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 2M)/(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 rotated by 180°, 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 xx (0.5)^(1/3)`

= 14 × 0.794

= 11.1 cm

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