Question

The horizontal component of earth's magnetic field at a place is $\frac{1}{\sqrt{3}}$ times the vertical component. Determine the angle of dip at that place.

Answer 1

Let the vertical component of magnetic field be $B_V$.

Thus horizontal component of magnetic field $B_H=\frac{1}{\sqrt{3}}{B}_V$

Angle of dip $\tan \delta =\frac{B_V}{B_H}$

$\therefore$ $\tan \delta =\sqrt{3}$

$⟹$ $\delta ={60}^o$