this way nav a
current flowing this way both of them
are the same distance from the observer
because you can see that they are
symmetrically placed so the distance
from this element to R is the same as
this element to R so it is the same r12
but as a points in different directions
but because r12 is the same when I do
integration I can add up these two J's
integration is nothing but sums and if I
add these two J's vectorially I get IJ
that is horizontal and in this
coordinate system a horizontal J
corresponds to a J's Phi so this J had
both in the X direction and in the Y
direction so did this but the net was
only in the Y direction and in spherical
coordinates that y direction would be
nothing but the Phi coordinate since for
every one of these elements as a
corresponding element I can take them in
pairs I can add them up so the final
answer is e is equal to something that
is only along the Phi direction so if I
want something only along the Phi
Direction obviously only J Phi can
contribute with equal to MU naught over
4 pi integral J Phi over r12
TV Prime now this GFI is actually coming
out of a current I which is only present
on this loop so I can replace this
volume integral by mu naught over 4 pi
integral 0 to 2 pi e D Phi prime that is
a DL times a current I but this current
if this is if this is angle Phi Prime
this current is flowing not in this
direction but at some other angle so how
much of the current is flowing in the
Phi direction well all of it is flowing
if Phi is 0 none of it is flowing Phi is
90 so I can put down of course Phi Prime
all right so what I have got divided by
R 1
what I have got is a simplified
expression for calculating the vector
potential if I can calculate this vector
potential then I can take the curl to
get the magnetic field it may seem that
this is not terribly simple and it isn't
but it is a great deal simpler than
calculating this expression this
expression is a is torture
all right so now I want to calculate
what R 1 2 is now if you look at this
point and I draw a line joining the
place where the current element is to
the observer this is my R distance
because that is nothing but the length
of the observers vector this is this
length is a because the loop has a
radius a and this is our 1/2 so it is a
triangle and I know that if I want our
1/2 I can write down our 1/2 square is
equal to R squared plus a squared minus
2a our cause sigh because psy is
referring to this angle side the angle
between the R vector and the a vector
this is what is called the triangle
relation you would have derived it in
school for a right angle triangle this
goes away and you get Pythagoras's
theorem for a straight line you will get
this becomes unity and then that just
says that r12 is R plus a or R minus a
but in general if you have any triangles
and if this angle is Phi this is our
this is a this is our 1/2 this is the
relation you must satisfy the problem in
there is very straightforward except we
need to know cos Phi if we can calculate
cos Phi we can substitute it in there
and we can try to write down the
integral but what is cos Phi cos Phi is
the angle between this radius vector and
this this observer vector and the radius
vector of the loop so in order to
calculate cos Phi what I am going to do
is I am going to say cause sy is equal
to unit vector along r dot unit vector
along e i know that because if i take
the dot product of these two it is
magnitude of this unit vector times
magnitude of this unit vector times the
COS of the angle between the
let me write down what these are this
our hat is nothing but I have taken the
Raider the observer position itself
written in cylindrical coordinates
actually and divided by Rho square plus
Z square
what about e well is nothing but this
vector which can be written in terms of
Rho hat and in terms of Phi hat so this
becomes Rho hat cos Phi Prime remember
this is Phi prime plus Phi hat sine Phi
Prime
maybe if I draw a picture we'll make it
clearer
this direction is X it's also the
direction if I have it is also the angle
Phi for the observer observer is in the
XZ plane I have rotated the coordinates
so that he is next plane I am talking
about a current element that is at an
angle Phi Prime so if I want to know
what this direction is I want unit
vector on along the radial line in that
direction I can build it out of these
two vectors that is the component along
X and the component along Y so I can
write this as XI hat is equal to X hat
cos Phi prime plus y hat sine Phi prime
but this X hat is nothing but the
cylindrical coordinate R direction and Y
hat is nothing but this direction which
is the Phi hat direction so that is why
I have written it in this fashion now
the purpose of doing all this is that I
when I take the dot product there is
only one common direction in all this
the Z hat appears only here and the Phi
hat appears only there so the dot
product is only between these two terms
so I get it is equal to Rho over square
root of Rho Square + H square times cos
Phi Prime
because there is no Z hat component in
that vector there is no Phi hat
component in this vector this Rho over
Rho square plus Z squared what is that
it is this length divided by this length
it is nothing but sine theta cos Phi
Prime it's a very standard result is
just that I have thought is better to
derive it so the angle between the
observer vector and this radius vector
this sigh is sine theta cos Phi Klein
let me put that in and let us see where
we get
what I want is the value of a the vector
potential is is a five PI hat which is
equal to mu knot I over 4 pi integral 0
to 2 pi cos Phi prime sends a C Phi
prime divided by r1 to what is R 1 2 is
equal to R square plus a squared minus
2a r and sine theta cos Phi to the power
of half I have taken this cos Phi and
substitute it sine theta cos Phi prime
so that is the expression we have this
is R 1 2 and this is DL that is a very
complicated integral except that
wherever we are interested in this
integral we know that e is much much
much less than R we are talking about
loops there are the radius of an atom so
there angstroms in size we are talking
about distances that are of the order of
macroscopic distances so a is tiny
compared to R and so we can expand this
denominator if we expand it I can pull
an R out I am going to do it in place to
save some space one cell divided by R
squared divided by R squared and pull
out an R so you know that 1 over square
root of 1 plus something is equal to 1
minus that same something divided by 2
that is 1 over square root of 1 plus X
is equal to approximately 1 minus x over
2 this is your binomial expansion for
square 1 over square root
so if I use that there I get that is
equal to MU knot I over 4 PI R integral
0 to 2 pi and I'll put this as
approximate I will pull this a out cos
Phi prime times 1 plus now this is a
very small number so I am NOT going to
keep it I am only going to keep this
term to a R over R squared with a 1/2
sine theta cos Phi plus g5
so I have taken this and I have applied
the assumption this is small and brought
it to the numerator the minus sign has
become a plus sign and I've got a half
now what happens I am doing an integral
from 0 to 2 pi integral of 0 to 2 pi of
cos Phi prime D Phi prime is 0 so this
first term cannot give me anything
but the second term involves cos squared
Phi Prime and I know that integral 0 to
2 pi of cos squared Phi prime D Phi
prime is equal to this is the average
half since equal to PI where is integral
0 to 2 pi cos Phi prime D Phi prime
equals 0 so since I have this I can
simplify this integral completely I get
this is equal to MU naught I a over 4 PI
R
this is going to give me pi sine theta
times this factor e arcs over R square
the factor of two cancels out let me
collect terms so I get mu naught over 4
pi that is still my normalization
constant for the magnetic field then I
have PI e squared I and then I have a
sine theta divided by R square so I have
taken care of mu naught over 4 pi PI a
squared I sine theta and there is an R
cubed in the denominator and R in the
numerator so I am left with R square
this quantity is the area of the loop PI
a squared multiplied by the current and
it is called the magnetic moment is
given the symbol M it is a vector in the
sense that if I have a loop with a
current I and an area a the sense in
which the current is rotating I take my
right hand screwdriver and I tighten
that screw driver in the direction that
the current is rotating and see how that
screwdriver is moving that direction
gives me the direction of my vector the
magnitude is the current into the area
so the final answer is the vector
potential is equal to K Phi along the
Phi direction it is equal to mine or
Malaysian constant mu naught over 4 pi
times the vector times the magnitude of
that magnetic moment times sine theta
over R square
this can be written in a simpler form
conceptually simpler form which is it is
mu naught over 4 pi times the vector
magnetic moment cross gradient of 1 over
R that is because we know that gradient
of 1 over R is nothing but minus R over
R square and if you take Z cross minus R
over R squared you will end up with the
desired expression I am Not sure of
signs so you can check the sign of that
result all right so now I have a result
for this loop the loop is at origin the
observer is also at some point in the X
Z plane and for this observer with the
current I I have so the vector potential
a is equal to PI hat mu naught over 4 pi
M cross of M into sine theta over R
square but this is only the vector
potential
I want the magnetic field and there is
no point calculating the vector
potential if calculating the magnetic
field after that is even harder the
magnetic field B is equal to this curl
of E
and we work this out but let me just do
it again it never hurts to repeat it is
R square sine theta times R hat R theta
hat R sine theta pi hat del Del R del
del theta del Del Phi e is only along
Phi so zero zero mu naught over 4 pi M
sine theta over R square and x are signs
that is my same R sine theta that comes
into H H Phi now this it has to be seen
that I cannot have any derivative of Phi
because there is nothing here the only
term that is present is a Phi so I get
derivatives with respect to R and
derivatives with respect to theta so it
will be it is straightforward to write
down what the curl is is 1 over R square
sine theta R hat times del Del theta of
this
I will call it as h phi d phi and n plus
r theta hat x minus del Del R of H Phi C
1 H Phi a Phi is nothing but this term
so whatever I am I have just expanded
along the first row R hat has del Del
Phi of this minus 0 or theta hat has 0
minus del Del R of this and Phi hat has
0 so there are only two terms that I
have to worry about when I take the
theta derivative I am taking theta
derivative of sine squared theta let me
write this out del Del theta of sine
square theta is equal to twice sine
theta cos theta
is that all right because I first I can
apply my chain rule so I D D sine theta
of sine square theta is twice sine theta
times D is d theta of sine theta which
is cos theta so the first term has a two
sine theta cos theta coming out of it so
it is one over R squared sine theta
times again mu naught over 4 pi times M
over R times 2 sine theta cos theta and
this is along our hat the second term
has a minus sign and there is a factor
of R now I want the are derivative of H
Phi a Phi this is 1 over R because there
is AR in the numerator and R squared in
the denominator take del Del R of 1 over
R it is minus 1 over R square so you
just write out mu naught over 4 PI M
sine theta times minus 1 over R square
and this is along theta
now the R squared sine theta the sine
theta part cancels out in both these
cases so I will remove the sine theta
sine theta sine theta R squared + R make
it R cubed and here you have an R in the
numerator and R to the power of 4 so
again R cubed so if we combine
everything the final answer you get is
that the magnetic field B is equal to R
hat times let me pull out the common
factors mu naught over 4 pi times M
times R hat
so I get cos theta over R cubed with a
factor of two mine plus theta hat sine
theta over R the factor of two comes out
here here there is a cos theta there
there is a there should be a sign this
is sine squared theta so one sine theta
goes other sine theta comes here the R
cubed is common so this is the
expression for the magnetic field and
let me recapitulate how we got there we
started with a loop of current this loop
of current this picture here is
producing a magnetic field far away now
we can directly try to calculate the
magnetic field but that is very
difficult the reason why it's very
difficult is you have to do a cross
product inside the integral and it is
bad enough doing a curl but doing the
cross product and then integrating is
much worse so instead we work with the
vector potential which directly
integrates current and because it
directly integrates current we can add
up symmetric components of current and
conclude that there is no component of
the vector potential along X this is all
the component is along Y and in in in
spherical geometry the Y Direction is
nothing but the Phi direction so we
calculate a 5 and a 5 involves mu knot I
over 4pi it involves this cos Phi prime
simply because the current is not
pointing along Phi it's pointing in a
slightly different angle that takes K
that is taken care of by cos Phi Prime
and then there is a 1 over R using the
triangle inequality we worked out what 1
over R should be
and we got this expression which is R
squared plus a squared minus 2a R cos
Phi cos I and by you dot product in two
directions we were able to obtain that
cause sighs scientist sine theta cos Phi
prime expanding the denominator we get
the answer what is the answer tell us
let me write down the vector potential
as well
these these expressions are only for
very tiny loops when you look at the
field very far away now what the vector
potential tells us is that once again it
is the R over R cubed factor that is
present you can see that the magnetic
the vector potential is proportional to
1 over R square which means the magnetic
field is proportional to 1 over R cubed
this is very similar to the dipole if
you go back to your electrostatics nodes
and look at the electric field due to a
dipole it will be exactly the same
exactly the same dependence so in a
certain sense if you properly change the
multiplying constants the magnetic loop
is exactly like an electrostatic dipole
and that is why if you look at a bar
magnet and you have a North Pole and a
South Pole and you look and you looked
at the fields due to that magnetic due
to that magnet it is essentially the
same field that you would have got if
you put a charge Q and the charge minus
Q this is an electrostatic dipole this
is a magnetic dipole they are the same
structure and now what we are finding is
that if you took a loop
and put that loop in the middle of the
magnet and remove the magnet you will
still get the same pattern so a loop of
current or a electrostatic dipole or a
bar magnet produces the same field far
away the field decays faster decays as 1
over R cube so you won't see it once you
go very far away now there is one other
very interesting thing that you can see
directly from this I am sure you do not
want to see it but you can take the curl
of B this is probably going to sound
very painful to you because we have been
doing a lot of vector analysis but let
us do it H 1 H 2 H 3 is 1 over R squared
sine theta R hat R theta hat R sine
theta PI hat del del R del del theta del
Del Phi and then these two components
there is a constant so I can pull the
constant out so I can call this mu
naught over 4 pi times M
so what's down here is twice cos theta
over R cubed and in the theta component
is R times sine theta over R cubed and
then there is zero
now if I want to take this determinant
it is important to realize this term and
this term are not functions of Phi so
this term goes away there is no
derivative with respect to Phi that is
possible here because the field does not
depend on how you on which Phi angle you
measure it because it is a loop it looks
the same at all angles so it is only
these two derivatives that can make any
difference if you look at the r
component del del theta X 1 0 and del
Del Phi is not present so there is no
component along R if you look at the
component along theta del Del Phi cannot
act and del Del R are acts on 0 so you
are left only with the Phi component so
it is equal to MU naught over 4 PI M
over R square sine theta times R sine
theta and Phi and then you have these
two terms which is del Del R of sine
theta over R squared minus del Del theta
of twice cos theta divided by R cube
delta theta of cos theta is minus sine
theta so i will do it well i am going to
erase the middle of this board
delta theta of cos theta is minus sine
theta del Del R 1 over R squared is
minus 2 over R cubed so it becomes this
common factor x minus 2 sine theta over
R cubed - twice - sine theta divided by
R cubed so derivative of 1 over R
squared is minus 1 over R cubed
derivative of cos theta is minus sine
theta well it is 2 sine theta or R cubed
in both places there is a minus sign
here there is a plus sign here this is
so not only do we have these two
expressions we've also proved something
very very important if you take the
curve of the magnetic field due to a
loop far away that curl is zero now why
is this important
supposing I take any general current
distribution I know that I am looking at
divergence J equal to zero which means
the current must go somewhere current
cannot just vanish since the current
cannot just vanish I can build this
current out of little little loops so
for example any general current can be
built out of little loops at every point
this current and this current cancel
except at the outside it Stokes theorem
in Reverse we did the exact same
construction for Stokes theorem so any
any current distribution can be built
out of tiny loops and we know the
magnetic field due to a tiny loop and
that magnetic field has zero curl
what that means is that if I have any
general current distribution J away from
that current distribution curl of B is
equal to zero what happens inside the
current distribution well if I am
looking here whatever currents are there
outside where I am looking curl of B is
zero but where I am looking itself curl
of B will not be zero it takes a little
bit more mathematics to prove the final
statement the final statement is if you
consider current
everywhere curl of B is zero for
currents far away and it is equal to J
itself nearby and this is called amperes
law there is a mathematical way of
proving it and that is again given in
appendix in your textbook what you do is
you say curl of B this curl of curl of e
apply a few vector identities then do a
lot of mathematical gymnastics and you
can prove the same result now we can't
actually prove that is equal to J but
what we can prove is it is not equal to
J or at any other point at most it is
equal to J at that point and we know
that B is derived from J so it must be a
function of J in some way so the purpose
of doing this exercise was just to
motivate the answer that it makes sense
that curl of B should be related to J at
the place itself it cannot be related to
J anywhere else because we know that
away from the loop curl of B is 0
alright so we have done this rather
painful derivation but I think the
lesson to be learned from this
derivation is that the vector potential
is useful when you first encounter the
vector potential the question that
should occur to you even though I didn't
mention it is that I am having a vector
field B and I am saying B is curl of
another vector field E and then I say
what a great thing I've done I am saying
I can get this vector field from another
vector field that is hardly anything I
mean the electric field I got from a
scalar
so instead of three components as
dealing with one component but when I
went to the vector field three
components came from three components so
I do not see any saving by working with
the vector potential rather than working
with the vector field with the magnetic
field
well the saving is there the way to look
at it is not to count the number of
components rather what you have to do is
you have to look at what is inside the
integral and if you do that here is what
you see the electric field is 1 over 4
pie epsilon-not
volume integral Rho R over R cubed R 1
to L say DV prime whereas potential was
equal to 1 over 4 PI epsilon naught
integral Rho over r12 DV Prime and
because of this construction we had that
divergence of E was equal to Rho over
epsilon naught ok actually there should
be mu naught you now B is equal to MU
naught over 4 pi volume integral J cross
R over R cubed R 1 to D V prime E is
equal to MU naught over 4 PI J over R 1
to DV Prime curl of B is equal to MU
naught J look at the symmetry of these
equations the electric field is derived
from a charge density the potential
involves only that source divided by r1
2 and if you take the divergence of the
electric field it is related to the
source at that same location at the
observers location
the magnetic field is derived from a
cross product of its source the current
density and R over R 1 or R cubed the
vector potential simplifies that by
being just the source divided by r 1 -
no cross product no vector operation at
all and the curl of B is connected to
the source at the observers location so
there is a complete symmetry between
electric field and magnetic field it is
only that we are very used to thinking
of vectors that keep going in the
direction that we push them in so we are
very happy with the electric field
because if the electric field is in a
certain direction the force is in that
direction whereas the magnetic field if
it's in a certain direction the force is
90 degrees to that direction the source
of the magnetic field is again 90
degrees - it's the direction in which it
points but that complication all came
from the fact that we had that force was
equal to i dl 1 cross i dl 2 cross r 1/2
divided by r 1 2 cubed once you had this
kind of structure and i 1 cross i to
cross r gradient of 1 over r you have to
have these kinds of 90 degree
relationships it forces i dl cross b
therefore already b has to be 90 degrees
to force b itself is i dl cross r over r
cubed which means b is 90 degrees to its
own source so that is why you keep
seeing cross products everywhere but
cross products notwithstanding there is
a complete symmetry between electric and
magnetic fields and as long as you do
not get scared by the cross products you
can in fact use all the techniques that
are used in electrostatics in
statics and we will do that next time we
will basically use Laplace's equation to
solve some simple magnetic problems