H at centre of square loop

The loop lies in the plane x = 0 (the yz-plane), centred at the origin. Vertices: A(0, −2, 2), B(0, 2, 2), C(0, 2, −2), D(0, −2, −2). Each side is 4 m long, so the centre O is d = 2 m from each side. I = 20 A, direction A→B→C→D→A.

Formula for a finite straight segment at perpendicular distance d:

H = (I / 4πd)(sin α₂ − sin α₁)

For a symmetric segment, the ends subtend ±45° from the perpendicular at the centre (half-length 2 m, d = 2 m), so sin α₂ = sin 45°, sin α₁ = −sin 45°.

Step 1: magnitude due to one side.

H_side = 20 / (4π × 2) × (0.7071 + 0.7071)
       = 0.7958 × 1.4142
       = 1.1254 A/m

Step 2: direction of each side (use dl × a_R, with a_R from the element to O).

Side AB: current +a_y at z = 2, a_R = −a_z     → a_y × (−a_z) = −a_x
Side BC: current −a_z at y = 2, a_R = −a_y     → (−a_z) × (−a_y) = a_z × a_y = −a_x
Side CD: current −a_y at z = −2, a_R = +a_z    → (−a_y) × a_z = −a_x
Side DA: current +a_z at y = −2, a_R = +a_y    → a_z × a_y = −a_x

All four add up in the −a_x direction (current is clockwise when viewed from the +x side).

Step 3: total.

H = 4 × 1.1254 (−a_x) = −4.50 a_x  A/m

Exact form: H = −(10√2/π) a_x.

Answer: H = −4.50 a_x A/m at the centre of the loop.