A telephonic cable at a place has four long straight horizontal wires carrying a current of 1.0 amp. in the same direction east to west. The earth's magnetic field at the place is 0.39 G and the angle of dip is 35°. The magnetic declination is almost zero. What are the resultant magnetic fields at points 4.0 cm below and above the cable?

Given,
Earth’s magnetic field, B_e = 0.39 G
Angle of dip, δ = 35°
Number of wires, n = 4
declination = 0
distance , r = 4 cm = 4 $\times {10}^{-2} \mathrm{m}$ 
Current carried by the cable, I = 1 A

∴ Horizontal component of earth’s magnetic field is given by,

B_H = B_e cos δ   
B_H = 0.39 cos 35°
     = 0.3195 G 

Vertical component of earth's magnetic field, B_v = B_e sin δ
        B_v = 0.39 sin 35°
             = 0.224 G 

Magnetic field produced by telephone cable having 4 wires is, 

 $\mathrm{B} = \left(\frac{{\mathrm{\mu }}_0\mathrm{I}. 2}{4\mathrm{\pi r}}\right)\times 4$
$$\begin{gathered} \mathrm{B}\text{'} = {10}^{-7}\times \frac{2\times 1.0\times 4}{4\times {10}^{-2}} \\ = 0.2 \times {10}^{-4}\mathrm{T} \\ = 0.2 \mathrm{G} \end{gathered}$$  

Now,
We will find the resultant field below the cable.
As per the right hand thumb rule, the direction of B’ will be opposite to B_H at a point below the cable.
Therefore at a point 4 cm below the cable, resultant horizontal component of earth’s field
R_H = B_H – B’
     = 0.3195 – 0.2
     = 0.1195 G 

Resultant vertical component of earth’s field
R_v = B_v = 0.224 G (unchanged) 

∴ Resultant of earth’s field 

$$\begin{gathered} \mathrm{R} = \sqrt{{\mathrm{R}}_{\mathrm{H}}^2+{\mathrm{R}}_{\mathrm{V}}^2} \\ = \sqrt{(0.1195{)}^2+(0.224{)}^2} \\ = \sqrt{0.0143+0.0500} \\ = \sqrt{0.0643} \\ = 0.254 \mathrm{G} \end{gathered}$$ 

Now,
Resultant field above the cable is given as per right hand thumb rule which says that, at a point above the cable, B’ will be in the same direction as B_H.
Hence, at a point 4 cm above the cable,
                       $$\begin{gathered} {\mathrm{R}}_{\mathrm{H}} = {\mathrm{B}}_{\mathrm{H}}+\mathrm{B}\text{'} \\ = 0.3195+0.2 \\ = 0.5195 \mathrm{G} \\ {\mathrm{R}}_{\mathrm{V}} = {\mathrm{B}}_{\mathrm{V}} \\ = 0.224 \mathrm{G} \end{gathered}$$ 

$\therefore$Resultant magnetic field is, 

   $\mathrm{R}=\sqrt{{\mathrm{R}}_{\mathrm{H}}^2+{\mathrm{R}}_{\mathrm{V}}^2}$
     $= \sqrt{(0.5195{)}^2+(0.224{)}^2}$ 
  $\mathrm{R} = 0.566 \mathrm{G}.$