A short bar magnet has a magnetic moment of `0*48JT^-1`. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of `10cm` from the centre of the magnet on (i) the axis (ii) the equatorial line (normal bisector) of the magnet.

Explanation

Given, magnetic moment `M = 0.48 J T^(-1)`
(i ) On an axial line of a short bar magnet, the magnetic field at a point distant d from the centre of magnet is
`B = (mu_0)/(4pi) (2M)/(d^3)`
here , d = 10 cm = 0.1 m
` rArr B = 10^(-7) xx (2 xx 0.48)/(10^(-3))`
`B = 0.96 xx 10^(-4) T = 0.96G` along S-N direction parallel to the magnetic moment.
(ii) On an equatorial line of a short bar magnet, the magnetic field at a point distant d from the centre of magnet is
`B = (mu_0)/(4pi) (M)/(d^3)`
here , d = 0.1m
` rArr B = 10^(-7) xx (0.48)/(10^(-3) ) `
B = 0.48 G, along N-S direction antiparallel to the magnetic moment.