Magnetism Question 246
Question: A bar magnet having a magnetic moment of $ 2\times 10^{4}J{{T}^{-1}} $ is free to rotate in a horizontal plane. A horizontal magnetic field $ B=6\times {{10}^{-4}}T $ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $ 60{}^\circ $ from the field is
Options:
A) 12 J
B) 6 J
C) 2 J
D) 0.6 J
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Work done $ =MB(\cos ,{\theta _{1}}-\cos ,{\theta _{2}}) $
$ =MB(\cos ,0^{o}-\cos ,60^{o}) $
$ =MB( 1-\frac{1}{2} )=\frac{2\times 10^{4}\times 6\times {{10}^{-4}}}{2}=6,J $
BETA
