Magnetism Question 97
Question: A small bar magnet A oscillates in a horizontal plane with a period T at a place where the angle of dip is 60o. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be
[MP PMT 1992]
Options:
A) $ \frac{T}{\sqrt{2}} $
B) $ T $
C) $ \sqrt{2}T $
D) $ 2T $
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Answer:
Correct Answer: A
Solution:
$ T=2\pi \sqrt{\frac{I}{MB}}\Rightarrow \frac{T}{{{T}’}}=\sqrt{\frac{{{B}’}}{B}}=\sqrt{\frac{B}{B _{H}}} $
$ \Rightarrow \frac{T}{{{T}’}}=\sqrt{\frac{1}{\cos \varphi }}=\sqrt{\frac{1}{\cos 60^{o}}}=\sqrt{2} $
$ \Rightarrow {T}’=\frac{T}{\sqrt{2}} $
BETA
