Magnetism Question 279
Question: A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $ {{2}^{5/4}} $ seconds. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is
Options:
A) $ {{2}^{1/4}} $
B) $ {{2}^{1/2}} $
C) 2
D) $ {{2}^{3/4}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Initially magnetic moment of system $ M _{1}=\sqrt{M^{2}+M^{2}}=\sqrt{2M} $ and moment of inertia $ I _{1}=I+I=2I $ . Finally when one of the magnet is removed then $ M _{2}=M $ and $ I _{2}=I $ So, $ T=2\pi \sqrt{\frac{I}{M,,,B _{H}}} $
$ \frac{T _{1}}{T _{2}}=\sqrt{\frac{I _{1}}{I _{2}}\times \frac{M _{2}}{M _{1}}}=\sqrt{\frac{2I}{I}\times \frac{M}{\sqrt{2}M}} $
$ \Rightarrow ,,,T _{2}=\frac{{{2}^{5/4}}}{{{2}^{1/4}}}=2,\sec $
BETA
