A horizontal overhead power line is at a height of 4 m from the ground and carries a current of 100 A from east to west. The magnetic field directly below it on the ground is (take \( \mu_0=4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5\times10^{-6}\,\mathrm{T} \), southward
The power line acts as a long straight wire. Its field at a distance \( r \) is \( B=\mu_0 I/2\pi r \). The direction of the field is that of \( d\vec{l}\times\vec{r} \), where \( \vec{r} \) runs from the wire to the point (the right-hand rule).
The point on the ground is \( r=4\,\mathrm{m} \) below the wire:
\( \begin{aligned} B &= \frac{\mu_0 I}{2\pi r} \\ &= \frac{4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\times100\,\mathrm{A}}{2\pi\times4\,\mathrm{m}} \\ &= 5\times10^{-6}\,\mathrm{T} \end{aligned} \)
For the direction, take east as \( +\hat{\imath} \), north as \( +\hat{\jmath} \) and vertically up as \( +\hat{k} \). The current flows west, along \( -\hat{\imath} \), and the point lies below the wire, along \( -\hat{k} \):
\( \begin{aligned} d\vec{l}\times\vec{r} &\propto (-\hat{\imath})\times(-\hat{k}) \\ &= \hat{\imath}\times\hat{k} \\ &= -\hat{\jmath} \end{aligned} \)
Therefore, the field on the ground directly below the line points along \( -\hat{\jmath} \), that is, southward.