A) \[0.12\times {{10}^{-4}}\]
B) \[0.24\times {{10}^{-4}}\]
C) \[0.40\times {{10}^{-4}}\]
D) \[0.62\times {{10}^{-4}}\]
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 \[{{B}_{V}}={{B}_{H}}\tan \varphi \] \[=0.36\times {{10}^{-4}}\times \tan {{60}^{o}}=0.623\times {{10}^{-4}}\ Wb/{{m}^{2}}\]
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