in this example. this figure shows a beam
balance. at one end of which a current carrying
coil c with number of turns n, cross sectional
area ay and current i is attached. which is
kept between two pole pieces, on other end.
a pan is kept, in which a counter weight.
of mass capital m is kept. it is saying if
the. beam is, horizontal in equilibrium state
we are required to find the magnetic induction
due to these pole pieces in which the coil
is kept in equilibrium. here we can see that
magnetic induction will exist in the direction
from north pole to south pole. and. the magnetic.
torque acting on the coil must balance the
clockwise torque, produced by the weight of
this mass on the right term of the beam balance.
so in this situation its magnetic moment must
exist in leftward direction such that m cross
b will produce a torque in, anticlockwise
manner to balance the torque of the counter
weight. now in this situation, we can directly
write. in equilibrium. magnetic torque. on
c, must be equal to torque of the. weight
of m. so magnetic torque on c can be directly
written as m cross b and as the angle between
magnetic moment and magnetic induction is
90 degree it can be simply written as m-b.
and magnetic moment of the. coil can be simply
written as the product of current area and
number of turns, so magnetic torque on c we
can write as, b i n- a. which must be balanced
by the torque of the counter weight m. so
in this situation it’ll be given as m g
multiplied by l-2 because the weight acting
on it will be m-g and on the right term of
the beam. the torque will be mg. multiplied
by, the. arm length that is l-2. so on simplifying
we can directly get the value of magnetic
induction required to keep, this equilibrium.
it can be given as m-g l-2 divided by, i n-
a. that will be the answer to this problem.
