Question

Question 5.18: A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10° north of east. The magnetic meridian of the place happens to be 10º 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.)

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-202

Answer 1

The neutral point can be achieved at a location above cable , where magnetic field of cable is balanced by earth's magnetic field $B_H$
$B=\frac{\mu_0}{4\pi}\frac{2I}{r}=B_H$
Here, I = 2.5 A , $B_H$=0.33G
so, 0.33 × 10^-4 T = (10^-7 × 2 × 2.5)/r
=> r = (10^-3 × 5)/0.33 = 15 × 10^-3 m
so, r = 15mm