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 ` cod alpha = cos (90 ^(@) - theta )= sin theta = (R^(2))/( (R^(2)+x^(2))^(1/2))`
let `dB_(x)` be the horizontal component of dB .
Applying Biot - Savart.s law ,
We write `d vecB = ((mu_(0))/(4pi ))(I(d vecl xx vec r))/(r ^(3))`
`i.e., dB =((mu_(0))/(4pi))(Irdl)/(r^(3)) sin theta . ` where ` theta . = 90 ^(@)`
`i.e., dB =((mu _(0))/(4pi)) (i rdl )/(r^(3))=((mu_0)/(4pi))(idl)/(r^(2))`
component of dB along the horizontal is `dB_X = dBcos alpha `
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))(2pi R )`
Where ` int dl = 2 pi R or `
` B_x =((mu_0)/(4 pi))((2 pi IR ^(2))/((R^(2)+x^2)^(3/2)))` tesl and
` vecb = B_(x) hati = ((mu_(0))/(4pi )) ((2pi IR ^(2))/((R^(2)+x ^(2))^(3/2)))hati , dB _(y)=0 `
For a circular loop , and at the centre , `x = 0 , `
` vec B _(0 ) =((mu_0)/(4 pi))((2pi l)/(R ))hati `
for a circular conductor containing n turns ,
`B_x hati =((mu_0)/(4pi ))(2pi n IR ^2)/((R^2+x^2)^(3/2))hati ` and at the centre ,
` B_0 hati =((mu_0)/(4 pi))((2pi n l)/(R )) hati`.