Search any question & find its solution
Question:
Answered & Verified by Expert
A long straight horizontal cable carries a current of $2.5 \mathrm{~A}$ 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 $0.33 \mathrm{G}$, 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 magnetic field.)
Solution:
2671 Upvotes
Verified Answer
The neutural point can be achieved at a location above cable, where magnetic field of cable is balanced by earth magnetic field $B_H$
$$
\begin{aligned}
&B=\frac{\mu_0}{4 \pi} \frac{2 I}{r}=B_H \\
&10^{-7} \times \frac{2 \times 2.5}{r}=0.33 \times 10^{-4} \\
&\text { or } r=15.15 \mathrm{~mm}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.