Derive the expression for magnetic field at a point on the axis of a circular current loop.

Explanation

Consider a circular loop carrying a steady current I. The loop is placed in YZ plane with its centre at the origin O and has radius R. The X-axis is the axis of the loop. We want to find the magnetic field at .P. and is a distance x from .O..
Let us consider an element Idl on the loop and which produces a tiny magnetic field dB at P.
ie., `dB=(mu_(0)I|dlxxr|)/r^3=(mu_(0))/(4pi)(Idlrsin theta)/r^3`
`mu_0/(4pi)(Idlsin theta)/r^2`
Since `dlbotr " " dB = (mu_(0))/(4pi)(Idl)/r^2" " [:. sin9=1]`
But `r^(2)=x^(2)+R^2`
`implies dB = mu_0/(4pi)(Idl)/((x^2+R^2))`
The direction of dB is `theta` as shown in fig and it has X - component `dB_(x)` and Y - component `dB_(bot)`
But `sumdB_(bot)= sumdB sin theta = 0 " " ` [ Since , each `dB_(bot)` due to diagonally opposite Idl vanish ]
Thus, the net magnetic field at P is
`sum dB_(x) = sumdB cos theta`
But , `cos theta=R/r=R/((x^(2)+R^(2))^(1//2))`
`sumdB_(x) = sum(mu_(0))/(4pi)Idl(R)/((x^(2)+R^(2))^(3//2))`
`B = mu_0/(4pi)(IR)/((x^2+R^(2))^(3//2))sumdl" "` [ where .B. is the total magnetic field ]
The summution of element dl over loop yields `2piP` , the circumference of the loop Thus ,
`B=(mu_0)/(4pi)(IR)/((x^(2)+R^(2))^(3//2))xx2piR`
`B = (mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2)`