Question

A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on

1. the axis,
2. the equatorial lines (normal bisector) of the magnet.

Answer 1

Given: Magnetic moment of the bar magnet, M = 0.48 J T⁻¹

Distance, d = 10 cm = 0.1 m

(a) The magnetic field at a distance d from the centre of the magnet on the axis is given by the relation:

B = `μ_0/(4pi) (2M)/d^3`

Where,

μ₀ = Permeability of free space = 4π × 10⁻⁷ T mA⁻¹

∴ B = `(4pi xx 10^-7 xx 2 xx 0.48)/(4pi xx (0.1)^3)`

= 0.96 × 10⁻⁴ T

= 0.96 G

The magnetic field is along the S-N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

B = `(μ_0 xx M)/(4pi xx d^3)`

= `(4pi xx 10^-7 xx 0.48)/(4pi (0.1)^3)`

= 0.48 G

The magnetic field is along the N-S direction.