H due to side of triangular loop

Assumptions (the figure is described in words): the loop lies in the z = 0 plane with corners at O(0,0,0), (2,0,0) and (0,1,0). Side 1 is the segment on the x-axis from O to (2,0,0). The current of 10 A is taken to flow counter-clockwise when seen from +z, so in side 1 it flows from O toward (2,0,0), i.e. along +a_x. (If the figure shows the opposite sense, the answer just reverses sign.)

Method: Biot-Savart law, H = ∫ I dl × R / (4π R³), where R is from the source element to the field point P(0, 0, 5).

Step 1: set up for side 1. Source point (x, 0, 0), 0 ≤ x ≤ 2.

dl = dx a_x
R  = (0 − x) a_x + 0 a_y + (5 − 0) a_z = −x a_x + 5 a_z
|R| = √(x² + 25)

Step 2: cross product.

dl × R = dx a_x × (−x a_x + 5 a_z) = 5 dx (a_x × a_z) = −5 dx a_y

Step 3: integrate.

H₁ = (I / 4π) ∫₀² (−5 dx) / (x² + 25)^(3/2) a_y

Use ∫ dx/(x² + a²)^(3/2) = x / (a² √(x² + a²)), with a = 5:

∫₀² 5 dx / (x² + 25)^(3/2) = 5 × [ x / (25 √(x² + 25)) ]₀²
                           = 5 × 2 / (25 × √29)
                           = 10 / (25 × 5.385)
                           = 0.07428

Step 4: result.

H₁ = −(10 / 4π) × 0.07428 a_y
   = −0.7958 × 0.07428 a_y
   = −0.0591 a_y   A/m

Answer: H₁ ≈ −59.1 a_y mA/m at (0, 0, 5) due to side 1.