Magnetism Question 278
Question: If the period of oscillation of freely suspended bar magnet in earth’s horizontal field H is 4 sec. When another magnet is brought near it, the period of oscillation is reduced to 2s. The magnetic field of second bar magnet is
Options:
A) 4H
B) 3H
C) 2H
D) $ \sqrt{3}H $
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Answer:
Correct Answer: A
Solution:
[a] Thus, $ \frac{T}{T’}=\frac{2\pi \sqrt{\frac{I}{MH}}}{2\pi \sqrt{\frac{I}{MH’}}} $ Given, $ T=4,\sec $ , $ T’=2\sec $ , So, $ \frac{4}{2}=\sqrt{\frac{H’}{H}}\Rightarrow H’=4H $
BETA
