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 + a= 90^@`, so that `alpha = 90^@ – 0` and `cosalpha = cos (90^@ – 0)`
`sintheta = 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=dBcosalpha`
or `dB_x =dB sin theta`
`=(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=int dB_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))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)))` tesla
and `vecB=B_x hati=(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=(mu_0/(4pi))((2piI)/R)hati`
For a circular conductor containing n turns ,
`B_x hati=(mu_0/(4pi)) (2pinIR^2)/((R^2+x^2)^(3/2))hati`
and at the centre ,
`B_0hati=(mu_0/(4pi)) ((2pinI)/(R))hati`