in this example we are given that 2 long wires.
2 long parallel wires carry current of equal
magnitude but in opposite directions. and
these wires have suspended from ceiling by
4 strings of same length l we can see here.
and we are given that the mass per unit length
of the wire is lambda and we are require to
determine the value of theta.
that is the angle by which the string are
diverging.
assuming it to be very small.
now in solution, if we just draw the cross
sectional view of the situation here we can
see.
these are the wires through which the 2.
wires are suspended, these are the threads
to which 2 wires are suspended.
and. in this situation the length of these
strings or threads l. and 1 wire carries a
current i say here in outward direction and
other is carrying a current i in inward direction.
if we consider the separation between the
wires is l, the 2 will ripple each other with
the magnetic force f-m. and here, we can write
repulsive.
magnetic force.
on wires.
per meter that is per unit length we can write
as this magnetic force is mu not, i square
by 2 pie r. here i have taken the length of
wire to be unity because we are given with
mass per unit length.
so in downward direction it’ll experience
its weight that will be lambda g. because
we are talking about per unit length so it’ll
be lambda g and.
the string will carry, tensions.
t and t say if. both of these strings, will
experience a tension t so we can write in
this direction it’ll be 2 t here also it’ll
be 2 t. if this angle is theta. this will
also be theta.
that is the angle made by string from vertical.
and here we can write for.
iquilibrium of wires.
per unit length.
we can simply balance out the forces like,
in this situation 2 t sine theta will be equal
to the magnetic force which is mu-not i square
by 2 pie r. and vertical it’ll be.
2 t coz theta which will be balanced by the.
weight of the wire per unit length that is
lambda g. so, in this situation.
if we divided the 2.
equations, this will give us tan theta is
equal to.
mu not i square.
by, 2 pie.
lambda. here you can see it is mu-not i square
by 2 pie lambda r g. now in this situation
this r we can simply substitute as.
twice of l sine theta.
so because this half of i is l sine theta
we can write it 2 l sine theta. or.
sine theta we can approximate as theta only
so it can be written as 2 l theta.
now in this situation.
if we substitute the value of r as 2 l theta
and tan theta. we can also substitute approximately
equal to theta.
so you can see here.
theta i can take on lefthand side so this
will become theta square is equal to mu not
i square.
divided by. this 2 l will be left over here
so this will be 4 pie.
lambda g l. now using this we are getting
the value of theta as, under the root.
mu not i square by 4 pie lambda g l. that
will be the answer to this problem.
