Magnetism Question 65
Question: The value of the horizontal component of the earth’s magnetic field and angle of dip are $ 1.8\times {{10}^{-5}}Weber/m^{2} $ and 30° respectively at some place. The total intensity of earth’s magnetic field at that place will be
[MP PET 1996]
Options:
A) $ 2.08\times {{10}^{-5}}Weber/m^{2} $
B) $ 3.67\times {{10}^{-5}}Weber/m^{2} $
C) $ 3.18\times {{10}^{-5}}Weber/m^{2} $
D) $ 5.0\times {{10}^{-5}}Weber/m^{2} $
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Answer:
Correct Answer: A
Solution:
Horizontal component $ B _{H}=B\cos \varphi $ Total intensity of earth magnetic field $ B=\frac{B _{H}}{\cos \varphi } $
$ =\frac{1.8\times 10^{5}}{\cos 30^{o}}=\frac{1.8\times {{10}^{-5}}}{\sqrt{3}/2}=2.08\times {{10}^{-5}}Wb/m^{2} $
BETA
