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Current in outer portion of wire

Short answer

Outer ring area = (3/4)πR² ≈ 9.42×10⁻⁶ m². Current I = J × area ≈ 1.9 A, assuming uniform current density J = 2×10⁵ A/m².

The current density in a cylindrical wire of radius R = 2 mm and uniform cross-sectional area is given by J = 2×10⁵ A/m². What is the current through the outer portion of the wire between radial distances R/2 and R?

Current through an area is I = ∫ J dA. The question gives J = 2×10⁵ A/m² for the wire of uniform cross-section, so J is taken as uniform and perpendicular to the cross-section.

Given: R = 2 mm = 2×10⁻³ m, J = 2×10⁵ A/m², region R/2 to R.

Area of outer ring = π[R² − (R/2)²] = (3/4) π R²
                   = 0.75 × 3.1416 × (2×10⁻³)²
                   = 0.75 × 3.1416 × 4×10⁻⁶
                   = 9.425×10⁻⁶ m²

I = J × A = 2×10⁵ × 9.425×10⁻⁶
  = 1.885 A

Answer: I ≈ 1.9 A (three quarters of the total current 2.51 A flows in the outer ring).

✓ Answer checked by Claude Opus 5.5 (high effort)

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