Magnetism Question 179
Question: A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. The torque required to maintain the needle in this position will be
[KCET 1994; MNR 1991; MP PET 1996; [AIEEE 2003; UPSEAT 2000; BHU 2004; Pb PET 2004]
Options:
A) $ \sqrt{3} W $
B) W
C) $ \frac{\sqrt{3}}{2}W $
D) 2W
Show Answer
Answer:
Correct Answer: A
Solution:
$ W=MB(\cos {\theta _{1}}-\cos {\theta _{2}})=MB(\cos 0^{o}-\cos 60^{o}) $
$ =MB ( 1-\frac{1}{2} )=\frac{MB}{2} $ and $ \tau =MB\sin \theta =MB\sin 60^{o}=MB\frac{\sqrt{3}}{2} $
$ \therefore \ \tau =( \frac{MB}{2} )\sqrt{3}\Rightarrow \tau =\sqrt{3} W $
BETA
