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

Explanation

Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Consider a ber magnet. NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength `q_(m)` and separated by a distance of 2l. The magnetic field at a point C (lies along the equatorial line ) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole `(q_(m) C = 1 A m )` at C. The force e`xx`perienced by the unit north pole at C due to pole strength N-S can be computed 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)) = - F_(N) cos theta hat(i) + F_(N) sin theta hat(j) `
Where `F_(N) = (mu_(0))/(4pi ) (q_(m))/(r.^(2))` The force of attraction ( in free space ) between south pole of the bar magnet and unit north pole at point C is
`vec(F_(S)) = - F_(S) cos theta hat(i) - F_(S) sin theta hat(j)`
where , `vec(F_(S)) = (mu_(0))/(q_(m))(q_(m))/(r.^(2))`
From equation (1) and equation (2) , the net force at poin C is `vec(F) = vec(F_(N)) + F_(S)`. This net force is equal to the magnetic field at the point C .
`vec(B) - (F_(N) + F_(S)) cos theta hat(i)`
since, `F_(N) = F_(S)`
`vec(B) = - (2mu_(0))/(4pi) (q_(m))/(r.^(2)) cos theta hat(i) = (2 mu_(0))/(4pi ) ( q_(m))/((r^(2) + l^(2))) cos theta hati`
In a right angle triangle NOC as shown in the Figure l
cos `theta= ("adjacent")/("hypotenuse") = (l)/(r.) = (l)/((r^(2) + l^(2))^((1)/(2)))`
Substituting equation 4 in equation 3 we get
`vec(B) = - (mu_(0))/(4pi) (q_(m) xx (2l))/((r^(2) + l^(2))^((3)/(2)) )`
Since, magnitude of magnetic dipole moment is `|vec(P_(m))| = P_(m) = q_(m). 2l` and substituting in equation (5 ). The magnetic field at a point C is
`vec(B)_("equatorial") = - (mu_(0))/(4pi) (P_(m))/((r^(2) + l^(2))^((3)/(2)) )`
If the distance between two poles in a bar magnet are small (lools like short magnet ) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r `gt gt ` l, then ,
`(r^(2) + l^(2))^((3)/(2)) approx r^(3)`
Therefore, using equation (7) in equation (6), we get
`vec(B)_("equatorial") = - (mu_(0))/(4 pi ) (P_(m))/(r^(3)) hat(i)`
Since, `P_(m) hat(i) = vec(P_(m)) . ` in general, the magnetic field at equatorial point is given by
`vec(B_("equatorial")) = - (mu_(0))/(4 pi ) (P_(m))/(r^(3))`
Note that magnitude of `vec(B_("axial")) `is twice that of magnitude of `B_("equatorial")` and the direction of `B_("axial ")and B_("equantorial") ` are opposite .