A long straight horizontal cable carries a current of 2.5 A in the direction from `10^(@)` in 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 angle of dip is zero . Locate the line of neutral point . (ignore the thickness of the cable .)

Explanation

As angle of dip , `theta=0^(@)`
`therefore" " B_(H)=Bcostheta=0.33xx1=0.33G`
If a is the distance of the neutral point , then
`B_(H)=(mu_(0))/(4pi).(2I)/(a)`
`[therefore` magnetic field at neutral point = horizontal component of earth's magnetic field ]
`therefore" " a=(mu_(0))/(4pi).(2I)/(B_(H))=(10^(-7)xx2xx2.5)/(0.33xx10^(-4))m=1.5cm`
Applying thumb rule it is seen that the direction of magnetic field is upwards from the cable . thus , the neutral point is situated on a line above the cable at a distance 1.5 cm and parallel to the cable .