A long straight vertical wire carries a current of 10 ampere flowing upwards through it at a place where the horizontal component of the earth’s magnetic induction is 0.3 gauss. Then the total magnetic induction at a point 5 cm due north of the wire is

 $\mathrm{magnetic} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{the} \mathrm{straight} \mathrm{conductor}$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}}=\frac{2\times {10}^{-7}\times 10}{5\times {10}^{-2}}=4\times {10}^{-5}=0.4 \mathrm{gauss}$

$\mathrm{its} \mathrm{direction} \mathrm{will} \mathrm{be} \mathrm{towards} \mathrm{west}$ on the north side of wire.

$$\begin{gathered} \mathrm{earth} \mathrm{magnetic} \mathrm{field} \mathrm{at} \mathrm{any} \mathrm{point} is \mathrm{towards} \mathrm{north} \mathrm{hence} \mathrm{they} \mathrm{are} \mathrm{perpendicular} \\ \mathrm{net} \mathrm{magnetic} \mathrm{field} \sqrt{0.3^2+0.4^2}= 0.5 \mathrm{G} \end{gathered}$$