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 (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

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

Ans (a).

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 = μ0 /4π x (2M/d³)

Where, μ0  = Permeability of free space =4π x 10^-7 T m A^-1

Therefore ,B =  (4π x 10^-7  x 2 x 0.48)/[4π x (0.1)³]

The magnetic field is along the S – N direction.

Ans (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 = μ0 /4π x (M/d³)

=> B =   (4π x 10^-7  x 0.48)/[4π x (0.1)³]

= 0.48 G

The magnetic field is along the N – S direction.