Answer –
The magnetic field at a distance d1 = 14 cm on the axis of the magnet can be written as –
${{B}_{1}}=\frac{{{\mu }_{0}}2M}{4\pi {{d}_{1}}^{3}}=H$
Where, M is the magnetic moment and
\[{{\mu }_{0}}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]
H is the Horizontal magnetic field component at d1
The neutral point will be on the equatorial line if the bar magnet is turned through 180 degrees.
As a result, the magnetic field at a distance d2 on the magnet’s equatorial line can be represented as
\[\]${{B}_{2}}=\frac{{{\mu }_{0}}2M}{4\pi {{d}_{2}}^{3}}=H$
Equating the magnetic fields B1 and B2, we get –
$\frac{2}{d_{1}^{3}}=\frac{1}{d_{2}^{3}}$
$\frac{d_{2}^{3}}{d_{1}^{3}}=\frac{1}{2}$
$\therefore {{d}_{2}}={{d}_{1}}\times {{\left( \frac{1}{2} \right)}^{\frac{1}{3}}}=14\times 0.794$
${{d}_{2}}=11.1cm$
So,the new null points will be located 11.1 cm on the normal bisector.