Answer :
Magnetic moment of the bar magnet is given by M = 5.25 × 10–2 J T–1
Magnitude of earth’s magnetic field at a place is given by H = 0.42 G = 0.42×10-4 T
- The magnetic field on the ordinary bisector at a distance R from the centre of the magnet is given by –
$B=\frac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}$
Where \[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]
It is known that B = H when the resultant field is inclined at 45° with earth’s field. So, we have –
$\therefore \frac{{{\mu }_{0}}M}{4\pi {{R}^{3}}}=H=0.42\times {{10}^{-4}}$
\[{{R}^{3}}=\frac{{{\mu }_{0}}M}{0.42\times {{10}^{-4}}\times 4\pi }=\frac{4\pi \times {{10}^{-7}}\times 5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}\times 4\pi }\]
\[{{R}^{3}}=12.5\times {{10}^{-5}}\]
\[\therefore R=5cm\]
- The magnetic field at a distance ‘R’ from the centre of the magnet on its axis can be given as follows –
${B}’=\frac{{{\mu }_{0}}2M}{4\pi {{d}_{1}}^{3}}=H$
Since, the resultant field is inclined at 45° with earth’s field.
Where, M is the magnetic moment and
\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]
$\therefore {{R}^{3}}=\frac{{{\mu }_{0}}2M}{4\pi \times H}$
\[{{R}^{3}}=\frac{4\pi \times {{10}^{-7}}\times 2\times 5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}\times 4\pi }=25\times {{10}^{-5}}\]
\[\therefore R=6.3cm\]