Magnetism Question 233
A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $ 60^\circ $. The torque required to maintain the needle in this position will be
Options:
A) $ \sqrt{3}W $
W
C) $ \frac{\sqrt{3}}{2}W $
2W
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ W=MB,\cos ,60^{o} $ and $ \tau =MB,\sin ,60^{o} $
$ \
Therefore ,,,,,\tau =\sqrt{3}W $ .
BETA
