in this example we are given that a beam of
singly ionized charged particles. having ki-netic
energy thousand electron volt contain particles
of masses 8 into 10 to power minus 27 kilogram.
and 1 point 6 into 10 power minus 26 kilogram
emerge from one end of an accelerated tube.
and it is saying there is a plate, at a distance
point zero 1 meter from end of the tube. placed
perpendicular to it. and we are required to
find the minimum magnetic field in region
so that beam will not strike the plate. to
understand the situation lets draw the pictures
say if this is accelerated tube from which
the charge particles are emitting out. and
if the magnetic field, is in. outward direction
we can see. these charge particles will experience
a magnetic force in inward direction. in this
downward direction if magnetic force is there.
with respect to this centre these will follow
a circular path. and in front of the tube
if, a metal plate is kept at a distance exactly,
equal to the radius the particle, will just
graze and will not strike the plate. so this
is the situation. which we wish to frame and
we need to calculate the value of this magnetic
field. for which these will not strike for
which if the separation between this, accelerated
tube and the plate is d this d should be more
then r. so here we can use that. the radius
of particles in magnetic field r we write
that it is given m v over q-b. and here we
are given with ki-netic energy this momentum
we can write as root 2 m-k k is the ki-netic
energy by q-b. this is the radius of the circular
path. and we can see radius is, dependent
on mass of particle. so for these heavy particles
the radius will be large. so here we can write.
larger radii. larger radius is for. heavy
particles. so. for not to strike. the plate.
in this situation. the radius of this circular
path. must be. less then or equal to the separation
d because even at equality these particles
will just grace will not exactly hit the plate.
now in this situation we can write it as root
2 m k by, q-b. should be less then or equal
to d. this implies here we are getting the
value of magnetic induction. should be more
then or equal to. this root 2, m k. by q-d.
this is the value of magnetic induction minimum
value of magnetic induction which is needed
here. so on calculation we are getting it
is root of, twice of. mass of particles. here
we are going to use the heavy one. that will
be. the heavy particle will be 1 point 6 into
10 power minus 26 kilogram because here it
is 10 to power minus 27 kilogram. so it’ll
be 1 point 6 into 10 to power minus 26. multiplied
by. we are given that ki-netic energy here
is thousand electron volt this will be thousand
into 1 point 6 into 10 to power minus 19.
divided by the charge of particle. again we
can use as it is singly ionized particle it
is 1 point 6 into 10 to power minus 19 multiplied
by, the separation we are given here is zero
point zero 1. on simplifying this calculation,
you will get the result as root 2 tesla that
will be the answer to this problem.
