Question

Two long parallel wires separated by distance ‘d’ carry currents I₁ and I₂ in the same direction. They exert a force F on each other. Now the current in one of the wire is increased to three times and its direction is made opposite. The distance between the wires is doubled. The magnitude of force between them is

• A. \(\frac {F}{2}\)
• B. \(\frac {3F}{2}\)
• C. \(\frac {2F}{3}\)
• D. 3F

Answer 1

The Correct Option is D

Force between two parallel current-carrying wires:
F = \(\frac {μ_₀ \times I_1 \times I_2 \times d}{2π}\)
When the current in one wire is increased to three times and its direction is reversed, the new current becomes -3I₁.The new distance between the wires is 2d.The new force can be calculated:
F' = \(\frac {μ_₀ \times(-3 I_1) \times I_2 \times 2d}{2π}\) 
F' = -3 x \(\frac {μ_₀ \times I_1 \times I_2 \times d}{2π}\)
F' = -3F
Therefore, the magnitude of the force between the wires after the changes is 3F.