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 carryinf a steady current I as shown in the figure. 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 `vec(dl)` at Q is given by the Biot-Savart law
`dB = (mu_(0))/(4pi)(I|vec(dl) xx vec(r)|)/(r^(3)) " "rarr (1)`
Any element `(vec(dl))` of the loop is perpendicular to the displacement vector `(vec(r))`
`therefore |vec(dl) xx vec(r)| = dl r sin 90^(@) = dl r`
Equation (1) becomes `dB = (mu_(0))/(4pi)(I r dl)/(r^(3))`
`dB = (mu_(0))/(4pi)(I dl)/(r^(2)) " "rarr (2)`
The direction dB is shown in the figure. It is perpendicular to the plane formed by `vec(dl)` and `vec(r)`. The magnetic field dB is resolved into two components along X- and Y-axes. 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 = Sigma dB cos theta`
`vec(B) = Sigma(mu_(0))/(4pi)(I dl)/(r^(2)) cos theta`
`B = (mu_(0))/(4pi) (I)/(r^(2)) cos theta Sigma dl " "rarr (3)`
From the triangle POQ, `cos theta = (R)/(r)` and `Sigma dl = 2pi R` circumference of the loop
Equation (3) becomes, `B = (mu_(0))/(4pi)(I)/(r^(2))(R)/(r) 2pi R`
`B = (mu_(0) I R^(2))/(2 r^(3)) " "rarr (4)`
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))) " "rarr (5)`