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A long straight horizontal cable carries...

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.]

Text Solution

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Here, `i=2.5amp`, `R=0.33G=0.33xx10^-4T`,
`delta=0^@`
Horizontal component of earth's field
`H=Rcosdelta=0.33xx10^-4cos 0^@=0.33xx10^-4tesla`
Let the neutral points lie at a disance r from the cable.
Strength of magnetic field on this line due to current in the cable `=(mu_0i)/(2pir)`
At neutral piont, `(mu_0i)/(2pir)=H`
`r=(mu_0i)/(2piH)=(4pixx10^-7xx2.5)/(2pixx0.333xx10^-4)=1.5xx10^-2m=1.5cm`
Hence neutral points lie on a straight line parallel to the cable at a perpendicular distance of `1.5cm` above the plane of the paper.
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