A) \[2.08\times {{10}^{-5}}Wb/{{m}^{2}}\]
B) \[3.67\times {{10}^{-5}}Wb/{{m}^{2}}\]
C) \[3.18\times {{10}^{-5}}Wb/{{m}^{2}}\]
D) \[5.0\times {{10}^{-5}}Wb/{{m}^{2}}\]
Correct Answer: A
Solution :
Horizontal component \[\therefore \] \[H=1.8\times {{10}^{-5}}Wb/{{m}^{2}}\] Angle of dip is the angle between horizontal component and total magnetic intensity. \[\therefore \] \[I\cos \theta =H\] \[(\because \,\,\theta ={{30}^{o}})\] \[\Rightarrow \] \[I=\frac{H}{\cos \theta }\] \[=\frac{1.8\times {{10}^{-5}}}{\cos {{30}^{o}}}\] \[=2.08\times {{10}^{-5}}Wb/{{m}^{2}}\]
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