What is the magnetic field along the axis and equatorial line of a bar magnet ?

Explanation

Magnetic field at a point along the axial line of the magnetic dipole (bar magnet) :
Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength `q_(m)` and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet at a distance from the geometrical center O of the bar magnet can be compute by keeping unit north pole (`q_(MC)` = 1 A m ) at C. the force e`xx ` perienced by the unit north pole at C due to pole strength can be compute using Coulomb.s law of magnetism as follows :
The force of repulsion between north pole of the bar magnet and unit north pole at point C (in free space ) is
`vec(F_(N)) = (mu_(0))/(4 pi ) (q_(m))/((r-l)^(2)) hat(l)`
where r + l is the distance between north pole of the bar magnet and unit north pole at C
The force of attraction between south Pole of the bar magnet and unit North Pole at point C (in free space ) is
`vec(F_(S)) = (mu_(0))/(4 pi ) (q_(m))/((r-l)^(2)) hat(l)`
where r + l is the distance between south pole of the bar magnet and unit north pole at C.
Magnetic field at a point along the axial line due to magnetic dipole
From equation (1) and (2) , then net force at point C is `vec(F ) = vec(F_(N)) + vec(F_(S))`. From definition, this net force is the magnetic field due to magnetic dipole at a point C `(vec(F) = vec(B))`
`vec(B) = (mu_(0))/(4 pi (r - l )^(2)) hat(i) + ((mu_(0))/(4pi)(q_(m))/((4 + l)^(2)) hat(i) ) = (mu_(0)q_(m))/(4pi) ((1)/((r - l)^(2)) - (1)/((r + l)^(2)) ) hat( i) `
`vec(B ) = (mu_(0)2r)/(4 pi ) ( (q_(m). (2l))/((r^(2) - l^(2))^(2)) )hat(i)`
Since, magnitude of magnetic dipole moment is `|vec(P_(m))| = p_(m) = q_(m).2l ` then magnetic field point C equation (3) can be written as
`vec(B_("axial" )) = (mu_(0))/(4 pi ) ( (2rp_(m))/((r^(2) - l^(2))^(2)) )hat(i)`
if the distance between two poles in a bar magnet are small (looks like short magnet ) compared to the distnace between geometical centre O of bar magnet and the locatio of point C i.e.,
`r gt gt `l then, `(r^(2) - l^(2))^(2) approx r^(4)`
Therefore, using equation (5) in equation (4) , we get
`vec(B_("axial") ) = (mu_(0))/(4 pi ) ( (2rp_(m))/((r^(3))) )hat(i) = (mu_(0))/(4 pi ) (2)/(r^(3)) vec(p_(m))`
Where `vec(P_(m)) = p_(m) hat(i)`.