Question

A long telephone cable at a place has four long straight horizontal wires carrying a current of $1.0\text{A}$ in the same direction east to west. The earth's magnetic field at the place is $0.39\text{G}$, and the angle of dip is ${35}^{\circ }$. The magnetic declination is nearly zero, What are the resultant magnetic fields at points $4.0\text{cm}$ above and below the cable?
$(\cos {35}^{\circ }=0.82\text{and}\sin {35}^{\circ }=0.57)$

• A. $0.56\text{G},0.25\text{G}$
• B. $0.25\text{G},0.56\text{G}$
• C. $0.56\text{G},0.56\text{G}$
• D. $0.25\text{G},0.25\text{G}$

Answer 1

The correct option is A $0.56\text{G},0.25\text{G}$

Given:
No. of. wires, $n=4$
Current, $I=1.0\text{A}$
Earth's magnetic field, $B=0.36\text{G}=0.39\times {10}^{-4}\text{T}$
Angle of dip, $\delta ={35}^{\circ }$
Angle of declination, $\theta =0^{\circ }$
$r=4\text{cm}=0.04\text{m}$

The magnetic field due to the long cable at $4\text{cm}$ above and below will be,

$B_{wire}=\frac{{\mu }_{0}nI}{2\pi r}$

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

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


$\text{MS}\Rightarrow$ Magnetic south pole
$\text{MN}\Rightarrow$ Magnetic north pole

And we know that the direction of earth's magnetic field is from $\text{MS}$ to $\text{MN}$.

Now, for the point $4\text{cm}$ above the cable, the net horizontal magnetic field is,

$B_{1H}=B\cos \delta +B_{\text{wire}}$

$=0.39\cos {35}^{\circ }+0.2=0.52\text{G}$

And net vertical component of field,

$B_{1V}=B\sin \delta$

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

The resultant magnetic field at the point $1$ is,

$B_1=\sqrt{(B_{1H}{)}^2+(B_{1V}{)}^2}$

$=\sqrt{0.22{)}^2+(0.52{)}^2}\approx 0.56\text{G}$

Now, for a point $4\text{cm}$ below the cable, the net horizontal magnetic field is,

$B_{2H}=B\cos \delta -B_{\text{wire}}$

$=0.39\cos {35}^{\circ }-0.2$

$=0.39\times 0.819-0.2\approx 0.12\text{G}$

The vertical component of earth's magnetic field is,

$B_{2V}=B\sin \delta$

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

The resultant field at the point is,

$B_2=\sqrt{(B_{2V}{)}^2+(B_{2H}{)}^2}$

$=\sqrt{(0.22{)}^2+(0.12{)}^2}=0.25\text{G}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.