A current of 5A flows downwards in a long straight vertical conductor and the earth’s horizontal flux density is 2×10^–7T then the neutral point occurs

$$\begin{gathered} \mathrm{As} \mathrm{the} \mathrm{current} \mathrm{is} \mathrm{flowing} \mathrm{downwards} \mathrm{the} \mathrm{magentic} \mathrm{field} \\ \mathrm{due} \mathrm{to} \mathrm{the} \mathrm{wire} \mathrm{is} \mathrm{towards} \mathrm{south} \mathrm{on} \mathrm{the} \mathrm{east} \mathrm{towards} \mathrm{west} \mathrm{on} \mathrm{the} \\ \mathrm{south} \mathrm{and} \mathrm{so} \mathrm{on}. \mathrm{So} \mathrm{null} \mathrm{point} \mathrm{will} \mathrm{be} \mathrm{on} \mathrm{the} \mathrm{east} \mathrm{of} \mathrm{the} \mathrm{conductor} \\ \mathrm{as} \mathrm{the} \mathrm{earth} \mathrm{magnetic} \mathrm{field} \mathrm{is} \mathrm{directed} \mathrm{form} \mathrm{south} \mathrm{to} \mathrm{north}. \\ \mathrm{At} \mathrm{null} \mathrm{point} \mathrm{B}={\mathrm{B}}_{\mathrm{H}} \\ \frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}={\mathrm{B}}_{\mathrm{H}} \frac{2\times {10}^{-7}\times 5}{\mathrm{r}}=2\times {10}^{-7} \\ \mathrm{r}=5 \mathrm{m} \end{gathered}$$