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

Explanation

Consider a circular loop of radius R carrying a steady current I as shown in the The loop is placed in the Y-Z plane with its centre at the origin O. The X-axis is the axis of the loop. Let P be a point on its axis, distant .x. from the centre O of the coil. The magnetic field is calculated at point P.
The magnetic field at P due to the conducting element `bar (dl)` at Q is given by the BIot-Savart law:
`dB = ( mu _(0))/( 4pi ) (I | bar (dl ) xx vec r |)/( r ^(3))`
Any element `(ber (dl))` of the loop is peerpendicular to the displacement vector `(vecr ) `
`therefore |vec (dl) xx vecr| =dl r sin 90^(@) = dl r `
Equation (1) becomes
`dB = ( mu _(0) )/( 4pi ) (Ir dl )/( r ^(2))`
`dB = ( mu _(0))/( 4pi ) ( I dl )/( r ^(2))`
The directio dB is shown is the figure. It is perpendicular to the plane formed by `vec(dl) and vecr.` The magnetic field dB is resolved into two components along X-and Y-axis. Y-components of dB `[dB _(bot)]` will be cancel out. But X-components of dB are added up.
Therefore magnetic field at P due to the loop is
`B = sum dB cos theta `
`B = sum (mu _(0))/( 4pi ) (Idl)/(r ^(2)) cos theta `
`B = ( mu _(0))/( 4pi ) (I )/( r ^(2)) cos theta sum dl `
From the triangle `POQ, cos theta = (R)/(r ) and sum dl = 2pi R,` circumference of the loop
Equation (3) become, `B = (mu _(0))/( 4pi ) ( I )/( r ^(2)) (R)/(r) 2pi R`
` B = ( mu _(0) IR ^(2))/( 2r ^(3))`
In the triangle `POQ, r ^(2) = R ^(2) + x ^(2) and r ^(3) = [R ^(2) + x ^(2) ] ^(3/2)`
Equation (4) becomes, `B = ( mu _(0) I R ^(2))/( 2 [ R ^(2) + x ^(2) ] ^(3/2)) `