A bar magnet of length 0.1m has a pole strength of 50Am. The magnetic field at a distance of 0.2m from its centre on its equatorial line 

Magnetic field induction at a point on its equatorial line is
                                            $$\begin{gathered} B=\frac{{\mu }_0}{4\pi }\frac{M}{{\left(d^2+l^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow B=\frac{{\mu }_0}{4\pi }\frac{m\left(2l\right)}{{\left(d^2+l^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow B=\frac{{10}^{-7}}{1}\times \frac{\left(50\right)\left(0.1\right)}{{\left[{\left(0.2\right)}^2+{\left(0.05\right)}^2\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow B=5.7\times {10}^{-5}T \end{gathered}$$