A short bar magnet of magnetic moment $1.28\times {10}^{-2}{\mathrm{Am}}^2$  is arranged in the magnetic meridian  with its south pole pointing geographic north. If $B_H=0.4$ G , the distance between the two null points  is 

When a north pole of  a bar magnet is facing south , null points are obtained on axial line
     $Given : B_a=B_H,{\text{ B}}_H=0.4G=4\times {10}^{-5}T , M=1.28\times {10}^{-2} A{m}^2$

$We know, at axis , B_H=\frac{{\mu }_0}{4\pi }\frac{2M}{d^3}$

$$\begin{gathered} \Rightarrow 4\times {10}^{-5}=\frac{{10}^{-7}\times 2\times 1.28\times {10}^{-2}}{d^3} \\ \Rightarrow d^3=\frac{2\times 1.28}{4}\times {10}^{-4} \end{gathered}$$

$$\begin{gathered} \Rightarrow d^3=64\times {10}^{-6} \\ \Rightarrow d=4\times {10}^{-2}m=4cm \end{gathered}$$
The distance between two null points 
         $2d=8cm$