According to the Biot-Savart law, magnetic field dB at point P due to current element idl in the above diagram is given by:

B = ∫(μ_o/4π)idlcosθ /x² = (μ_o/4π)∫idlcosθ /x²……………….(i)
dB = (μ_o/4π)idl sin(90°-θ) /x² = (μ_o/4π)idlcosθ /x²

Considering triangle ABN, cosθ = AN/dl Hence, AN = dl cosθ

Considering triangle ANP: sin(dθ)˜dθ  = AN/x Hence, AN = x(dθ)

Using the value of AN from the above 2 equations:

dlcosθ = xdθ……………………..(ii)

Considering triangle AOP:cosθ = r/x

x = r/cosθ…………………(iii)

Using the values of dlcosθ from eq.(ii) and x from eq.(iii) in eq.(i):

B = ∫(μ_o/4π)ixdθ/x² = ∫(μ_o/4π)idθ/x = ∫(μ_o/4π)i(cosθ)dx/r

B = (μ_o/4π)(sinθ₂ + sinθ₁)

For infinitely long wire (θ₁ = 90° θ₂ = 90°):The above equation becomes B = μ_oi/(2πr)