A long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ 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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Current in the wire, $I=2.5 \,A$

Angle of dip at the given location on earth, $\delta=0^{\circ}$

Earth's magnetic field, $H=0.33\, G=0.33 \times 10^{-4} \,T$

The horizontal component of earth's magnetic field is given as:

$H_{H}=H \cos \delta$

$=0.33 \times 10^{-4} \times \cos 0^{\circ}=0.33 \times 10^{-4} \,T$

The magnetic field at the neutral point at a distance $R$ from the cable is given by the relation:

$H_{H}=\frac{\mu_{0} I}{2 \pi R}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \,T\,m\, A ^{-1}$

$\therefore R=\frac{\mu_{0} I}{2 \pi H_{H}}$

$=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \pi \times 0.33 \times 10^{-4}}$$=15.15 \times 10^{-3} \,m =1.51 \,cm$

Hence, a set of neutral points lie on a straight line parallel to the cable at a perpendicular distance of $1.51\, cm ,$ above the plane of the paper.

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