A long straight horizontal cable carries a current of 2.5 A in the direction `10^(@)` south of west to `10^(@)` north of east. The magnetic meridian of the place happens to be `10^(@)` west of the geographic meridian. The earth's magnetic field at the location is 0.33 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?.

Explanation

Here I = 2.5 amp
`R = 0.33 G = 0.33 xx 10^(-4) T, theta = 0^(@)`
Horizontal component of earth's field
`H = R cos theta = 0.39 xx 10^(-4) cos 35^(@)`
`= 0.39 xx 10^(-4) xx 0.8192`
`= 3.9 xx 10^(-5)` tesla.
Vertical component of earth's field.
`H = R cos theta = 0.33 xx 10^(-4) cos 0^(@)`
`= 0.33 xx 10^(-4)` tesla.
Let the neutral points lie at a distance r from the cable
Strength of magnetic field on this line due to current in the cable `= (mu_(0)i)/(2pi r)`
At current point,
`(mu_(0)i)/(2pi r) = H`
`r = (mu_(0)i)/(2pi H) = (4pi xx 10^(-7) xx 2.5)/(2pi xx 0.33 xx 10^(-4)) = 1.5 xx 10^(-2)m`
`= 1.5 cm`
Hence neutral points lie on a straight line parallel to the cable at a perpendicular distance of 1.5 cm above the plane of the paper.