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Q.
If $B_H= \frac{1}{\sqrt{3}}B_V$ find angle of dip. (where symbols have their usual meanings)
Magnetism and Matter
Solution:
Magnetic dip or magnetic inclination is given $\not{b}y,$
$\tan\, \delta=\frac{B_V}{B_H} ...(i)$
where $ B_V $ and $B_H$ are vertical and horizontal components of earth's magnetic field, respectively.
Given $B_H=\frac{1}{\sqrt{3}}B_V$
$\therefore \, \frac{B_V}{B_H}=\sqrt{3} ...(ii)$
From Eqs. (i) and (ii), we get
$ \tan\delta=\sqrt{3}$
$\therefore \, \delta=60^\circ$