Video: Magnetic Field on the Axis of a Circular Current Loop Video Lecture | Physics for EmSAT Achieve

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18 so now we will calculate the magnetic
field at some point from a
current-carrying circular do so now let
us suppose you have a circular loop as
you can see here this blue colored blue
which is in this plane XZ it is in the
XZ plane okay and we have to find out
the magnetic field due to this current
carry carrying blue at some point let us
call this point as P which is located at
a distance X from the centre of the loop
on the y axis that means this point P is
lying on the other axis other than the
plane of them it is not lying on the
plane of the circular loop so until a
loop is in the XZ plane and you have
this point P on the y axis right and let
us suppose that the radius of the
circular loop this capital R so you have
to find it out at this point so let us
suppose I consider I consider a small
element that is DL I consider a small
element let us say a B of length DL so
this element is of length DL right so
what I will do I will first according so
what is my current element here so the
current element will be I DL right so
now according to by its seven law so as
per by observer law the small magnetic
field due to this current element DL at
Point P will be what if this will be
equal to MU naught by 4 PI I DN sine
theta by r square what
sine theta theta is nothing but the
angle which DL that is the length
element mates with the position vectors
position vector is this one so let us
draw the position vector so this is the
position vector right so this will be I
DL into sine theta so what is Theta
theta is the angle between this DL and
this line now if you look at it the DL
that is length element would be somewhat
like this because current is flowing in
this direction so this is DF therefore
the angle which it makes with this
position vector is nothing but 90 degree
right because you try to imagine this
what be the best way to understand this
is take a bangle in your hand and
imagine that the bangle is this circular
loop which is carrying some current so
if you consider a small element say DL
so the direction of that DL will be
tangential to the loop now if you
consider another point P which is not
lying in the plane of the loop then the
line joining the point P and this small
element DL they will be at 90 degrees to
each other than they will be
perpendicular to each other so in this
case your theta will be 90 degree this
is sine 90 degree divided by R square R
is nothing but this position vector so
what is the value of this this is
nothing but this is 90 degree this is
odd this is X so this will be root over
R square plus X square according to
Pythagoras theorem so this will be equal
to root over R square plus X square
whole square right so DV becomes equal
to mu naught by 4 pi i dl divided by r
square plus X square right so this is
the small value of DV now what about the
direction of DB when you look at the
direction you again apply the right hand
rule so try to apply right hand rule so
DV is governed by DL cross R so take all
your fingers towards DL whereas aunt
iris towards the right so if you take it
towards the right you
would see what is the direction of DB so
the direction of DB comes out to be
somewhat like this this becomes the
direction of DB give you if you get
trouble getting this direction try it in
this way you align your hand in the
direction of DL I have drawn the DL
vector right keep your hand like that
and try to fold your fingers towards
this art that is root over R square plus
X square so what do you get so it will
be somewhat like this that means first
you keep your hand here like you keep
your four fingers like this and this is
your thumb right so now you move your
hand in such a way so that you can roll
it roll your fingers towards I okay so
basically you move your entire hand
somewhat in this direction because as
your R is somewhat towards that
direction
therefore your thumb goes in this
direction and that becomes the angle of
that that becomes the direction of V B
so now you look at it as you consider
more small elements like this that means
as you keep considering elements here
here here here and so on gradually the
direction of V B also keeps changing now
it is here again it will be here here
here and so on so this D D can be
resolved into two components that is D B
cos theta and sine theta it is suppose
if this angle is Theta and that is this
angle will also be theta right so if
this is DB you can resolve DB into two
components that is DB if this is Theta
this is also theta so what will be this
angle this will be nothing but 90 minus
theta so this will be 90 minus heat up
so now if you resolve DB into two
components this is DB so this will be D
because 90 minus theta that
DB sine theta and this will be DB sine
90 minus theta that is DV cos theta so
these are the two components now what do
we see by symmetry we will see that you
will have all other DVDs but for all of
them
they're horizontal component will be T V
sine theta but their vertical component
for some of them will lie in this
direction so they will all cancel out so
by symmetry summation of DB cos theta
will be equal to 0 because gradually if
as you keep considering your small
elements DL you will keep getting more
DV and again you will resolve all those
DV into DB sine theta and cos theta all
sine Thetas will come in this direction
but half of the COS Thetas will come in
this direction and half of the cos theta
will come in the opposite direction so
your summation of all DB cos theta will
be 0 but summation of DB sine theta will
not be equal to 0 therefore what would
be the net magnetic field that will be
integration of DB sine theta because DB
sine theta is the only nonzero one so
now you put the value of DB here so what
you get you get u naught by 4 pie IDL
divided by R squared plus X square into
sine theta so now you apply simple
trigonometry in triangle a o P in this
triangle sine theta is perpendicular by
hypotenuse so you can write this sine
theta as perpendicular that is R divided
by root over R square plus X square so
you can write sine theta like this and
this remains the same that is mu naught
by 4 pie IDL divided by R squared plus x
squared
right so this becomes equal to so this
becomes equal to mu naught by 4 pi I R
divided by r square plus X square to the
power 3 by 2 into integration of DL so
DL will be integrated from there to
where this DL is basically a small
component of the entire circumference so
it will be integrated from 0 to the
circumference of the circle that is 2 pi
I so what do you get you get B is equal
to MU naught I R divided by 4 PI into X
square plus R square to the power 3 by 2
into 2 pi I so this phi and phi will get
cancelled this 2 will get cancelled with
this there fully get B is equal to MU
naught I R square divided by 2 into X
square plus R square to the power 3 by 2
so this is the value of magnetic field
at some point P due to a
current-carrying circular loop right so
now in this you have seen that the most
important part to understand here is the
direction of the magnetic field so I
will try to explain that to you very
clearly in this slide that is why I have
kept a separate slide for this even
though here also by calculating the when
I took a small element what I did I
considered that if that element is here
so what would be the direction of
magnetic field the direction of magnetic
field at this point was somewhat here
right
similarly if you consider this point
what would be the direction of magnetic
field here also your direction would be
somewhat like this so more you come this
side your direction will move this side
as you keep coming like this to this
point so if you will also keep reaching
the direction like this so a direction
will also keep changing like this as
your point keeps changing so now if you
are given a circular loop and you are
asked that what would be the net
direction of magnetic field so what did
you conclude from this discussion you
observe that for each direction for each
small element BL you have a particular
direction of magnetic field which can be
resolved into two components one is this
one is the component along this
direction and the other one is a
component along this direction so
whatever is along this direction will
get cancelled out in the opposite
direction so you are net magnetic field
so your net magnetic field will act
along this direction so this would be
the direction of net magnetic field if a
circular loop is carrying a current-i in
this direction so in this case we use
the right-hand thumb rule we call this
as I can come through this is used to
calculate or to determine the direction
of net magnetic field for a circular
current time loop how do you apply this
rule you fold all the fingers all the
four fingers of your hand in the
direction of current let us suppose in
this case the current is flowing like
this that means that if you look look at
this loop very carefully make a bangle
in your hand and you imagine that the
cup the bangle is kept in front of you
not as a circle it is looking like an
over to you right and the current is
flowing like this try to imagine it
properly the current is flowing like
this right so you take all the four
fingers of your hand this way one two
three oh these are the four fingers of
your hand and this is your thumb so you
keep your thumb like this you keep your
hand in front of you this is how your
you should keep your hand in front of
your face right now you curl your
fingers the way you are curling your
thing
is there be the current is flowing
through the wire now if you turn your
finger what will happen here is suppose
this is how it is this is the circular
loop right in which the current is
flowing like this
so if you curl your fingers the fingers
are denoting the direction of current
and this thumb which is pointing in this
direction is telling you the direction
of net magnetic field due to this
current which is flowing in the circular
loop so wherever you have a circular
loop you can use this right-hand thumb
rule to find out the direction of
magnetic field because if you use the
previous right hand rule you can
calculate the magnetic field for each
individual small elements but then you
will have to determine the net magnetic
field because of each of those elements
so this is a an easier way to know the
direction of net magnetic field due to a
circular current carrying loop so I hope
you understand see it is very important
I tried my best to explain this diagram
to you but it is very important that you
understand this diagram properly I mean
this circular loop how the circular loop
is placed please try to understand that
the loop is basically a circle but it is
placed in front of you in such a way
that the circle is not visible to you
this if you look the sideways then you
can see the circle otherwise the loop is
placed like this so you keep your hand
in front of your face in this position
cut all your fingers that is the
direction of the current which is
flowing through the wire and your thumb
points towards the direction of the net
magnetic field right so now that we have
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