5.15 A short bar magnet of magnetic movement 5.25 × 10–2 J T–1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

The magnetic moment of the bar magnet,
The magnitude of earth's magnetic field at a place,
a) The magnetic field at a distance R from the center of the magnet on the normal bisector is given by the relation:
$\mathrm{B}=\frac{{\mathrm{\mu }}_{0}\mathrm{M}}{4{\mathrm{\pi R}}^{3}}$
Where, o =Permeability of free space
When the resultant field is inclined at 45$°$ with earth's field, B = H
$\begin{array}{l}\therefore \frac{{\mu }_{0}M}{4\pi {R}^{3}}=H=0.42×{10}^{-4}\\ {R}^{3}=\frac{{\mu }_{0}M}{0.42×{10}^{-4}×4\pi }\\ =\frac{4\pi ×{10}^{-7}×5.25×{10}^{-2}}{4\pi ×0.42×{10}^{-4}}=12.5×{10}^{-5}\\ \therefore R=0.05\mathrm{m}=5\mathrm{cm}\end{array}$
(b) The magnetic field at a distance ${\mathrm{R}}^{1}$ from the center of the magnet on its axis is given as
${\mathrm{B}}^{\text{'}}=\frac{{\mathrm{\mu }}_{0}2\mathrm{M}}{4{\mathrm{\pi R}}^{3}}$The resultant field is inclined at 45° with the earth's field.
$\therefore \mathrm{B}\text{'}=\mathrm{H}$
$\begin{array}{l}\frac{{\mu }_{0}2M}{4\pi {\left({R}^{\mathrm{\prime }}\right)}^{3}}=H\\ {\left({R}^{\mathrm{\prime }}\right)}^{3}=\frac{{\mu }_{0}2M}{4\pi ×H}\\ =\frac{4\pi ×{10}^{-7}×2×5.25×{10}^{-2}}{4\pi ×0.42×{10}^{-4}}=25×{10}^{-5}\\ \therefore {R}^{\mathrm{\prime }}=0.063\mathrm{m}=6.3\mathrm{cm}\end{array}$