Magnetism Question 267
Question: At a place on earth, horizontal component of earth’s magnetic field is $ B _{1} $ and vertical component of earth’s magnetic field is $ B _{2} $ . If a magnetic needle is kept vertical, in a plane making angle $ \alpha $ with the horizontal component of magnetic field, then square of time period of oscillation of needle when slightly distributed is proportional to
Options:
A) $ \frac{1}{\sqrt{B _{1},\cos ,\alpha }} $
B) $ \frac{1}{\sqrt{B _{2}}} $
C) $ \frac{1}{\sqrt{{{({B _{1,}}\cos ,\alpha )}^{2}}+B _{2}^{2}}} $
D) infinite
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Resultant magnetic field in the plane, $ B=\sqrt{{{(B _{1},\cos ,\alpha )}^{2}}+B _{2}^{2}} $ Time period, $ T=2\pi \sqrt{\frac{I}{MB}} $
BETA
