A) \[0.3\text{ }G\]and \[\delta ={{30}^{o}}\]
B) \[0.4\text{ }G\]and \[\delta ={{40}^{o}}\]
C) \[0.5\text{ }G\]and \[\delta ={{50}^{o}}\]
D) \[0.6\text{ }G\]and \[\delta ={{60}^{o}}\]
Correct Answer: D
Solution :
\[{{B}_{E}}=\sqrt{B_{H}^{2}+B_{V}^{2}}\] and \[\tan \,\,\delta =\frac{{{B}_{V}}}{{{B}_{H}}}\] Given, \[{{B}_{H}}=0.39G,\,\,{{B}_{V}}=0.52G\] \[{{B}_{E}}=\sqrt{{{(0.30)}^{2}}+{{(0.52)}^{2}}}=\sqrt{0.36}=0.6\] and \[\delta ={{\tan }^{-1}}\frac{{{B}_{V}}}{{{B}_{H}}}={{\tan }^{-1}}\frac{0.52}{0.3}={{60}^{o}}\]
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