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Current through surface, continuity numerical

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.

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.

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.

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

At ρ = 2 m:

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

At t = 10 ms = 10⁻² s:

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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