Skip to main content

Current crossing surface, spherical J

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.

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.

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.

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electromagnetism chapter 2 questions →