H due to finite current filament

The filament lies on the z-axis from z = −3 to z = 3 and carries I = 12 A in a_z. By the Biot-Savart law, the field of a finite straight filament is

H = (I / 4πρ) (sin α₂ − sin α₁) a_φ

where ρ is the perpendicular distance from the filament to the point, and α₁, α₂ are the angles (measured from the perpendicular through P) to the lower and upper ends.

Step 1: geometry. P(−1.5, −4, 3), so the point is at height z = 3.

ρ  = √((−1.5)² + (−4)²) = √18.25 = 4.272 m
lower end:  z₁ = −3 − 3 = −6 m (below P)
upper end:  z₂ =  3 − 3 =  0 m (level with P)
sin α₁ = −6 / √(4.272² + 6²) = −6 / 7.366 = −0.8146
sin α₂ =  0 / 4.272          =  0

Step 2: magnitude.

Hφ = 12 / (4π × 4.272) × (0 − (−0.8146))
   = 0.2236 × 0.8146
   = 0.1821 A/m

Step 3: direction. a_φ = a_z × a_ρ, with a_ρ = (−1.5 a_x − 4 a_y)/4.272 = −0.3511 a_x − 0.9363 a_y.

a_φ = a_z × a_ρ = 0.9363 a_x − 0.3511 a_y

Step 4: vector form.

H = 0.1821 (0.9363 a_x − 0.3511 a_y)
  = 0.1705 a_x − 0.0639 a_y   A/m

Answer: H ≈ 0.17 a_x − 0.064 a_y A/m (magnitude 0.182 A/m).