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

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

1. Distance, d = 10 cm = 0.1 m
   The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation:
   B = `μ_0/(4pi) (2"M")/"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.
2. 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.