Question

A long straight horizontal cable carries a current of $2.5A$ in the direction ${10}^{\circ }$ south of west to ${10}^{\circ }$ north of east. The magnetic meridian of the place happens to be ${10}^{\circ }$ west of the geographic meridian. The earth’s magnetic field at the location is $0.33G$, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points, magnetic field due to a current carrying the cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answer 1

Step 1: Draw a rough diagram of the given situation.


Step 2: Horizontal component of earth’s magnetic field.

Given,

Current carries by cable, $I=2.5A$

The earth’s magnetic field at a location,$B=0.33G=0.33\times {10}^{-4}T$

Angle of dip is zero, $\delta =0^{\circ }$

If $B$ is the earth’s magnetic field at the point, then, horizontal component of the earth’s magnetic field,
$B_H=B\cos \delta$

Substituting the values, we get

$B_H=0.33\times {10}^{-4}\cos 0^{\circ }$

$B_H=0.33\times {10}^{-4}T$

Step 3: Find the distance at neutral point lie.

Let the neutral points lie at a distance $r$ from the cable.

Magnetic field due to a current carrying conductor,
$B=\frac{{\mu }_{0}I}{2\pi r}$

At neutral point $B=B_H$
$\frac{{\mu }_{0}I}{2\pi r}=B_H\Rightarrow r=\frac{{\mu }_{0}I}{2\pi B_H}\Rightarrow r=\frac{4\pi \times {10}^{-7}\times 2.5}{2\pi \times 0.33\times {10}^{-4}}$

$r=1.5\times {10}^{-2}m$

$r=1.5\text{cm}$

Final Answer: Parallel to and above the cable at a distance at $1.5\text{cm}$