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

Concept: The force between two long, straight, parallel current-carrying wires is given by the formula:
F = \(\frac{μ₀ \cdot I_1 \cdot I_2 \cdot d}{2π}\)
Where:
• F is the magnetic force per unit length
• μ₀ is the permeability of free space
• I₁ and I₂ are the currents in the wires
• d is the distance between the wires

Given Change: The current in one wire is increased to three times its original value and its direction is reversed, so the new current becomes −3I₁. The distance between the wires is also changed to 2d.

New Force Calculation:
The new force F' between the wires is calculated as:
F' = \(\frac{μ₀ \cdot (-3I_1) \cdot I_2 \cdot 2d}{2π}\)
Simplifying the above expression:
F' = −3 × \(\frac{μ₀ \cdot I_1 \cdot I_2 \cdot d}{2π}\)
So, F' = −3F

Conclusion: The negative sign indicates that the direction of the force has reversed (i.e., from attraction to repulsion or vice versa), but the magnitude of the force becomes three times greater than the original force.

Final Answer: The magnitude of the force between the wires after the changes is 3F.