A magnetic dipole of magnetic moment $0.72 \sqrt{2} A-m^2$ is placed horizontally with the north pole pointing towards east. Find the distance of the neutral point  from the centre of the dipole  if the horizontal component of the earth's magnetic field is $18 \mu T$.

Given  $M = 0.72\sqrt{2} A-m$ 

From the figure  

                           $\alpha +\theta = 90°$

                           $\alpha = 90°-\theta$

We have,           $\tan \alpha = \frac{\tan \theta }{2}$

       or        $\tan (90°-0) = \frac{\tan \theta }{2}$

       or                 $cot\theta = \frac{\tan \theta }{2}$

       or                  ${\tan }^2\theta = 2$

                                   $\theta = {\tan }^{-1}\left(\sqrt{2}\right)$

Let neutral point is at a distance$r$, then

                                   $B_H = B$

                                          $= \frac{{\mu }_0}{4\pi }\frac{M\sqrt{3{\cos }^2\theta +1}}{r^3}$

         or          $18\times {10}^{-0.6} = \frac{{10}^{-7}\times 0.72\sqrt{2}\sqrt{3{\displaystyle \frac{1}{\sqrt{3}}}+1}}{r^3}$

On solving    

                                 $r = 0.20m$