Question

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

Answer 1

For a point $4\text{cm}$ below the cable

Step 1: Find magnetic field due to straight carrying conductors.

Number of horizontal wires in the telephone cable, $n=4$

Current in each wire, $I=1.0A$

Earth’s magnetic field at a location, $B=0.39G=0.39\times {10}^{4}T$
Angle of dip at the location, $\delta ={35}^{\circ }$

Angle of declination, $\theta \sim 0^{\circ }$

Distance, $r=4cm=0.04m$

Let the magnetic field produced due to four straight carrying conductors be, $B_1$, then,

$B_1=4\times \frac{{\mu }_{0}I}{2\pi r}$

Where, ${\mu }_0$ = permeability of free space $4\pi \times {10}^{-7}Tm/A$
Substituting the values, we get

$B_1=4\times \frac{4\pi \times {10}^{-7}\times 1}{2\pi \times 0.04}$

$B_1=0.2\times {10}^{-4}T=0.2G$

Step 2: Find resultant of horizontal magnetic field.

Now the horizontal component of resultant magnetic field can be written as,

$B_H=B\cos \delta -B_1$

Putting the values,
$B_H=0.39\cos {35}^{\circ }-0.2$

$=0.39\times 0.819-0.2\approx 0.12G$
The vertical component of the resultant field is the same as the vertical component of earth’s magnetic field i.e.

$B_V=B\sin \delta$

$=0.39\times \sin {35}^{\circ }=0.22G$

Step 3: Find the angle made by the field.

The angle made by the field with its horizontal component is given by the relation,

$\theta ={\tan }^{-1}\left(\frac{B_V}{B_H}\right)={\tan }^{-1}1.83={61.39}^{\circ }$

The resultant field at this point is given by,
$B_{net}=\sqrt{(B_H{)}^2+(B_V{)}^2}$

$B_{net}=\sqrt{(0.12{)}^2+(0.22{)}^2}$

$B_{net}=0.25G$