---
title: "Current through surface, continuity numerical"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:continuity-current-through-surface-rho"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/continuity-current-through-surface-rho
site: BCT Engineering
---

# Current through surface, continuity numerical

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/continuity-current-through-surface-rho
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Let J = e^(−10⁴t)/ρ² a_ρ A/m² be the current density in a given region. At t = 10 ms, calculate the amount of current passing through the surface ρ = 2 m, 0 ≤ z ≤ 3 m, 0 ≤ φ ≤ 2π.

## Short answer

Current I = ∫J·dS over the cylindrical surface ρ=2, 0≤z≤3, 0≤φ≤2π gives I = 3π e^(−10⁴t) A. At t = 10 ms, I = 3π e^(−100) ≈ 3.5×10⁻⁴³ A.

## Answer

**Given:** J = e^(−10⁴t) / ρ² a_ρ A/m². Find the current through the cylindrical surface ρ = 2 m, 0 ≤ z ≤ 3 m, 0 ≤ φ ≤ 2π at t = 10 ms.

**Method:** The current through a surface is the flux of J through it: I = ∫ J · dS. On the cylinder, dS = ρ dφ dz a_ρ, so J · dS = J_ρ ρ dφ dz.

```text
I = ∫₀³ ∫₀^(2π) [ e^(−10⁴t) / ρ² ] ρ dφ dz
  = e^(−10⁴t) · (1/ρ) · (2π)(3)
```

At ρ = 2 m:

```text
I = e^(−10⁴t) × (1/2) × 6π = 3π e^(−10⁴t)   A
```

At t = 10 ms = 10⁻² s:

```text
10⁴ t = 10⁴ × 10⁻² = 100
I = 3π e^(−100) = 9.425 × 3.72 × 10⁻⁴⁴ = 3.5 × 10⁻⁴³ A
```

**Answer: I = 3π e^(−100) ≈ 3.5 × 10⁻⁴³ A** (outward, along a_ρ). The current is essentially zero at this time, as the exponential factor has decayed. By the continuity equation, this is the rate at which the charge enclosed by the surface decreases.

## Check yourself

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

1. How is the current through a surface found from J?
   A. I = ∫ J dl
   B. I = ∫ J · dS
   C. I = ∇ · J
   D. I = ∫ J dv
   Answer: B. Current is the flux of J through the surface. (Key point: Current as flux)

2. On the cylinder ρ = 2 m, the surface element is:
   A. dS = ρ dφ dz a_ρ
   B. dS = dρ dz a_φ
   C. dS = ρ dρ dφ a_z
   D. dS = ρ² dφ dz a_ρ
   Answer: A. The curved surface of a cylinder has dS = ρ dφ dz a_ρ. (Key point: Area element)

3. At ρ = 2 m, the current expression is:
   A. 6π e^(-10⁴t) A
   B. 12π e^(-10⁴t) A
   C. (3/π) e^(-10⁴t) A
   D. 3π e^(-10⁴t) A
   Answer: D. I = e^(-10⁴t)(1/ρ)(2π)(3) = 3π e^(-10⁴t) at ρ = 2. (Key point: Current expression)

4. At t = 10 ms the current is about:
   A. 9.4 A
   B. 0.94 A
   C. 3.5×10⁻⁴³ A, essentially zero
   D. 3π A
   Answer: C. 10⁴t = 100, so I = 3π e^(-100) ≈ 3.5×10⁻⁴³ A. (Key point: Value at 10 ms)

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