A) 0.15 G
B) 0.15 \[\sqrt{3}\] G
C) 0.15\[\sqrt{2}\]G
D) 0.60 G
Correct Answer: D
Solution :
\[{{B}_{H}}=0.30G\]Angle of dip \[=60{}^\circ \]\[B=?\]Earths magnetic field is horizontal only at the magnetic equator. At any other place, the total intensity can be resolved into horizontal component\[({{B}_{H}})\]and vertical component\[({{B}_{V}})\] Also,\[{{B}_{H}}=B\cos \phi \]and\[{{B}_{V}}=B\sin \phi \] where,\[\phi \]is angle of inclination or dip \[\therefore \] \[0.30=B\text{ }cos\text{ }60{}^\circ \] \[0.30=\frac{B}{2}\] \[\Rightarrow \] \[B=0.60G\]
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