Let us consider a circular loop of radius a with centre C . Let the plane of the coil be perpendicular to the plane of the paper and current I be flowing in the direction shown. Suppose P is any point on the axis at direction r from thecentre.
Let us consider a current element dl on top (L) where, current comes out of paper normally whereas at bottom (M) enters into the plane paper normally.
LP⊥dl
Also MP ⊥dl
LP=MP=√r2+a2
Now, magnetic field at P due to current element at L according to Biot-Savart Law,
dB=‌

µ0

4π

⋅‌

Idlsin‌90∘

(r2+a2)

Where, a= radius of circular loop.
r=‌ distance of point ‌P‌ from centre ‌
along the axis.
dB‌cos‌ϕ components balance each other and net magnetic field is given by
B=∮dBsin‌ϕ
‌=∮‌

µo

4π

[‌

Idl

r2+a2

]⋅‌

a

√r2+a2

‌[∴sin‌ϕ=‌

a

√r2+a2

‌ In ‌∆PCM]
=‌

µo

4π

‌

Ia

(r2+a2)‌ 3 2

‌∮dl
B‌=‌

µo

4π

‌

Ia

(r2+a2)‌ 3 2

‌∮dl
B‌=‌

µoI2a2

2(r2+a2)‌ 3 2

(2πa)
For n turns, B=‌

µonI2a2

2(r2+a2)‌ 3 2