Question

When plane of dip circle is in magnetic meridian it shows dip angle 30, if plane of dip circle is turned by 30, the new dip angle is

• A. ${\tan }^{-1}\left(\frac{3}{2}\right)$
• B. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• C. ${\tan }^{-1}2$
• D. ${\tan }^{-1}\left(\frac{5}{3}\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{2}{3}\right)$

Let's consider,

$B_V=$ Vertical component of earth magnetic field

$B_H=$ Horizontal component at magnetic meridian

The angle of dip at the magnetic meridian is given by

$tan{30}^0=\frac{B_V}{B_H}$. . . . . . . .(1)

When the plane of dip circle is rotated by ${30}^0$, then the vertical component is the same but the horizontal component of the earth magnetic field becomes $B_{H}cos{30}^0$.

$tan\theta =\frac{B_V}{B_{H}cos{30}^0}=\frac{B_V}{B_H}.\frac{1}{cos{30}^0}$

$tan\theta =tan{30}^0.\frac{1}{cos{30}^0}$ (from equation 1)

$tan\theta =\frac{tan{30}^0}{cos{30}^0}=\frac{2}{3}$

$\theta =ta{n}^{-1}\left(\frac{2}{3}\right)$.