# 5.18. A long straight horizontal cable carries a current of 2.5 A in the direction  south of west to  north of east. The magnetic meridian of the place happens to be  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.)

Given,

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,  = 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:

,

We know, Horizontal component of earth’s magnetic field,  =

Also, at neutral points,

⇒

Required distance is 1.515 cm.

## Related Chapters

### Preparation Products

##### Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
##### Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
##### Foundation 2021 Class 9th Maths

Master Maths with "Foundation course for class 9th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
##### Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-