in this example. the figure shows a large
horizontal coil of radius r, which is carrying
a current i. and another small coil of radius
small r. which is very less compare to radius
of this larger coil. which is also carrying
a current small i and total n turns is placed
at the centre with its plane at an angle theta
from the axis of this larger coil. and we
are required to find the torque experienced
by the smaller coil in this situation. here
we can see that smaller coil is placed in
the magnetic induction of the larger coil.
at some angle theta to the direction of magnetic
induction so here we can see. it will experience
some magnetic torque. in that situation if
we calculate the magnetic moment. of smaller
coil. we know that magnetic moment m can be
given as, the product of current multiplied
by the area enclosed by the smaller coil which
is pie r square multiplied by the total number
of turns. so it’ll be n, i pie r square
which is the magnetic moment. and the direction
of this magnetic moment depending on the direction
of small i we can see it will exist in this
direction this is the direction of m vector.
which is always normal to the plane of coil.
and the direction is given by righthand thumb
rule by circulating fingers along the direction
of current. now if we calculate the magnetic
induction due to. large coil. at its centre.
then, this large coil, will. produce a magnetic
induction b which can be given as mu not i
by 2-r. which we know this is the magnetic
induction due to circular coil at its centre.
and here we can see. the direction of this
magnetic induction will be. in this direction,
that is b vector in upward direction, here
according to the direction of flowing current
in the large coil. now in this situation if
we see the angle between magnetic induction
vector and magnetic moment vector it is pie
by 2 plus theta. so, theta between. b vector
and, m vector we can write this is equal to
pie by 2 plus theta. and, here we can. directly
write the torque. on smaller coil. here we
can see it is given as torque is equal to
m cross b. so in this situation direction
of m cross b we can calculate again by righthand
thumb rule it’ll be in outward direction.
so the torque vector will be acting in outward
direction or the smaller coil will experience
an anticlockwise torque. the magnitude of
torque we can write as m b, sine of the angle
between the 2 which is pie by 2 plus theta.
so in this situation in magnitude we can write
the value of torque is, n i pie r square.
multiplied by the, magnetic induction which
is given as mu not i by 2 r. multiplied by
for magnitude we take coz theta. so the result
of net torque on the smaller coil can be written
as mu not i capital i, n. pie r square by,
2 r coz theta that will be the answer to this
problem.
