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

Explanation

Consider a current loop of radius R carrying a steady current I. let .P. be a point at a distance .Y. from the center of the current loop on its axis.
Let .dl. be the current element as shown. The magnetic field due to it is given by Biot- savart.s law.
`dvecB=mu_0/(4r) (I|vec(dl)xxvecr|)/r^3`
But `r^2=x^2 +R^2`
The displacement vector `vecr` from `vec(dl)` to the axial point P is in the X-Y plane. Hence
`|vec(dl)xxvecr|=rdl`

`therefore dB=mu_0/(4pi) (I|vec(dl)xxvecr|)/r^3 =mu_0/(4pi)(Ir.dl)/r^3 =mu_0/(4pi) (I. dl)/r^2 =mu_0/(4pi) (I dl)/(x^2+R^2)`
The component of magnetic field along X-direction is `dB_x = dB cos theta`
`cos theta = R/r = R/(x^2+R^2)^(1//2)`
`dB=mu_0/(4pi) (I dl)/r^2 R/(x^2+R^2)^(1//2) = mu_0/(4pi) (I dl) /((x^2+R^2)) R/(x^2+R^2)^(1//2)`
`therefore dB=(mu_0 IR)/(4pi(x^2+R^2)^(3//2)dl`
Therefore the total magnetic field at P due to the loop
`B=(mu_0. IRdl)/(4pi(x^2+R^2)^(3//2))`
`therefore` As `int dl = 2piR " " B=(mu_0 IR^2)/(2(x^2+R^2)^(3//2))=mu_0/2(IR^2)/((x^2+R^2))^(3//2)`