Magnetism Question 64
Question: The earth’s magnetic field at a certain place has a horizontal component 0.3 Gauss and the total strength 0.5 Gauss. The angle of dip is
[MP PMT 1995]
Options:
A) $ {{\tan }^{-1}}\frac{3}{4} $
B) $ {{\sin }^{-1}}\frac{3}{4} $
C) $ {{\tan }^{-1}}\frac{4}{3} $
D) $ {{\sin }^{-1}}\frac{3}{5} $
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Answer:
Correct Answer: C
Solution:
$ B^{2}=B _{V}^{2}+B _{H}^{2},\Rightarrow ,B _{V}=\sqrt{B^{2}-B _{H}^{2}}=\sqrt{{{(0.5)}^{2}}-{{(0.3)}^{2}}}=0.4 $ Now $ \tan \varphi =\frac{B _{V}}{B _{H}}=\frac{0.4}{0.3}=\frac{4}{3} $
Therefore $ \varphi ={{\tan }^{-1}}( \frac{4}{3} ) $ .
BETA
