End-on position
Consider a bar magnet of length 2l and pole strength m. Suppose a point P on the axis of the magnet at a distance d from its center. (d –l) is the distance of P form the N-pole of the magnet. The magnetic field intensity at P due to the north-pole of the magnet is
which is directly away from N-pole. Since the south of the magnet is at a distance r = d + l from P, so magnetic field intensity at P due to *-pole is
which is direct towards, the *-pole of the magnet. The magnetic field intensity B at P is the resultant of these two fields,
where M = m x 2l, the magnetic moment of the bar magnet. So, the magnetic field at a point on the axis of a bar magnet is
If the length of the magnet is very small, d>>I and the magnetic field intensity is
Broadside of position
Suppose a point P is on the equatorial line of the bar magnet. The equatorial line of the magnet is the line perpendicular to the axis of the magnet which bisects the magnet. Let d be the distance of the point P from the centre of the magnet and P due to the North Pole is
directed away from N-pole. the magnetic field B2 at P due to *-pole is
directed towards *-pole. These fields have different directions, but the same magnitude as shown in the figure. Let ∠PSO = θ and by symmetry, ∠PNO = θ. The angle between B1 and B2 is then 2θ. The resultant magnetic field, B at P is given by
The direction of B is parallel to the axis of the magnet, from north to south pole. If the magnet is very short, d>>l, and the magnetic field at P is
