5.14 If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?

The magnetic field on the axis of the magnet at a distance
d1=14 cm, can be written as:
B1=μ02M4πd13=H                ...(i)
Where M = Magnetic moment
μ0 = Permeability of free space, H= Horizontal component
of the magnetic field at d1
If the bar magnet is turned through 180°,
then the neutral point will lie on the equilateral line.
Hence, the magnetic field at a distance d2, on the equatorial l in of the magnet can be written as
B2=μ0M4πd23=H                   ...ii
Equating equations i and ii, we get:
d2d13=12d2=d1×1213d2=14×0.794=11.1cmThe new null points will be located 11.1 cm on the normal bisector.