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

Consider a circular loop of radius `r` carrying current `I` ampere, held perpendicular to the plane of the paper.We want to find the magnetic field at point `P` which is at a distance `x` from the centre of the coil.Let us take a small element ab of length `dl`.From Biot-Savart's Law, the field at point `P` due to small elements is given by:The direction of the field is in a plane perpendicular to the plane containing `dl` and is at right angle to line `AP` Resolve this into two components `dB sin alpha` along the `x`-axis and `dB cos alpha` at right angle to the line `OP`.It is clear from the figure that the resultant field along `x`axis (i.e.,component `dB cos alpha` cancel each other)is zero.Field the total field at `P` due to all such elements is given by
`B=intdBsinalpha=int(mu_(0)l)/(4pi(r^(2)+x^(2)))dlsinalpha`
`=(mu_(0)Isinalpha)/(4pi(r^(2)+x^(2)))int dl=(mu_(0)I sin alpha)/(4pi(r^(2)+x^(2))) [thereforeintdl=2pir]`
`sinalpha=r/sqrt(r^(2)+x^(2))`substuting the values, we get
`B=(mu_(0)lr^(2))/(2(r^(2)+x^(2))^(3//2))`
If there are `n` turns in the coil then `B=(mu_(0)nIr^(2))/(2(r^(2)+x^(2))^(3//2))`
The magnetic field lines due to a circular current carrying loop are shown below: