If the bar magnet in exercise 5.13 is turned around by 180º, 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 = μ0 /4π x [2M/(d₁)³] =H   ———–Eq 1

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 equatorial line.

Hence, the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:

B₂ = μ0 /4π x [2M/(d₂)³] =H   ———–Eq 2

Equating equations (1) and (2), we get:

2/(d₁)³= 1/(d₂)³

=> (d₂/d₁)³ = 1/2

=> d₂  =  d₁ x (1/2)^1/3

=> d₂  = 14 x 0.794 = 11.1 cm

The new null points will be located 11.1 cm on the normal bisector.