Q. A bar magnet 20 cm in length is placed with it south pole towards geographic north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth’s field $= 3.2 \times 10^{- 5}$ T, then pole strength of magnet is
A) 5 AM
B) 10 AM
C) 45 AM
D) 20 AM
Explanation
(c) Here, $2l = 20\ cm \Rightarrow l = 10\ cm,\ d = 40\ cm.$
As neutral point, $H = B = \frac{\mu_{0}}{4\pi}\frac{2\ Md}{\left( d^{2} - l^{2} \right)^{2}}$
$3.2 \times 10^{- 5} = \frac{10^{- 7} \times 2M(0.4)}{15 \times 15 \times 10^{- 4}}$
$\therefore M = \frac{3.2 \times 15 \times 15 \times 10^{- 4} \times 10^{- 5}}{0.8 \times 10^{- 7}} = 9$
$m = \frac{M}{2l} = \frac{9}{0.2} = 45\ $A-m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.