A long straight horizontal cable carries a current of `2.5amp`. In the direction `10^@` south of west to `10^@` north of east, figure. The magnetic meridian of the place happens to be `10^@` 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 fied due to a current cable is equal and opposite to the horizontal component of earth's magnetic field.]

Explanation

Current through the cable, I = 2.5 A
Earth.s field , `B = 0.33 xx 10^(-4) T`
Angle of dip, `delta = 0^@`
Horizontal component, `B_H = B cos delta`
`B_H = B cos 0^@ = B = 0.33 xx 10^(-4)T`

` therefore ` Vertical component,` B_V = 0`
Let the neutral point lie at, at a distance r from the current carrying cable. At a neutral point, intensity of magnetic field = `B_H`
` (mu_0 I)/(2pi r) = B_H`
` r = (mu_0 I)/(2pi B_H) = (4pi xx 10^(-7) xx 2.5)/(2pi xx 0.33 xx 10^(-4) ) `
` = 1.5 xx 10^(-2) m`
r = 1.5 cm
The locus of the neutral points is a straight line parallel to the cable at a perpendicular distance of 1.5 cm above the cable.