Question

Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)

Answer 1

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 q_m and separated by a distance of 21. 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 experienced 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

  Where \(\vec F_S\) = \(\frac{ μ_0 }{4π} \) \(\frac{q_m}{r'^2}\). The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is

From equation (1) and equation (2), the net force at point C is \(\vec F\) = \(\vec F_N\) + F_{S .} This net force is equal to the magnetic field at the point C.

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,

Therefore , using equation (7) in equation (6) ,we get

\(\vec B_{equatorial} = - \frac{μ_0 p_m}{4π r^3} \hat i\)

Since in genaral , the magnetic filed at equatorial points is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatoraland the direction of B_axial and B_equatoral are opposite.