Magnetic field in Solenoid - Moving Charges & Magnetism Video Lecture - Class 12

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ahead with part 25 let us now determine
the strength of magnetic field in case
of a solenoid let us suppose this is the
solenoid let us suppose we have a
solenoid like this and I want to
calculate the magnetic field of a
solenoid so I consider a very small
element let us suppose I take a very
small element like this okay the
thickness of this element is DX so I
consider a small element from this
solenoid and let us suppose that I want
to calculate the elect of the magnetic
field inside the solenoid right inside
the solenoid is somewhere inside this
only way I want to calculate let us
suppose at some point P which is at a
distance X I want to calculate a
magnetic field so the line which joins
this small element with P is this line
and let us say that this angle is Theta
and let us suppose that the radius of
the solenoid is capital I so this line
represents the radius right this will
represent the radius of the circular
loop so this is the radius for the
solenoid now we will apply it now from
our previous lessons we know that for a
circular coil we derived the magnetic
field for a circular coil right so for a
circular coil at any point P which is at
a distance X from the circular coil the
strength of magnetic field is mu knot I
R square divided by 2 into R square plus
X square to the power 3 by 2 so this is
the strength of magnetic fields now in
this case there are n turns because the
solenoid is made up of end such circular
coils so because of this n turns the
magnetic field will become n this is
capital
n that is n number of turns we will not
I are square divided by 2 into R square
plus X square to the power 3 by 2 so now
for the small element DX so for the
small element DX what would be the small
amount of magnetic field the small
amount of magnetic field would be the
total number of terms over this element
now let us suppose that small n is the
number of turns per unit length so if
this is the number of turns per unit
length then how many turns are there in
this length DX so number of turns in the
length DX will be nothing but n into DX
that is number of turns per unit length
into the length over which I want to
calculate the number of turns so this
will be my number of terms mu naught I R
square divided by 2 into R square plus X
square to the power 3 by 2 let me call
this equation as equation 1 now from the
figure we can say that so from the
figure which I have drawn I can say that
tan theta is equal to ad by X right so
we can say that X is equal to R cot
theta so from this we can say that DX by
D theta is equal to minus R cosec square
theta right because d by d theta of God
theta is minus cosec square theta so
from this we can write DX is equal to
minus R cosec square theta into D theta
so now we put this value of DX we put
this DX in equation 1 so what do we get
we get D B is equal to minus n R cosec
square theta mu naught I R Square D
theta
/ - into R square plus instead of X
square I can write R square cot square
theta so plus R square cot square theta
to the power 3 by 2 so this becomes
equal to minus n R - n R cube mu knot I
cosec square theta divided by 2 into R
cube into cosec cube theta so this R
cube + R cube will get cancelled this
cos x square + + X square will get
cancelled so we are left with so we are
left with minus n mu knot I divided by 2
into cosec theta D theta so this can be
written as minus n mu knot I divided by
2 into sine theta D theta so this is the
small value of the magnetic field due to
the small element DX therefore B will be
equal to integration of DB so that will
be equal to integration of minus n mu
knot I by 2 sine theta D theta so this
will be equal to minus n mu knot I by 2
into cos theta from theta 1 to theta 2
so we integrate this over theta from
some initial theta 1 to some final theta
2 so this becomes equal to minus n mu
knot I divided by 2 so here it was minus
so this will be minus cos theta 2 plus
cos theta 1 right so now here we assume
now here if the coil is in finite so if
the coil is infinite what would be your
theta 1 and
Cheeta - what is theta 1 and theta 2
basically theta 1 and theta 2 is nothing
but the angle which they subtank at that
particular point so if the coil is
infinite in that is your theta 1 will
start theta 1 that is the initial angle
so the initial angle had to be 0 and
theta 2 would be 180 degree therefore
your P will become minus n mu knot I by
2 cos 180 minus cos 0 so this will be
equal to minus n mu knot I by 2 into
minus 2 so this will get cancelled
therefore you get magnetic field is
equal to n mu knot I so please make a
note of this point here that this n here
does not represent total number of turns
this is the number of turns per unit
length so please make a note of this
because fading at times people make
mistake they think that n mu not either
n is the number of turns but here n is
not the number of turns it is the number
of turns per unit length right so I hope
it is clear to you now please make a
note of this theta 1 and theta 2
sometimes you don't understand this so
please have a look at the figure what is
theta 1 and theta 2 theta 1 is nothing
but the angle which this small element
will extend will a subtend at the point
where we want to calculate the magnetic
field so now if this wire is in finite
that means this theta will start from
this end which will be very far now as
the distance will increase this side
theta will decrease so even if it goes
to infinity theta can be 0 it cannot be
less than 0 right so in this side if it
is infinity so theta will be 0 similarly
this side it is infinity that means I
mean let us suppose if that if DX starts
from here
when DX is here let us suppose theta is
0 as it starts coming this side theta
keeps increasing
it becomes theta when it reaches DX
again DX keep shifting this side theta
keeps increasing
increasing increasing increasing and
when it goes to infinity the maximum
theta can be 180 degree it cannot go
beyond this so that is why for infinite
solenoid we have taken theta 1 as 0 and
theta 2 as 180 degrees right okay so
this is your magnetic field inside the
solenoid now what is the magnetic field
outside the solenoid now a solenoid is
considered ideal if it does not produce
any magnetic field just outside the
solenoid when current is flowing through
it so we assume that the magnetic so we
assume the ideal scenario right how
every real life it is not like that
there is some magnetic field even just
outside the solenoid but for an ideal
solenoid what happens in case of ideal
solenoid magnetic field just outside is
equal to zero so magnetic field just
outside the solenoid when current is
flowing through it is zero in that case
we see that the solenoid is an ideal
solenoid right so that is all about
solenoid so you understood what the
solenoid what was the need to construct
the solenoid what are the practical uses
of solenoid and how do we determine the
magnetic field of the solenoid so till
now we talked about the magnitude of the
magnetic field in a solenoid what about
the direction when it comes to direction
the rule will still remain the same we
will still apply the right-hand thumb
rule the same rule which we applied in
case of circular loop now let us suppose
in this case if current is flowing in
this direction then you take all the
food fingers and curl them in the
direction of current you are turn all
the whole fingers like this in the
direction of current as you see here in
that case your kumble point in this
direction so that means the magnetic
field is in this direction and current
is in this direction right so whenever
you have the little the same thing as in
case of a circular loop the direction of
current is
by folding your fingers in that
direction and the direction of your
thumb will tell us the direction of the
magnetic field thank you please visit
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