Q. A short magnet of moment $6.75\ Am^{2}$ produces a neutral point on its axis. If horizontal component of earth’s magnetic field is $5 \times 10^{- 5}Wb/m^{2}$, then the distance of the neutral point should be
A) 10 $cm$
B) 20 $cm$
C) 30 $cm$
D) 40 $cm$
Explanation
(c) At neutral point
$\left| \begin{matrix}
Magnetic\ field\ due \\
to\ magnet
\end{matrix} \right| = \left| \begin{matrix}
Magnetic\ field\ due \\
to\ earth
\end{matrix} \right|$
$\frac{\mu_{0}}{4\pi}.\frac{2M}{d^{3}} = 5 \times 10^{- 5} \Rightarrow 10^{- 7} \times \frac{2 \times 6.75}{d^{3}} = 5 \times 10^{- 5}$
$\Rightarrow d = 0.3\ m = 30\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.