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A long straight horizontal cable carries a current of $2.5 A$ in the direction $10 °$ south of west to $10 °$ north of east. 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 the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)

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$15.15 mm$

$30.15 mm$

$35.15 mm$

$40.15 mm$

answer is A.

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Detailed Solution

The neutral point can be achieved at a location above cable, where magnetic field of cable is balanced by earth's magnetic field $B_{H} .$

$B = \frac{μ_{0}}{4 π} \frac{2 I}{r} = B_{H}; 10^{- 7} \times \frac{2 \times 2.5}{r} = 0.33 \times 10^{- 4}$

$r = 5 \times 10^{- 3} m = 15.15 \times 10^{- 3} = 15.15 mm$

or $r = \frac{5 \times 10^{- 3} m}{0.33} = 15.15 \times 10^{- 3} = 15.15 mm$

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