Q. A long magnet is placed vertically with its $S -$pole resting on the table. A neutral point is obtained 10 cm from the pole the geographic north of it. If $H = 3.2 \times 10^{- 5}$T ,then the pole strength of magnet is
A) $8$ ab-A-$cm^{- 1}$
B) 16 ab-A-$cm^{- 1}$
C) 32 ab-A-$cm^{- 1}$
D) 64 ab-A-$cm^{- 1}$
Explanation
(c) As the magnet is long, we assume that the upper north pole produces no effect, $B$ due to south pole of the magnet is equal and opposite to horizontal component of earth’s magnetic field $ie,$
$B = \left( \ \frac{\mu_{0}}{4\pi} \right)\frac{m}{r^{2}} = H$
In CGS system, $\ \frac{\mu_{0}}{4\pi} = 1$
$\therefore 1 \times \frac{m}{10^{2}} = 3.2 \times 10^{- 5} \times 10^{4}$ (gauss)
$m = 32\ $ab-amp-cm.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.