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A long straight horizontal cable carries...

A long straight horizontal cable carries a current of `3*3A` in the direction `10^@` south of west to `10^@` north of east. The magnetic meridian of the plane 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 degree. Locate the positions of neutral points?

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The correct Answer is:
A set of neutral points parallel to and above the cable are located at distance of `2cm` from cable

Here, `I=3*3A`, `B=0*33G=0*33xx10^-4T`, `delta=0^@`. Horizontal component of earth's magnetic field `H=Bcos delta=0*33xx10^-4cos0^@=0*33xx10^-4T`. The neutral points will be parallel and above the cable. Let r be the distance of neutral point from the cable. Then
`H=(mu_0)/(4pi)(2I)/(r)` or `r=(mu_0)/(4pi)xx(2I)/(H)`
`r=10^-7xx(2xx3*3)/(0*33xx10^-4)=2xx10^-2m=2cm`
Thus the set of neutral points will lie parallel to cable, and above the cable at distance of `2cm` from cable.
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