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 in the wire, `I=2.5A`
Angle of dip at the given location on earth, `delta=0^(@)`
Earth's magnetic field, H=0.33G=0.33`xx10^(-4)T`
The horizontal component of earht's magnetic field is given s:
`H_(H)=H cosdelta`
The horizontal component of earths magnetic field is given as:
`H_(H)=H cos delta`
`H_(H)=(mu_(@)I)/(2piR)`
where
`mu_(0)=` permeability of free spac e`=4pixx10^(-7) TmA`
`thereforeR=(mu_(@)l)/(2piH_(H))`
`=(rpixx10^(-7)xx2.5)/(2pixx0.33xx10^(-4))=15.15xx10^(-3)m=1.51cm`
Hence, a set of neutral points parallel to and above the cable are located at normal distance of 1.51cm.