Magnetic moment of the bar magnet is given by M = 0.48 J T^–1

• Distance, d = 10 cm = 0.1 m

The magnetic field at distance d from the magnet’s centre on the axis can be calculated using the following formula –

$B=\frac{{{\mu }_{0}}2M}{4\pi {{d}^{3}}}$

Where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 0.48}{4\pi \times {{(0.1)}^{3}}}$

$B=0.96G$

The magnetic field is along the S-N direction.

(b)                                                                         

The magnetic field at a distance 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given by the following relation –

$B=\frac{{{\mu }_{0}}M}{4\pi {{d}^{3}}}$

Where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

$B=\frac{4\pi \times {{10}^{-7}}\times 0.48}{4\pi \times {{(0.1)}^{3}}}$

$B=0.48G$

The magnetic field is along the N – S direction.