Question

A long straight horizontal cable carries a current of $2.5\text{A}$ 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 $0.33\text{G}$, and the angle of dip is zero. Then neutral points lie on a straight line (ignore the thickness of the cable).

• A. perpendicular to the cable, below the plane of paper at a perpendicular distance of $1.51\text{cm}.$
• B. parallel to the cable at a perpendicular distance of $3.02\text{cm}$.
• C. perpendicular to the cable at a perpendicular distance of $3.02\text{cm}$.
• D. parallel to the cable, above the plane of paper at a perpendicular distance of $1.51\text{cm}$.

Answer 1

The correct option is D parallel to the cable, above the plane of paper at a perpendicular distance of $1.51\text{cm}$.

Given,

Current through the wire, $I=2.5\text{A}$

Angle of dip, $\delta =0^{\circ }$

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

The horizontal component of earth's magnetic field is ,

$B_H=B\cos \delta$

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

The magnetic field at the null point at a distance $R$ from the long horizontal cable is,

$B_{\text{cable}}=\frac{{\mu }_{0}I}{2\pi R}$

At null point, $B_{\text{cable}}=B_H$

$\Rightarrow \frac{{\mu }_{0}I}{2\pi R}=0.33\times {10}^{-4}\text{T}$

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


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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.