Question

If the horizontal component of the magnetic field is two times that of vertical component of the earth’s magnetic field then, the angle of dip is
(Given: $\tan {27}^{\circ }=0.5,\tan {37}^{\circ }=0.75,\tan {47}^{\circ }=1.07,\tan {17}^{\circ }=0.3$)

• A. ${27}^{\circ }$
• B. ${37}^{\circ }$
• C. ${47}^{\circ }$
• D. ${17}^{\circ }$

Answer 1

The correct option is A ${27}^{\circ }$

Given: $B_H=2B_V$, $\delta =?$

We know that, $\tan \delta =\frac{B_V}{B_H}$
And $B_H=2B_V$, then,

$\tan \delta =\frac{B_V}{2B_V}=\frac{1}{2}$
So, $\delta ={\tan }^{-1}\left(\frac{1}{2}\right)$

$\delta ={27}^{\circ }$

Option (a) is the correct answer.