Magnetism Question 57
Question: At a place the earth’s horizontal component of magnetic field is $ 0.36\times {{10}^{-4}}weber/m^{2} $ . If the angle of dip at that place is 60o, then the vertical component of earth’s field at that place in weber/m2 will be approximately
[MP PMT 1985]
Options:
A) $ 0.12\times {{10}^{-4}} $
B) $ 0.24\times {{10}^{-4}} $
C) $ 0.40\times {{10}^{-4}} $
D) $ 0.62\times {{10}^{-4}} $
Show Answer
Answer:
Correct Answer: D
Solution:
From the relation $ B _{H}=B\cos \varphi $ and $ B _{V}=B\sin \varphi $
$ \frac{B _{V}}{B _{H}}=\tan \varphi $ or $ B _{V}=B _{H}\tan \varphi $
$ =0.36\times {{10}^{-4}}\times \tan 60^{o}=0.623\times {{10}^{-4}}\ Wb/m^{2} $
BETA
