---
title: "Current crossing surface, spherical J"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:continuity-current-crossing-surface-theta"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/continuity-current-crossing-surface-theta
site: BCT Engineering
---

# Current crossing surface, spherical J

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Asoj, 3 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-crossing-surface-theta
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

In a certain region, J = 3r² cosθ a_r − r² sinθ a_θ A/m². Find the current crossing the surface defined by θ = 30°, 0 < φ < 2π, 0 < r < 2.

## Short answer

Current I = ∫ J·dS. Only the a_θ component contributes. I = ∫_0^{2π}∫_0^2 (−r² sinθ)(r sinθ) dr dφ = −2π ≈ −6.28 A, meaning 6.28 A flows across the cone in the −a_θ direction.

## Answer

The current crossing a surface is the flux of J through it: I = ∫ J·dS.

On the cone θ = 30° the surface element is dS = r sinθ dr dφ a_θ (normal along increasing θ). Only the a_θ component of J contributes, so J·dS = J_θ · r sinθ dr dφ.

```text
J_θ = −r² sinθ
I = ∫(φ=0→2π) ∫(r=0→2) (−r² sinθ)(r sinθ) dr dφ
  = −sin²θ ∫₀²ᵖⁱ dφ ∫₀² r³ dr
  = −sin²(30°) · (2π) · [r⁴/4]₀²
  = −(0.25)(2π)(4)
  = −2π
  = −6.283 A
```

The cos θ term of J is along a_r and is tangential to the cone, so it carries no current across the surface.

**Answer: I = −2π ≈ −6.28 A**, that is, 6.28 A flows across the cone in the −a_θ direction (towards the z-axis side, decreasing θ).

## Check yourself

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

1. On the cone θ = 30°, the surface element is:
   A. r² sinθ dθ dφ a_r
   B. r sinθ dr dφ a_θ
   C. r dr dθ a_φ
   D. r sinθ dθ dφ a_r
   Answer: B. A constant-θ surface has dS = r sinθ dr dφ a_θ. (Key point: Surface element)

2. Which component of J contributes to the current crossing the cone?
   A. Only J_θ, since J_r is tangential to the cone
   B. Only J_r
   C. Both J_r and J_θ
   D. Neither
   Answer: A. Only the component normal to the surface (a_θ) carries current across it. (Key point: Contributing component)

3. The current is:
   A. +2π A
   B. −π A
   C. −4π A
   D. −2π ≈ −6.28 A
   Answer: D. I = −sin²30° (2π)(2⁴/4) = −0.25 × 2π × 4 = −2π. (Key point: Result)

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