Q. A very long magnet is placed vertically with one pole on the table. A neutral point was found at 20 cm from the pole. What is the pole strength if the vertical component of earth’s field is $0.4 \times 10^{- 4}Wbm^{- 2}$?
A) 16 A-m
B) 8 A-m
C) 4 A-m
D) None of these
Explanation
(a) At neutral point, $\frac{\mu_{0}}{4\pi} \times \frac{m}{d^{2}} = V$
$10^{- 7} \times \frac{m}{(20/100)^{2}} = 0.4 \times 10^{- 4}$
$m = \frac{0.4 \times 10^{- 4}}{25 \times 10^{- 7}} = 16$ A-m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.