A) 0.32 G
B) 0.42 G
C) 4.2 G
D) 3.2 G
Correct Answer: A
Solution :
(a) 0.32 G Here, \[{{H}_{E}}=0.16\,G,=0.16\times {{10}^{-4}}T,\] dip angle \[\delta =60{}^\circ \] Then, magnitude of Earth's magnetic field, \[{{B}_{E}}=\frac{{{H}_{E}}}{\cos \delta }=\frac{0.16\times {{10}^{-4}}}{\cos \,60{}^\circ }T\] \[\Rightarrow \,\,\,\,\,\,{{B}_{E}}=\frac{0.16\times {{10}^{-4}}}{1/2}=0.32\times {{10}^{-4}}T=0.32\,G\]You need to login to perform this action.
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