NEET2026PhysicsChapterActual
Two long, straight, parallel transmission cables, X and Y, are separated by a distance of 20 cm . Cable X carries a steady current of 20 A and cable Y carries a steady current of 30 A in the opposite direction. What is the magnitude and nature of the magnetic force per unit length experienced by cable Y? (Take ₀ = 4 10⁻⁷ T m A ⁻¹ )
Options
- A6 10⁻⁴ N m ⁻¹ , repulsive
- B6 10⁻⁴ N m ⁻¹ , attractive
- C3 10⁻⁴ N m ⁻¹ , repulsive
- D6 10⁻⁶ N m ⁻¹ , attractive
Correct answer
A. 6 10⁻⁴ N m ⁻¹ , repulsive
Step-by-step solution
The magnetic force per unit length between two long, straight, parallel current-carrying wires is given by the formula: F L = ₀ I₁ I₂ 2 d Given: I₁ = 20 A I₂ = 30 A d = 20 cm = 0.2 m ₀ = 4 10⁻⁷ T m A ⁻¹ Substituting the values into the formula: F L = 4 10⁻⁷ 20 30 2 0.2 F L = 2 10⁻⁷ 600 0.2 F L = 1200 10⁻⁷ 0.2 = 6000 10⁻⁷ = 6 10⁻⁴ N m ⁻¹ Since the currents in the two cables are flowing in opposite directions, the magnetic force between them is repulsive. Answer: 6 10⁻⁴ N m ⁻¹ , repulsive