Now, let us move on to magneto statics. In
magneto statics before developing the concept
of magnetic field let us understand what develops
a magnetic field just like electric field
was developed by charges magnetic fields are
developed by currents . So, what are currents?
That is very important to understand.
Current is nothing, but the flow of charges
along one direction and that gives us a current.
Now, let usconsider different kind of situations
for example, if we have 1 dimensional geometry
and charge carriers flow in this direction
we get a line current, if we have a surface
like this and the charge carriers move in
this direction we get a surface current.
So, line current is expressed as I surface
current is expressed as K and if we consider
a volume any arbitrary volume we are considering
a cylindrical shape and if a volume current
flows through this that is indicated as J
that isthat is the kind of a volume current
that we can understand.
Now,we can actually express the total current
from volume current and surface current where
by integrating it over the cross section in
case of a volume current and over the line
perpendicular to the flowing charges for a
surface current.
Now, let us consider the magnetic field . 
We will discuss about the origin of magnetic
field later now we are just assuming that
there exists a magnetic field B and what happens
to the charge carriers, what happens to acharged
particle under the influence of a magnetic
field that is what we are interested in finding
out now.
So, magnetic field exerts a force . 
If we write magnetic force as F mag of course,
this is a vector this magnetic field exerts
a force on moving charged particles, if the
amount of charge is Q and it moves with a
velocity v then the magnetic force on the
charged particle at a magnetic field B would
be given as Q times v cross B.
So, similarly we can write down the electromagnetic
force 
from our earlier discussion about electric
field and this discussion about magnetic field.
So, the total electromagnetic force would
become Q times electric field plus v cross
the magnetic field and this is also known
as the Lorentz force ; the total electromagnetic
force is also known as the Lorentz force.
Now, how does a charged particle move under
the influence of an electromagnetic field,
electric as well as magnetic field? Let us
find that out. If we consider only magnetic
field that will make the situation simpler.
So, let us start with that .
So, magnetic field as we have seen it its
direction is v cross B. So, magnetic field
does not act along the velocity of the particle,
it also does not act along the direction of
the magnetic field, its perpendicular to both.
And a particle if a particle is say moving
in this direction and the magnetic there is
a magnetic field that exerts a force along
this direction, then that force will act like
a centripetal force and it will curve the
motion of this particle.
For particle of mass m and charge Q 
it moves with speed v, the magnetic force
its magnitude would be 
Q v B and that will be the that will be it
provided the velocity is perpendicular to
the magnetic field and if we consider B to
be a uniform magnetic field , 
then this arrangement would give.
So, the magnetic field will provide a centripetal
acceleration, centripetal force and that way
we will realize a cyclotron motion . So, the
charged particle would move along a circle
andit will keep on doing that. Now, for circular
motion we need an amount of centripetal force
in order to maintain that motion over that
circle . So, the required amount of centripetal
force is 
for circular motion m v squared over R where
capital R is the radius of the circle .
So, for this kind of a trajectory we must
have Q v B being equal to m v squared over
R and; that means, the momentum of that particle
would be Q B R, momentum is expressed as p
here. Now, if we also include the electric
force into consideration.
So, this time we are considering motion of
a charged particle 
under electromagnetic force 
under the influence of electromagnetic fields.
So, as a simple example let us consider that
the magnetic field points along x direction
and electric field points along the z direction.
Let us draw the coordinate system this is
the z direction, y direction and x direction
. 
We have a particle with positive charge and
its released from the origin at rest , 
electric field is along this direction, magnetic
field is along this direction, we have a positively
charged particle at the origin and its released
from there at rest our job is to find out
what is going to happen to this particle.
Qualitatively we can think that initially
the particle is at rest. So, the magnetic
force is 0 on this, the electric force will
accelerate the particle in the z direction.
So, initially the motion of the particle will
be along the z direction and once the electric
field accelerates it, it will pick up a speed
and the magnetic force will start to apply
on it. And what will the magnetic field do?
It pulls the particle in the y direction.
And when the ; when the particle starts moving
once again the electric force under the influence
of electric force it slows down reducing the
magnetic force on it and it it this thing
gets repeated. So, initially the particle
moves along this direction and when it moves
along that direction that is z direction,
then we get a magnetic fieldmagnetic force
applicable on it, the direction of the magnetic
force would be z cross s. So, that is y direction
and that bends the particle like this.
Now, the velocity of the particle is at this
point along y direction and when that happens
we will have a force on the particle magnetic
force on the particle that isy cross x that
is minus z direction. So, that way the particle
will come back here and it willbe immobile
and there will be no magnetic force on it
again electric force will pull it up here,
magnetic force will pull it down there and
this thing will be repeated this kind of a
motion will be there of the particle.
So, after doing this qualitative estimate
let us try to write downthe equations of motion
and try to make a quantitative estimate of
whatever we have discussed so far . So, there
is no force in the x direction to begin with,
the position of the particle can be described
at any time by . So, there is no force along
x direction any time what, so ever throughout
this motion.
So, we candescribe the position of the particle
at any time t as 0 there is no displacement
along x direction, y that is a function of
t and z that is a function of t. And the velocity
can be described as there is no velocity along
x direction, y dot velocity along y direction
that is the time derivative of y and z dot
velocity along z direction that the times
derivative of z.
So, we have to calculate v cross B to find
out the magnetic force and that is x cap,
y cap, z cap, 0 y dot z dot B 0 0 this determinant
and that is B z dot y cap minus B y dot z
cap .
Now, we can write down the equation of motion
as using the second law of motion force equals
the charge times the electric field plus velocity
cross the magnetic field . That means, mass
times the acceleration equals this quantity
is also equal to mass times the acceleration
that is mass times y double dot in y direction
plus z double dot in z direction.
Now, if we have this then treating the components
separately we can write down we have already
found the expression of v cross B, now treating
the components separately we can write down
Q times B times z dot that equals m y double
dot and Q times the electric field that is
along z direction minus Q B y dot this quantity
equals m z double dot.
That means we have a second order differential
equation, second order in time and it these
are coupled. So, y double dot has z dot in
it and z double dot has y dot in it. Now,
let us consider a frequency omega for this
motion that equals Q B over m, if that happens
to be the frequency that happens to the let
us not call it frequency right now let us
just denote it as as omega, then these two
equations we can write as y double dot equals
omega times z dot and z double dot equals
omega times E over B minus y dot ok.
After getting this we can differentiate the
first equation and put the value of z double
dot from the second equation. So, we get the
third order time derivative with respect to
y and second order time derivative for z comes
here and if we put the value of z double dot
fromthe second equation, we get one equation
without cross terms; that means, only in terms
of y and.
So, in terms of y triple dot and y dot we
will one equation and if we solve that equation
the general solution for that equation would
be 
y as the function of time can be given as
C 1 cosine of omega t plus C 2 sin of omega
t plus E over B times t plus C 3 which is
the constant of the integral. And z t z as
a function of time can be expressed as C 2
cosine of omega t minus C 1 sin of omega t
plus C 4 that is also a constant of the integration.
With these two equations using the initial
condition, initially what we had? We had the
particle at rest at the origin . So, the initial
condition can be written down as y dot at
t equals 0 is 0 there is no velocity along
any direction. So, z dot at t equals 0 is
also 0 and the particle is at the origin.
So, y and z those are also 0at t equals 0.
Now, if we apply this initial condition on
the general solutions for the differential
equations, we will find the specific solution
to this problem y t equals E over omega B
omega t minus sin omega t and z t would be
given as E over omega B one minus cosine of
omega t.
Now, let us put this quantity E over omega
B as R 
let us consider this and let us also use the
identity that sin squared omega t plus cosine
squared omega t this will be 1 using these
we can write down that y minus R omega t squared
plus z minus R squared equals R squared. This
is the equation of a circle of radius R whose
centre is at 0 , R omega t R, circle of radius
capital R with centre at this position.
And this travels in y direction at a constant
speed . 
How much speed? u equals omega r; that means,
E over B it travels along the y direction
at a constant speed. So, what is the trajectory
of the particle, how are we finding this?
If we consider a circle like this 
and if there be a charge there be a particle
like this here ok.
Let us not start from there let us start from
here if this was the origin, then this is
the y direction we are rolling this circle
this way. So, with we are rolling in this
circle along this direction; that means, the
motion of this point will look like 
something like this and that is what we qualitatively
expected and after working out this problem
we found exactly the same solution to this
problem.
