Question

A long straight horizontal cable carries a current of $2.5$A in the direction ${10}^0$ south of west to ${10}^0$ north of east. The magnetic meridian of the place happens to be ${10}^0$ 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.)

• A. $15.15$ mm
• B. $30.15$ mm
• C. $35.15$ mm
• D. $40.15$ mm

Answer 1

The correct option is A $15.15$ mm

The neutral point can be achieved at a location above cable, where magnetic field of cable is balanced by earth's magnetic field $B_H$.
$B=\frac{{\mu }_0}{4\pi }\frac{2I}{r}=B_H$
${10}^{-7}\times \frac{2\times 2.5}{r}=0.33\times {10}^{-4}orr=\frac{5\times {10}^{-3}m}{0.33}=15.15\times {10}^{-3}=15.15mm$