A horizontal overhead power line is at a height of 4m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is $\left({\mathrm{\mu }}_0=4\mathrm{\pi }\times {10}^{-7}{\mathrm{TmA}}^{-1}\right)$

$$\begin{gathered} \mathrm{Magnetic} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{a} \mathrm{straight} \mathrm{conductor} \mathrm{according} \mathrm{to} \mathrm{Ampere}\text{'}\mathrm{s} \\ \mathrm{circuital} \mathrm{law} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}} \mathrm{and} \mathrm{direction} \mathrm{according} \mathrm{to} \mathrm{right} \mathrm{hand} \mathrm{grip} \mathrm{rule}. \\ \mathrm{A}s \mathrm{the} \mathrm{current} \mathrm{is} \mathrm{going} \mathrm{from} \mathrm{east} \mathrm{to} \mathrm{west} \mathrm{direction} \mathrm{of} \mathrm{magnetic} \mathrm{field} \\ \mathrm{below} \mathrm{the} \mathrm{conductor} \mathrm{is} \mathrm{along} \mathrm{south} \mathrm{and} \\ \mathrm{B}=\frac{2\times {10}^{-7}\times 100}{4}=5\times {10}^{-6}\mathrm{T} \end{gathered}$$