Question

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

Answer 1

The magnetic field on the axis of the magnet at a distance d₁ = 14 cm, can be written as:

`"B"_1 = (μ_0 2"M")/(4pi ("d"_1)^3) = "H"` .......(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:

`"B"_2 = (μ_0 "M")/(4pi ("d"_2)^3) = "H"` .............(2)

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

`2/("d"_1)^3 = 1/("d"_2)^3`

`("d"_2/"d"_1)^3 = 1/2`

∴ `"d"_2 = "d"_1 xx (1/2)^(1/3)`

= 14 × 0.794

= 11.1 cm

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