Question

MULTIPART-3

Two long parallel wires carry currents of 5 A and 7 A in the opposite direction. The wires are separated by 0.28 m.

a) Find the magnetic force per unit length between the two wires.

b) Find the magnetic field midway between the wires

c) Find the magnetic force on each wire. The wires are then suspended horizontally one above the other with the currents flowing in same direction.

d) If the upper wire is held securely without sagging, for what mass/length will the lower wire be suspended without sagging.

(Hint: there are two forces on lower wire that are exactly balanced)

PLEASE SHOW HOW TO WORK THE PROBLEM!!!

Answer 1

(A) We know that the force per unit length is
dF/dL = u_oi₁i₂ / (2*Pi*d)
where d is the distance between two
dF/dL = (4Pi*10^-7)(5*7) / (2*Pi*0.28)
dF/dl = 2.5*10^-5 N/m
(b) Magnetic field due to wire having current 5 A
B = u_oi₁ / (2Pi*(d/2))
B₁ = (4Pi*10^-7)*(5) / (2Pi*(0.14)) = 7.14*10^-6 T
Magnetic field due to wire having current 7 A
B₂ = (4Pi*10^-7)*(7) / (2Pi*(0.14)) = 1*10^-5 T
Net magnetic field will be
B₂ - B₁ = 1*10^-5 - 7.14*10^-6 = 2.86*10^-6 T
(c) Magnetic force will remain same as earlier only the nature will be attractive.
(d) The force on wire will be in upward direction is
  = 2*10^-5 N/m
therefore mass/length should be same to avoid sagging.