Total current through conical surface

The current through a surface is I = ∫ J·dS with J = ∇×H. The surface is the cone θ = 20° with 0 ≤ r ≤ 5 and 0 ≤ φ ≤ 2π. Its normal is a_θ and dS = r sin θ dr dφ a_θ.

Given: H_r = 0, H_θ = 3r²/sin θ, H_φ = 54 r cos θ.

Step 1: a_θ component of the curl (spherical coordinates).

(∇×H)_θ = (1/r) [ (1/sin θ) ∂H_r/∂φ − ∂(r H_φ)/∂r ]

H_r = 0, and r H_φ = 54 r² cos θ, so ∂(rH_φ)/∂r = 108 r cos θ:

(∇×H)_θ = (1/r) [ 0 − 108 r cos θ ] = −108 cos θ   A/m²

Step 2: integrate over the cone.

I = ∫_{φ=0}^{2π} ∫_{r=0}^{5} (−108 cos θ)(r sin θ) dr dφ
  = −108 cos θ sin θ × (r²/2)|₀⁵ × 2π
  = −108 × cos 20° × sin 20° × 12.5 × 2π
  = −108 × 0.9397 × 0.3420 × 12.5 × 6.2832
  = −2726 A

Step 3: check by the line integral (Stokes' theorem). The boundary of the cone is the circle r = 5, θ = 20°, with dl = r sin θ dφ a_φ.

|∮ H·dl| = ∫₀^{2π} 54 r cos θ × r sin θ dφ
         = 54 × 25 × cos 20° sin 20° × 2π = 2726 A

The right-hand rule for the normal a_θ (pointing away from the z-axis and downward) makes the circulation run in the −a_φ direction, so the sign is negative. Both methods agree.

Answer: I ≈ −2.73 kA (magnitude 2726 A, flowing opposite to a_θ).