Q. A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass 50 mg, and the needle becomes horizontal. If the strength of each pole is 98.1 ab-amp-cm and g=981 $cms^{- 2}$, then the vertical component of earth’s magnetic induction is
A) 0.50 G
B) 0.25 G
C) 0.005 G
D) 0.05 G
Explanation
(b) As shown in figure,
mass $M = 50\ mg = 50 \times 10^{- 3}$ g strength of each pole,
$m = 98.1\ $ab-amp-cm, g = 981 $cms^{- 2},V = ?$
[image:5_image61.emf]
In equilibrium, $mV \times 2l = Mg \times l$
$V = \frac{Mg}{2m}$
$V = \frac{50 \times 10^{- 3} \times 981}{2 \times 98.1} =$0.25 G
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.