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A long straight horizontal cable carries...

A long straight horizontal cable carries a current fo 2.5 A 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 0.33 G , and the angle of dip is zero. Locate the line of neutral points.

Text Solution

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Given : `I = 2.5 A, B_(E) = 0.33 G, theta_(D)=0`
We know that , `B_(H) = B_(E) ` for `theta _(D) = 0`
At neutral point , `(mu_(o))/(4pi) (2i)/(a) = B_(E )`
i.e., `" " B_(E) = (10^(-7)xx 2xx2.5 )/(a) = 0.33xx 10^(-4)` T
`:. " " a = (5xx10^(-3))/(0.33) = 15.15xx 10 ^(-3)` m
`a = (5xx 10^(-3))/(0.33) = 15.15xx 10^(-3)`m
= `1.515 xx 10^(-2) `m
i.e., `" " a = 1.5 cm`
Neutral point will be away from the cable and and above it at a height of 1.5 cm
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