---
title: "Current in outer portion of wire"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:current-density-wire-outer-portion"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/current-density-wire-outer-portion
site: BCT Engineering
---

# Current in outer portion of wire

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2071 Shrawan, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/current-density-wire-outer-portion
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

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?

## 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.

## Answer

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.

```text
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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When J is uniform and perpendicular to the area, the current through an area is:
   A. I = J / A
   B. I = J × A
   C. I = J × A²
   D. I = J × radius
   Answer: B. I = ∫J dA reduces to JA for uniform J. (Key point: Current from J)

2. The area of the ring between R/2 and R is:
   A. (3/4)πR²
   B. (1/4)πR²
   C. (1/2)πR²
   D. πR²
   Answer: A. π[R² − (R/2)²] = (3/4)πR². (Key point: Ring area)

3. With R = 2 mm and J = 2×10⁵ A/m², the current in the outer ring is about:
   A. 0.63 A
   B. 2.5 A
   C. 19 A
   D. 1.9 A
   Answer: D. I = 2×10⁵ × 9.425×10⁻⁶ ≈ 1.885 A. (Key point: Result)

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