Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius `R`.
Draw the magnetic field lines due to circular wire carrying current `I`.

Explanation

Magnetic field lines due to a circular current loop :

The magnitude of the magnetic induction `vecdB` at point `P` due to this element is given by
`vecdB_(x)=(mu_(0))/(4pi)(Idelta I sin alpha)/(r^(2))`
The direction of `vecdB` is perpendicular to the plane containing , `vecdl` and `vecr` and is given by right hand screw rule. As the angle between `I vecdl` and `r` is `90^(@)` the magnitude of the magnetic induction `vecdB` is given by ,
`vecdB=(mu_(0)I)/(4pi)(dl sin 90^(@))/(r^(2))=(mu_(0)Idl)/(4pir^(2))`
The component of `vecdB` along the axis,
`vec(dB)c=(mu_(0)Idl)/(4pir^(2))sin alpha`
But `sin alpha=(R )/(r )` and `r=(R^(2)+x^(2))^(1//2)`
`:. vecdB_(x)=(mu_(0)Idl)/(4pir^(2))*(R )/(r )=(mu_(0)IR)/(4pir^(3))dl=(mu_(0)IR)/(4pi(R^(2)+x^(2))^(3//2))dl`
Total field `vecB=oint(mu_(0)IR)/(4pi(R^(2)+x^(2))^(3//2))dl=(mu_(0)IR)/(4pi(R^(2)+x^(2))^(3//2))ointdl`
But, `ointdl=` length of the loop `=2piR`
Therefore, `B=(mu_(0)IR)/(2pi(R^(2)+x^(2))^(3//2))(2piR)impliesvecB=B_(x)i=(mu_(0)IR^(2))/(2(R^(2)+x^(2))^(3//2))hati`
If the coil contains `N` turns, then
`B=(mu_(0)NIR^(2))/(2(R^(2)+x^(2))^(3//2))`tesla.