H due to bent filament at P

Setup. A 5 mA current comes in from infinity along the +z-axis to the origin (direction −a_z), then goes out along the +x-axis to infinity (direction +a_x). Find H at P(0, 1, 0) using the Biot-Savart law for a straight filament:

H = (I/4πρ)(cosα₁ − cosα₂) a_φ         or  H = (I/4πρ)(sinβ₂ − sinβ₁) a_φ

For a semi-infinite wire whose end lies at the foot of the perpendicular from P, the factor is 1, so H = I/(4πρ).

   z
   ↑ ↓ I (from ∞)
   |
   |
   O ---------→ x   I (to ∞)
    \
     P(0,1,0)  at y = 1

Segment 1: z-axis from z = ∞ to 0 (current −a_z)

P lies at a perpendicular distance ρ = 1 m from the z-axis, and the foot of the perpendicular is the origin (the end of the segment).

Magnitude: H₁ = I/(4πρ) = 5×10⁻³/(4π × 1) = 0.398×10⁻³ A/m
Direction: dl × a_R : (−a_z) × (a_y) = +a_x

H₁ = 0.398 a_x  mA/m

(From Biot-Savart: with dl = −dz a_z and R = a_y − z a_z, dl × R = dz a_x; H₁ = (I/4π)∫₀^∞ dz/(1+z²)^(3/2) a_x = (I/4π) a_x.)

Segment 2: x-axis from 0 to ∞ (current +a_x)

P is a perpendicular distance 1 m from the x-axis, again with the foot at the origin.

Magnitude: H₂ = I/(4π × 1) = 0.398 mA/m
Direction: dl × a_R : a_x × a_y = +a_z

H₂ = 0.398 a_z  mA/m

Total

H = H₁ + H₂ = 0.398 a_x + 0.398 a_z  mA/m
|H| = 0.563 mA/m

Answer: H(0,1,0) = 0.398 a_x + 0.398 a_z mA/m (= 0.4 (a_x + a_z) mA/m).