5.18. A long straight horizontal cable carries a current of 2.5 A in the direction 10 \degree south of west to 10 \degree north of east. The magnetic meridian of the place happens to be 10 \degree west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, 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 cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Answers (1)


Current in the cable, I = 2.5 A

Earth’s magnetic field at the location, H = 0.33 G = 0.33 × 10-4T

The angle of dip, \delta = 0

Let the distance of the line of the neutral point from the horizontal cable = r m.

The magnetic field at the neutral point due to current carrying cable is:

H_{n} = \mu_{0}I/2\pi r ,

We know, Horizontal component of earth’s magnetic field, H_{E} = Hcos\delta

Also, at neutral points, H_{E}= H_n

 ⇒Hcos\delta\mu_{0}I/2\pi r

\Rightarrow 0.33 \times 10^{-4} T × cos0^o =\frac{4\pi \times 10^{-7\times2.5}}{2\pi r}

\Rightarrow r = 1.515 cm  

Required distance is 1.515 cm.