A telephone cable at a place has four long straight horizontal wires carrying a current of $1.0 \mathrm{A}$ in the same direction east to west. The earth's magnetic field at the place is $0.39\mathrm{G}$, and the angle of dip is $35°$. The magnetic declination is nearly zero. What are the resultant magnetic fields at point $4.0 \mathrm{cm}$ below the cable?

 Let us first decide the directions which can best represent the situation.

Here, $B_H=B\cos \delta =0.39\times \cos 35°\mathrm{G}$

$B_H=0.32\mathrm{G}$

and ${\mathrm{B}}_{\mathrm{V}}=\mathrm{Bsin\delta }=0.39\times \sin 35° \mathrm{G}$

${\mathrm{B}}_{\mathrm{V}}=0.22 \mathrm{G}$

Telephone cable carry a total current of $4.0 \mathrm{A}$ in direction east to west. We want resultant magnetic field $4.0 \mathrm{cm}$ below.

$$\begin{gathered} {\mathrm{B}}_{\text{wire }}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{I}}{\mathrm{r}}={10}^{-7}\times \frac{2\times 4}{4\times {10}^{-2}} \\ =2\times {10}^{-5} \mathrm{T}=0.2 \mathrm{G} \end{gathered}$$

Net magnetic field

$$\begin{gathered} {\mathrm{B}}_{\text{net }}=\sqrt{{\left({\mathrm{B}}_{\mathrm{H}}-{\mathrm{B}}_{\text{wire }}\right)}^2+{\mathrm{B}}_{\mathrm{V}}^2} \\ {\mathrm{B}}_{\text{net }}=\sqrt{(0.12{)}^2+(0.22{)}^2}=\sqrt{0.0144+0.0484}=0.25 \mathrm{G} \end{gathered}$$