A short bar magnet has a magnetic moment of $0.48 \mathrm{~J} \mathrm{~T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10 \mathrm{~cm}$ 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, $M=0.48 \mathrm{~J} \mathrm{~T}^{-1}$
(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:
$B=\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}$
Where,
$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$
$\therefore B=\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}}$
$=0.96 \times 10^{-4} \mathrm{~T}=0.96 \mathrm{G}$
The magnetic field is along the S − N direction.
(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:
$B=\frac{\mu_{0} \times M}{4 \pi \times d^{3}}$
$=\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi(0.1)^{3}}$
$=0.48 \mathrm{G}$
The magnetic field is along the N − S direction.