A) H
B) \[\sqrt{2}H\]
C) 2 H
D) \[{{\text{H}}^{\text{2}}}\]
Correct Answer: B
Solution :
The relation between angle of dip \[\theta \], and horizontal component \[(\mu )\] of earths magnetic field is given by \[H=B\cos \theta \] Given, \[\theta ={{45}^{o}}\], \[\therefore \] \[H=B\,.\,\cos {{45}^{o}}\] \[\Rightarrow \] \[H=\frac{B}{\sqrt{2}}\] \[\Rightarrow \] \[I=\sqrt{2}\,H\]You need to login to perform this action.
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