$A$ 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.)

(N/A) 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 by $H_{H} = H \cos \delta = 0.33 \times 10^{-4} \times \cos 0^{\circ} = 0.33 \times 10^{-4} \; T$.
The magnetic field $B$ at a distance $R$ from a long straight wire is $B = \frac{\mu_{0} I}{2 \pi R}$.
At neutral points,the magnetic field due to the wire must be equal and opposite to the horizontal component of Earth's magnetic field,so $B = H_{H}$.
Substituting the values: $0.33 \times 10^{-4} = \frac{4 \pi \times 10^{-7} \times 2.5}{2 \pi \times R}$.
Solving for $R$: $R = \frac{2 \times 10^{-7} \times 2.5}{0.33 \times 10^{-4}} = \frac{5 \times 10^{-7}}{0.33 \times 10^{-4}} \approx 1.515 \times 10^{-2} \; m = 1.51 \; cm$.
Thus,the neutral points lie on a straight line parallel to the cable at a perpendicular distance of $1.51 \; cm$ above the cable.