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Find J from H in cylindrical

Short answer

J = ∇×H. For H = (2/ρ) cos(0.2φ) a_ρ, only the z-component is nonzero: J_z = 0.4 sin(0.2φ)/ρ² A/m². At P_B(1.5, 90°, 0.5), sin(0.2×90°) ≈ 0.3090, so J_z = 0.0549 A/m². J = 0.055 a_z A/m².

Calculate the value of the vector current density J in cylindrical coordinates at P_B(1.5, 90°, 0.5) if H = (2/ρ)(cos 0.2φ) a_ρ A/m.

J is found from Ampere's law in point form, J = ∇×H. Given H = (2/ρ) cos(0.2φ) a_ρ A/m, so H_ρ = (2/ρ) cos 0.2φ and H_φ = H_z = 0.

Curl in cylindrical coordinates

(∇×H)_ρ = (1/ρ) ∂H_z/∂φ − ∂H_φ/∂z = 0
(∇×H)_φ = ∂H_ρ/∂z − ∂H_z/∂ρ = 0
(∇×H)_z = (1/ρ)[∂(ρH_φ)/∂ρ − ∂H_ρ/∂φ] = −(1/ρ) ∂H_ρ/∂φ

Only the z-component survives:

∂H_ρ/∂φ = (2/ρ)(−0.2 sin 0.2φ) = −0.4 sin(0.2φ)/ρ
J_z = −(1/ρ)(−0.4 sin 0.2φ / ρ) = 0.4 sin(0.2φ)/ρ²   A/m²

At P_B(ρ = 1.5, φ = 90°, z = 0.5)

φ = 90° = π/2 rad, so 0.2φ = 0.3142 rad and sin(0.3142) = 0.3090.

J_z = 0.4 × 0.3090 / (1.5)² = 0.1236/2.25 = 0.0549 A/m²

Answer: J = 0.055 a_z A/m² (J_ρ = J_φ = 0).

✓ Answer checked by Claude Opus 5.5 (high effort)

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