A long vertical wire in which a current is flowing produces a neutral point with the earth's magnetic field at a distance of 5 cm from the wire. If the horizontal component of the earth's magnetic induction is 0.18 gauss, then the current in the wire is

The magnetic induction B due to current I ampere in a long straight conductor at a distance r metre from it is given by

$\begin{array}{l}\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2\mathrm{I}}{\mathrm{r}}\\ \mathrm{Here},\mathrm{r}=5\mathrm{cm}=5\times {10}^{-2}\mathrm{m}\\ \therefore \mathrm{B}={10}^{-7}\times \frac{2\times \mathrm{I}}{5\times {10}^{-2}}\text{= 0 .04 I Gauss}\\ \end{array}$

At the neutral point, the magnetic induction due to earth's horizontal component is equal and opposite to that due to current in the conductor.

$\therefore 4\mathrm{I}\times {10}^{-2}=0.18\mathrm{or}\mathrm{I}=4.5\mathrm{A}$