Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength qm 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 (qm C = 1 A m) at C.
The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follow’s:
The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is
Magnetic field at a point along the equatorial line due to a magnetic dipole
The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is
From equation (1) and equation (2), the net force at point C is \(\vec F\) = FN + FS This net force is equal to the magnetic field at the point C.
\(\vec B\) = -(FN + FS ) cos θ \(\hat i\)
Since, FN = FS
In a right angle triangle NOC as shown in the Figure 1
Substituting equation 4 in equation 3 We get
Since, magnitude of magnetic dipole moment is \(|\vec P_m|\) = Pm = qm.2l and substituting in equation (5), the magnetic field at a point C is
If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r>> l, then,
\((r^2 + L^2)^{\frac{3}{2}}\)
Therefore, using equation (7) in equation (6), we get ≈ r3 ……… (7)
Since Pm \(\hat i\) = \(|\vec P_m|\)m, in general, the magnetic field at equatorial point is given by
Note that magnitude of Baxial is twice that of magnitude of Bequatorial and the direction of are opposite.
