Question

Two parallel electrical conducturs each leads a current of the same size. The distance between the center of the conductors can be varied from 100 mm to 1000 mm. The currents are going in the same directions in both of the conductors and have a value of I=19 A.

1. Calculate the force per meter (F/l) between the conductors in their plane and plot F/l against the distance between them (100 mm to 1000 mm) in a graph. Do not take to few points.
2. Show if there is a repelling or attractive force between the conductors.

Answer 1

Two parallel electrical conductors carrying current of I = 19 A

in the same direction
and the separation of the conductors may vary from 100mm to 1000 mm

We know that the moving chanrges can create magnetic field so both the conductos have the magnetic field due to current flowing in them

the magnetic field of the conductors let B1 is due to the first and B2 is due to second conductor

From Ampere's law
  
   B1 = mue0*I1/2pi*r

r is the separation of the conductors

and B2 = mue0*I2/2pi*r

now the magnetic force is F = Il B sin theta , here theta is 90 degrees so
the force per unit lenght on conductor 2 is
       F/l = I2*B1

       F1/l = I2*mue0*I1/2pi*r

here I1= I2 = I

       F1/l = mue0*I^2/(2pi*r)

same way force per unit lenght on conductor 1 is
       F/l = I1*B2

       F2/l = I1*mue0*I2/2pi*r

here I1= I2 = I

       F2/l = mue0*I^2/(2pi*r)

here the attractive force between the conductors takes place from right hand rule

that is conductor 1 attracts conductor2 with a force of F1 , conductor 2 attracts conductor1 with a force of F2

the magnitudes of the force per unit length is
for r = 100 mm

   F/l = 4pi*10^-7*19^2/(2pi*100*10^-3) N/m = 0.000722 N.m = 7.22*10^-4 N.m

and for r = 1000 mm is

   F/l = 4pi*10^-7*19^2/(2pi*1000*10^-3) N/m = 7.22*10^-5 N.m

so From the above values (F/l) proportional to (1/r)