Force between two long, parallel transmission-line conductors carrying equal and opposite currents Two conductors of a transmission line carry equal current I in opposite directions. How does the magnetic force on each conductor vary with current?

Introduction / Context:
Parallel current-carrying conductors exert magnetic forces on each other. This principle underpins busbar spacing, mechanical design of lines, and even the classic definition of the ampere. Understanding the magnitude’s dependence on current is essential in power and high-current engineering.

Given Data / Assumptions:

• Two long, straight, parallel conductors separated by distance d.
• Currents are equal in magnitude I, flowing in opposite directions.
• Quasi-static (DC or in-phase AC) and uniform medium (μ0) assumed.

Concept / Approach:

The magnetic field produced by conductor 1 at the location of conductor 2 is B1 = μ0 I/(2π d). The force per unit length on conductor 2 is f = I × B1 = μ0 I^2/(2π d). Thus, the magnitude of force scales with the square of current and inversely with spacing d. Opposite current directions result in a repulsive force; same directions produce attraction.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional analysis confirms [μ0 I^2/(d)] has units of N/m. This relationship is standard in electromagnetics and power engineering texts.

Why Other Options Are Wrong:

(a) Linear in I is incorrect; (c) the force decreases with distance (inverse), not increases; (d) I^3 is nonphysical here; (e) the force clearly depends on I.

Common Pitfalls:

Confusing direction (repulsive vs. attractive) with magnitude’s dependence; forgetting the 1/d factor; mixing field from a line (∝ 1/d) with that from a point or loop.

Final Answer:

Proportional to I^2