in this example.
the figure shows a long straight current carrying
wire with a current i-1. and a thin strip
of width b and length l is placed which is
parallel to it.
with a current i-2, as we can see here the
strip is placed at a distance r from the wire.
we are require to find the magnetic force
of interaction between these 2 currents i-1
and i-2.
now in the solution.
we can directly say as the 2 wires are parallel
to each other these will attract each other.
but i-2 is distributed within width b. so
here to calculate the magnetic force in strip
at a distance x from the wire we can consider
in, elemental.
wire which is of width d-x. if we calculate
the current.
In elemental wire.
then this current in elemental
wire d-i can be directly written as i-2 by
b into d x because current i-2 is distributed
in a width b. so within a width d-x current
will be i-2 by b into d-x. and if we calculate
the force. between. i-1 and d-i. is. d-f.
then this d-f can be given by the expression
mu not, i-1 d-I, l upon, 2 pie x. as the force
between two wires. separated by x is given
by mu-not i-1 i-2 l by 2 pie x. now in this
situation. total force. can be given as integration
of this d-f if we substitute the value of,
d-i and we take all constant out of the sign
of integration this will be mu-not. i-1 i-2
l upon 2 pie b. and. this will be integration
of d-x by x within limits of x from. r to
r plus b. so this will be the force which
is given by mu-not i-1 i-2 l upon 2 pie b.
and this integral is ellen x, which is. having
its limits from r to r plus b. which will
give us ellen r plus b minus ellen r and the
final result here will be mu not i-1 i-2 l
upon 2 pie b. ellen r plus b upon r. and this
will be the answer to this problem.
