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) \[5.0\times {{10}^{-5}}Weber/{{m}^{2}}\]
Correct Answer: A
Solution :
Horizontal component \[{{B}_{H}}=B\cos \varphi \] Total intensity of earth magnetic field \[B=\frac{{{B}_{H}}}{\cos \varphi }\] \[=\frac{1.8\times {{10}^{5}}}{\cos {{30}^{o}}}=\frac{1.8\times {{10}^{-5}}}{\sqrt{3}/2}=2.08\times {{10}^{-5}}Wb/{{m}^{2}}\]
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