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

Explanation

Let R be the radius of a current loop , carrying current I . Let R be a point on the axis of a conductor . Let dB be magnetic field at P , due to a current element .idl. .
From the figure , `theta+alpha =90^@`, so that `alpha=90^@-theta` and `cos alpha =cos (90^@-theta)`
`=sin theta=R^2/((R^2+x^2)^(1/2))`...(1)
Let `dB_x` be the horizontal component of dB.
Applying Biot-Savart.s law ,
we write
`dvecB=(mu_0/(4pi))(I(d vecl xx vecr))/r^3`
i.e., `dB=(mu_0/(4pi))(Irdl)/r^3 sin theta` . where `theta.=90^@`
i.e., dB=`(mu_0/(4pi))(Irdl)/r^3=(mu_0/(4pi)) (Idl)/r^2` ...(2)
Component of dB along the horizontal is `dB_x=dB cos alpha`
or `dB_x=dBsintheta=(mu_0/(4pi)) (Idl)/r^2 sin theta`.
By using ...(1)
we write
`dB_x=(mu_0/(4pi))(Idl)/r^2 R/r =(mu_0/(4pi)) (IdlR)/(R^2+x^2)^(3/2)`....(3)
or integrating
`B_x=intdB_x = int (mu_0/(4pi)) (IR)/((R^2+x^2)^(3/2))dl`
i.e., `B_x=(mu_0/(4pi))(IR)/((R^2+x^2)^(3/2))(2piR)`
where `intdl=2piR`
or
`B_x=(mu_0/(4pi))((2piIR^2)/((R^2+x^2)^(3/2)))hati, dB_y=0`
For a circular loop and at the centre , x=0 .
`vecB_0=(mu_0/(4pi))((2pil)/R) hati`
For a circular conductor containing n turns ,
`B_xhati=(mu_0/(4pi))(2pinIR^2)/((R^2+x^2)^(3/2))hati`
and at the centre ,
`B_0hati=(mu_0/(4pi))((2pinI)/R)hati`