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 magnetat a distance of 10 cm from the centre of the magnet on (a) the axis,(b) the equatorial lines (normal bisector) of the magnet.

Answer 1

Given,

Magnetic moment of the bar magnet, M=0.48  JT -1

Since magnetic field is produced at a distance of 10 cm from the centre. Thus, d=10 cm.

(i)

The magnetic field at its axis is,

B a = μ o 4π 2M d 3

Where μ 0 is the permeability of the free space and is equal to 4π× 10 −7   TmA -1

Thus,

B a = 4π× 10 −7 4π 2×0.48 ( 0.1 ) 3 =0.96× 10 −4  T =0.96 G

Thus, magnetic field at the axis is 0.96× 10 −4  T along S−N direction.

(ii)

Magnetic field along the equatorial line is given by the formula,

B e = μ o 4π M d 3

Thus

B e = 4π× 10 −7 4π 0.48 ( 0.1 ) 3 =0.48× 10 −4  T =0.48 G

Thus, the magnetic field at a distance 10 cmon the equatorial line is 0.48× 10 −4  T along N−Sdirection.