L15 : Numericals - Moving Charge & Magnetism, Physics, Class 12 Video Lecture

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magnetism part 15 is brought to you by
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please make sure that you have watched
all the videos - part 14 before going
ahead with part 15 so let us look at the
first problem the problem says what is
the radius of the part of an electron of
mass 92 10 to the power minus 31 kgs and
charge 1.6 into 10 to the power minus 19
coulomb moving at a speed of 3 into 10
to the power 7 meter per second in a
magnetic field of 6 into 10 to the power
minus 4 Tesla perpendicular to it that
means there is an electron which is
moving at a speed of this match that is
let us say this velocity is 3 into 10 to
the power 7 meter per second in a
magnetic field of 6 into 10 to the power
minus 4 Tesla perpendicular to it that
means B and B are perpendicular to each
other leg so how will the motion Li let
us suppose this is the magnetic field so
magnetic field is acting in the downward
direction and let us suppose this is my
electronic so this electron is moving
with some velocity let us say V it is
moving with some velocity V so what
would be the direction of the force you
apply your right hand rule V cross B so
take all your four fingers towards V put
it down so your thumb will point towards
this direction so this will be the
direction of force similarly the
electron goes here so velocity is in
this direction force will be in this
direction so you will end up getting a
circle like this so the electron will
move in a circular path so basically the
force experienced by the electron due to
the magnetic field that is force
experienced by electron due to the
magnetic field is nothing but this force
acts as the centripetal force because of
which the
Ronn moves in a circular path so what is
the force experienced by the electron it
is Q into V cross B and centripetal
forces MV square by I now V cross B is V
B sine theta and theta is equal to 90
degree so this will be Q V V is equal to
mV square by I so V and V will get
cancelled so you get R is equal to MV
divided by QD C if you understand the
concepts then there is no need of even
memorizing the formula it did it take a
long time and then take a long time
right so understanding is more important
so what does M M is nothing but mass of
electron that is 9 into 10 to the power
minus 31 V is the velocity of electron
which is 3 into 10 to the power 7
divided by Q is the charge of electron
1.6 into 10 to the power minus 19 B is
the magnetic field so this comes out to
be 26 centimeters so this is the radius
of the path traveled by the electron now
it is also asking to calculate the
frequency now for frequency also if you
go on similar lines you will get the
frequency of rotation is equal to we
know that this is Omega by 2 pi and what
is Omega Omega is nothing but V by ad so
this is B by 2 pi I write because from
where do we get this we know that V is
equal to R Omega right that is our
linear velocity and angular velocity is
related so from this we can write V is
equal to R into 2 pi nu right so from
this we can write nu is equal to V
divided by 2 pi R and here you can write
down the value of V V is given as P into
10 to the power 7 divided by 2 into Phi
into R where R is the radius which we
can chelated just now as 26 centimeters
so this comes out to be 2 mega Hertz so
this is the frequency
see and now we have to calculate the
energy so what is energy since the
particle is moving so energy refers to
the kinetic energy that is energy due to
motion so energy will be equal to half
MV square so that is half into M that is
9 into 10 to the power minus 31 do be
square what is V V is 3 into 10 to the
power 7 into 3 into 10 to the power 7 so
this comes out to be 2 point 5 kilo
electron volts so this is the energy in
kilo electron volt let us go ahead and
look at the next problem the problem
says in a chamber a uniform magnetic
field of 6.5 Gauss is maintained an
electron is shot into the field with a
speed of hole point 8 into 10 to the
power 6 meter per second normal to the
field explain why the part of the
electron is a circle so against again
you have a similar kind of problem let
us suppose that this is the magnetic
field which exists so an electron is
shot into the field with a speed of this
much let us suppose the speed is given
as 4 point 8 into 10 to the power 6
meter per second so let's just say this
is the electron so it is shot with this
speed say in some direction V so if you
use your right-hand thumb rule you will
see that the force acts in this
direction ship since force acts in this
direction it will push the electron a
little bit towards itself so let us say
the electron will come here instead of
going here again at this point the
electron will tend to go in this
direction but the magnetic but the
magnetic field is downward so by right
hand rule the force will again try to
push it in this direction so again the
electron will come somewhere here again
it will try to move in this direction
but by right hand rule the force will be
along this direction so now if you look
at these points and if you start joining
these points you will end up in a circle
so what does it show it shows that the
force that the magnetic force which acts
on the
electron it always act towards the
center
it acts inwardly so the magnetic force
acts like the centripetal force
therefore the electron moves in a
circular path that is why the path of
the electron is a circle so hey there
again you can find out the radius of the
circular path that is magnetic force is
equal to the centripetal force which is
MV square by r what is magnetic force
that is q vb because Q into V cross V
theta is 90 degrees so that will be Q VB
is equal to mV square by ad so from this
you get ad is equal to mV divided by QB
so now you can just put the values M is
mass of electron which is 9 point 1 into
10 to the power minus 31 V is 4 point 8
into 10 to the power 6 divided by Q is 1
point 6 into 10 to the power minus 19
and V is magnetic field that is 6 point
5 into 10 to the power minus 4 Tesla so
this gives us 4 point 2 centimeters so
this would be the radius of the circular
orbit now let us look at the next
problem in problem 3 obtain the
frequency of revolution of the electron
in its circular orbit does the answer
depend on speed of electron so frequency
of circular orbit we already know that
the frequency of a circular orbit is
given by Q B divided by 2 pi and we just
derived it a couple of times before
right so now in this if you put the
values it is 1.6 into 10 to the power
minus 19 into B that is magnetic field
which is 6 point 5 into 10 to the power
minus 4 divided by 2 into 5 into M that
is mass of electron which is 9 point 1
into 10 to the power minus 31 it comes
out to be 18 Nega Hertz right so this is
the frequency of revolution or does it
depend on speed if you look look at this
expression frequency depends
on the charge of electron mass of
electron and the magnetic field so this
frequency is independent of speed of
electron so it does not depend upon the
velocity of electron so it is
independent of it and this forms the
basis of your cyclotron right because
the frequency of revolution does not
depend upon what is the energy of the
electron right so even though the speed
of the electrons still remain the same
but since the radius keep increasing
your the particle gets accelerated again
and again and as a result its energy
keeps increasing even though the
frequency of revolution does not have an
effect because of that now let us look
at the last problem which is based on
the cyclotron this problem sees that a
cyclotrons oscillator frequency is 10
megahertz what should be the operating
magnetic field for accelerating protons
if the radius of its DS is 60
centimeters what is the kinetic energy
of the proton beam produced by the
accelerator so let us look at the first
part first what should be the operating
magnetic field for accelerating protons
now the whatever is the oscillator
frequency of the cyclotron you would
need that much frequency to be applied
to the cyclotron for accelerating the
charged particles right so that means
the oscillator frequency should be equal
to the cyclotron frequency right which
is given as 10 megahertz which is equal
to 10 into 10 to the power 6 Hertz right
now we have to calculate now we know
that the cyclotron frequency is given by
Q B divided by 2 pi M now here we know
the charge on a proton and the mass of a
proton so we can calculate the value of
the magnetic field so this comes out to
be 2 pi m into nu C divided
did bite you so you can put these values
m is mass of proton which is 1 point 6 7
into 10 to the power minus 27 you see is
the cyclotron frequency which is 10 to
the power 7 Hertz charge is the charge
on proton which is 1.6 into 10 to the
power minus 19 so this value comes out
to the zero point six six Tesla so this
is the value of the magnetic field which
has to be applied for oscillating for
accelerating the protons now here it
says that the radius of the DS is given
as 60 centimeter calculate the kinetic
energy now we I told you in that slide
what would be the kinetic energy of the
protons which will come out of the
cyclotron it will be half MV square what
is V V is the velocity which width with
which the protons will come out of the
cyclotron right so this velocity is
nothing but ad into Omega whole square
right so this we can write it as R into
Omega is nothing but 2 pi mu so the
square so now we know all the values
half into M is 1.6 into 10 to the power
minus 27 this into what is our radius is
given as 60 centimeter that is 0.6
meters into 2 into 3.14 into nu that is
the frequency which is 10 to the power 7
this whole square so this entire value
on calculation comes out to be 7 mega
electron volts so this would be the
kinetic energy of the protons with which
they will go out of the cyclotron right
so now that we have solved a couple of
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