Magnetism Question 28
Question: If a magnet is suspended at an angle 30o to the magnetic meridian, it makes an angle of 45o with the horizontal. The real dip is
Options:
A) $ {{\tan }^{-1}}(\sqrt{3}/2) $
B) $ {{\tan }^{-1}}(\sqrt{3)} $
C) $ {{\tan }^{-1}}(\sqrt{3/2}) $
D) $ {{\tan }^{-1}}(2/\sqrt{3}) $
Show Answer
Answer:
Correct Answer: A
Solution:
Let the real dip be f, then $ \tan \varphi =\frac{B _{V}}{B _{H}} $ For apparent dip, $ \tan {{\varphi }^{’}}=\frac{B _{V}}{B _{H}\cos \beta }=\frac{B _{V}}{B _{H}\cos 30^{o}}=\frac{2B _{V}}{\sqrt{3}B _{H}} $ or $ \tan 45^{o}=\frac{2}{\sqrt{3}}.\tan \varphi $ or $ \varphi ={{\tan }^{-1}}( \frac{\sqrt{3}}{2} ) $
BETA
