write equation the magnetic field due to current carrying loop as magnetic dipole on its axial point.

Explanation

1. The magnetic field on the axis of a circular loop of a radius R, carrying a steady current I, at x distance from centre is given by
`B=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3/2))" "...(1)`
2. And its direction is along the axis and the right hand thumb rule. Here, x is the distance along the axis from the centre of the loop.
3. For `xgtgtR`, we may neglect `R^(2)` term in the denominator.
Thus, `B=(mu_(0)IR^(2))/(2x^(2))" "...(2)`
4. Multiply right side of equation (2) with `pi` in denominator and numerator,
`thereforeB=(mu_(0)I(piR^(2)))/(2pix^(3))`
Area of coil A = `piR^(2)`
`thereforeB=(mu_(0)IA)/(2pix^(3))`
Here, IA = magnetic dipole moment = m
`B=(mu_(0)m)/(2pix^(3))`
`thereforeB=(2mu_(0)m)/(4pix^(3))` and In vector form,
`thereforevecB=mu_(0)/(4pi)((2vecm)/x^(3))`
which is formula for magnetic field on point of axis of coil.