Magnetism Question 89
Question: A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a place where dip angle is $30^ \circ$ and 15 oscillations per minute at a place where dip angle is $60^ \circ$. The ratio of total earth’s magnetic field at the two places is
[MP PMT 1991; BHU 1997]
Options:
A) $ 3\sqrt{3}:8 $
B) $ 16:9\sqrt{3} $
C) $ 4:9 $
D) $ 2\sqrt{3}:9 $
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Answer:
Correct Answer: B
Solution:
Given $ {\nu _{1}}=\frac{20}{60}=\frac{1}{3}{{\sec }^{-1}}\text{and}\ {\nu _{2}}=\frac{15}{60}=\frac{1}{4}{{\sec }^{-1}} $ Now $ \nu =\frac{1}{2\pi }\sqrt{\frac{MB _{H}}{I}}=\frac{1}{2\pi }\sqrt{\frac{MB\cos \varphi }{I}}\ \ ( \because B _{H}=B\cos \varphi ) $
$ \
Therefore \ \frac{{\nu _{1}}}{{\nu _{2}}}=\sqrt{\frac{B _{1}\cos {\varphi _{1}}}{B _{2}\cos {\varphi _{2}}}} $
Therefore $ \frac{B _{1}}{B _{2}}={{( \frac{{\nu _{1}}}{{\nu _{2}}} )}^{2}},{{( \frac{\cos {\varphi _{2}}}{\cos {\varphi _{1}}} )}^{2}} $
Therefore $ \frac{B _{1}}{B _{2}}={{( \frac{1/3}{1/4} )}^{2}}\frac{\cos 60{}^\circ }{\cos 30{}^\circ }=\frac{16}{9}\times \frac{1/2}{\sqrt{3}/2}=\frac{16}{9\sqrt{3}} $ .
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