in this example we are given that a slightly
divergent beam of charged particles, accelerated
by potential difference v. propagates from
a point ay along the axis of a solenoid. and
here we are given that the beam is focused
at a point at distance l, from ay by 2, magnetic
induction values b-1 and b-2, one after another.
and we are required to find the specific charge
of the particles. now in solution we can see
if this is a solenoid. and along its axis.
we know that magnetic induction everywhere
exist along the axis of solenoid. and say
from point ay slightly diversion beam of charge
particle is ejected out. so we know that all
these particles will follow, helical path.
and say, these are focused at a point. p which
is located at a distance l from the initial
point ay. we know that in helical path, after
every pitch. the particle reaches a point
which is just ahead of the starting point.
so we can say point p must be located at some
distance such that this l, is integral multiple
of the pitch of hellix, then only we can say.
all these, points will be brought into focus
at point p. or at every point after a pitch
all these particles which are ejected will
be focused. now here we are given that. the
focusing is done at 2 successive value of
induction these are b-1 and b-2. and we know
the pitch of hellix is given by, 2 pie m v
coz theta by q-b. and as we are given that
beam is slightly divergent we can take coz
theta as unity because then angle is very
small. so pitch of hellix in terms of the
ki-netic energy because we are also given
with potential difference. this can be written
as 2 pie momentum m v i can write as root
of. 2 m e v. divided by q-b that is pitch
of hellix. and in this situation we can say.
for b-1 and b-2 these are focused so we can
write. for magnetic induction b-1 we can say
the length l is. n times the pitch of hellix
so which can be written as 2 pie. root of
2 m-e-v, divided by q b-1. and for magnetic
induction b-2 as we know. for higher value
of induction pitch is less so the same length
can be written as n plus 1 because 1 pitch
must be increase then only focusing will take
place. so here it’ll be 2 pie root of 2
m-e-v by q b-2. from these 2 equations 1 and
2 here we subtract 1 from 2. by taking b-1
and b-2 on the other side you can see we are
getting. the result to be b-2 minus b-1 multiplied
by l. is equal to 2 pie root of, 2 m-e-v.
divided by q. and in this situation you can
see. the value of q by m can be obtained by
squaring, here you can also take the charge
q instead of e because. i have already taken
charge is q so i just replace e by q. so in
this situation squaring we are getting, b-2
minus b-1 whole square l square is equal to.
8 pie square. m, q v by q square when q gets
cancelled out here. and remaining q by m we
can directly calculate as 8 pie square v.
over, l square b-2 minus b-1 whole square,
that will be the answer to this problem.
