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

Magnetic moment of the bar magnet , `M=0.48J T^(-1)`
(a) Distance, d=10cm=0.1m
The magnetic field at distance d. from the centre of the magnet on the axis is given by the relation:
`B=(mu_(0))/(4pi)(2M)/(d^(3))`
where,
`mu_(0)=` permeability of free space`=4pixx10^(-7) TmA^(-1)`
`thereforeB=(4pixx10^(-7)xx2xx0.48)/(4pixx(0.1)^(3))`
`=0.96xx10^(-4) T=0.96G`
The magnetic field is along the S-N direction
(b) The magnetic field at a distance of 10 cm(i.e., d=0.1m) on the equatorial line of the magnet is given as:
`B=(mu_(0)xxM)/(4pixxd^(3))`
`=(4pixx10^(-7)xx0.48)/(4pi(0.1)^(3))`
=0.48G
The magnetic field is along the N-S direction.