The two wires shown carry currents of 3.0 A and 5.0 A in the direction indicated. (a) Find the direction and magnitude of the magnetic field at a point midway between the wires. (b) Find the magnitude and direction of the magnetic field at a point 20 cm above the wire carrying the 5.0 A current. (c) Calculate the force per unit length (including direction) that exists between the wires.
Answer
Magnetic Field and Force Between Parallel Current-Carrying Wires
🧲 Part (a): Magnetic Field at the Midpoint
- Distance to midpoint from each wire: 0.10 m
- Use: B = μ₀I / 2πr
B₁ (3A wire) = (4π × 10⁻⁷ × 3) / (2π × 0.10) = 6 × 10⁻⁶ T
B₂ (5A wire) = (4π × 10⁻⁷ × 5) / (2π × 0.10) = 1 × 10⁻⁵ T
B₂ (5A wire) = (4π × 10⁻⁷ × 5) / (2π × 0.10) = 1 × 10⁻⁵ T
Using the right-hand rule, the field from the 3A wire is upward, and from the 5A wire is downward at midpoint.
✅ Net magnetic field: 4 × 10⁻⁶ T downward
🧲 Part (b): Magnetic Field at Point P (20 cm above the 5A wire)
- Distance from 5A wire: 0.20 m
- Distance from 3A wire: √(0.20² + 0.20²) = 0.283 m
B₅ = (4π × 10⁻⁷ × 5) / (2π × 0.20) = 5 × 10⁻⁶ T (leftward)
B₃ = (4π × 10⁻⁷ × 3) / (2π × 0.283) ≈ 2.12 × 10⁻⁶ T
B₃ = (4π × 10⁻⁷ × 3) / (2π × 0.283) ≈ 2.12 × 10⁻⁶ T
Decompose B₃ into components:
B₃ₓ = 2.12 × 10⁻⁶ × cos(45°) ≈ 1.5 × 10⁻⁶ T (left)
B₃ᵧ = 1.5 × 10⁻⁶ T (down)
B₃ᵧ = 1.5 × 10⁻⁶ T (down)
Add horizontal components:
Bₓ = 5 × 10⁻⁶ + 1.5 × 10⁻⁶ = 6.5 × 10⁻⁶ T
Total magnetic field:
B = √(6.5² + 1.5²) × 10⁻⁶ = 7.12 × 10⁻⁶ T
✅ Magnetic field at point P: 7.12 × 10⁻⁶ T leftward and downward
🧲 Part (c): Force Per Unit Length Between the Wires
- Use formula: F/L = μ₀I₁I₂ / 2πr
F/L = (4π × 10⁻⁷ × 3 × 5) / (2π × 0.20) = 3 × 10⁻⁵ N/m
Since both currents are in the same direction, the force is attractive.
✅ Force per unit length: 3 × 10⁻⁵ N/m, attractive