Magnetism Question 190
Question: A bar magnet of magnetic moment 104J/T is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of 4×10?5 T to a direction 60° from the field will be
[MP PET 1995]
Options:
A) 0.2 J
B) 2.0 J
C) 4.18 J
D) 2 × 102 J
Show Answer
Answer:
Correct Answer: A
Solution:
Magnetic moment of bar $ M=10^{4}A\cdot m^{2} $
$ B=\frac{{\mu _{0}}}{4\pi }.\frac{2\sqrt{2}M}{d^{3}} $ Hence work done $ W=\overrightarrow{M}.\overrightarrow{B} $
$ =10^{4}\times 4\times {{10}^{-5}}\times \cos 60^{o}=0.2\ J $
BETA
