Answer:
(a)The current in the second wire = 9.43 A
(b) The two current are in the opposite direction.
Step-by-step explanation:
The force per unit length (F/l) on a current carrying conductor = μ₀I₁I₂/2πr
Making I₂ The subject of the equation,
I₂ = [2πr(F/l)]/μ₀I₁.................... Equation 1
Where I₁ = Current in the first wire, I₂ = current in the second wire, μ₀ = permeability of vacuum, r = distance of separation between the two wires.
Given: F/l = 3.30 × 10⁻⁵ N/m, I₁ = 0.56 A, r = 3.20 cm = 0.032 m.
Constant: μ₀= 4π × 10⁻⁷ Am⁻¹.
Substituting these values into equation 1
I₂ = (2π × 0.032 × 3.30 × 10⁻⁵ )/(4π × 10⁻⁷×0.56)
I₂ = (0.032 × 3.3 × 100)/1.12
I₂ = 10.56/1.12
I₂ = 9.43 A
Therefore the current in the second wire = 9.43 A
(b) The two current are in the opposite direction, because the force on each wire repel each other.