Magnetism Question 220
Question: A magnet of length 0.1 m and pole strength $ {{10}^{-4}}A.m. $ is kept in a magnetic field of $ 30 Wb/m^{2} $ at an angle 30°. The couple acting on it is ____ $ \times {{10}^{-4}}Nm. $
[MP PET 2005]
Options:
A) 7.5
B) 3.0
C) 1.5
D) 6.0
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tau =MB\sin \theta =m\times (2l)\times B\sin \theta $
$ ={{10}^{-4}}\times 0.1\times 30\sin 30{}^\circ $
$ =1.5\times {{10}^{-4}}Nm $
BETA
