Magnetism Question 184
Question: A small bar magnet of moment M is placed in a uniform field H. If magnet makes an angle of 30° with field, the torque acting on the magnet is
[CPMT 1989]
Options:
A) MH
B) $ \frac{MH}{2} $
C) $ \frac{MH}{3} $
D) $ \frac{MH}{4} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tau =MH\sin \theta =MH\sin 30^{o}=\frac{MH}{2} $
BETA
