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In the magnetic meridian at a certain place the horizontal component of earths magnetic field is 0.32 G and the angle of dip is 60°. The total magnetic field of the earth at this location, is: |
0.52 G 0.64 G 0.54 G 0.32 G |
0.64 G |
The correct answer is Option (2) → 0.64 G $H = 0.32 G$ $\delta=60^{\circ}$ $H=R \cos \delta$ $R=\frac{H}{\cos \delta}$ $R = \frac{0.32}{\cos 60^\circ}$
$R = \frac{0.32}{0.5}$ = 0.64 G
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