Magnetism Question 225
Question: A bar magnet of magnetic moment $ 4.0,A-m^{2} $ is free to rotate about a vertical axis through its center. The magnet is released from rest from east-west position. Kinetic energy of the magnet in north-south position will be $ (H=25\mu T) $
Options:
A) $ {{10}^{-2}}J $
B) $ {{10}^{-4}}J $
C) $ {{10}^{-6}}J $
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Loss in P.E. = gain in K.E.
$ \
Therefore ,,,E _{k}=U _{i}-U _{j}=-MB,\cos ,90^{o}-(-MB,cos,0^{o}) $
$ =4\times 25\times {{10}^{-6}}={{10}^{-4}}J $
BETA
