Physics - Electrostatics Question with Solution | TestHub

Here, length ofmagnet $=10\mathrm{cm}=10\times 10^{-2}\mathrm{m}$, $r=15\times 10^{-2}\mathrm{m}$
Since, at the neutral point, magnetic field due to the magnet is equal to $B_H$, $$\begin{array}{l}{B}_H=\frac{{\mu }_0}{4\pi }\cdot \frac{M}{{\left(O{P}^2+A{O}^2\right)}^{3/2}}\\ 0.4\times 10^{-4}=10^{-7}\times \frac{M}{{\left(200\times 10^{-4}+25\times 10^{-4}\right)}^{3/2}}\\ \Rightarrow \frac{0.4\times 10^{-4}}{10^{-7}}\times {\left(225\times 10^{-4}\right)}^{3/2}=M\\ \Rightarrow 0.4\times 10^3\times 10^{-6}(225{)}^{3/2}=M\\ \Rightarrow M=1.35\mathrm{A}-\mathrm{m}\end{array}$$