Question

The dip at a place is $\delta .$ For measuring it, the axis of the dip needle is perpendicular to the magnetic meridian. If the axis of the dip needle makes angle $\theta$ with the magnetic meridian, the apparent dip will be given $\tan {\delta }_1$ which is equal to

• A. $\tan \delta cose\theta$
• B. $\tan \delta \sec \theta$
• C. $\tan \delta \sin \theta$
• D. $\tan \delta \cos \theta$

Answer 1

The correct option is B $\tan \delta \sec \theta$

apparent angle of dip
let $B_V$ = vertical component of earth’s field ( it remains same in every vertical plane )

$B_H$= horizontal component of earth’s field

then horizontal component of field in the plane at angle $\theta$ with magnetic meridian = $B_H\cos \theta$

here $\delta and{\delta }_1$ are the angle of dip in magnetic meridian and another plane at angle $\theta$ with magnetic meridian respectively

in magnetic meridian,
$\tan \delta =\frac{B_V}{B_H}....\left(1\right)$

Now apparent angle of dip in ${\delta }_1$ (in another plane)

$\tan {\delta }_1=\frac{B_V}{B_H\cos \theta }$
using (1)

$\tan {\delta }_1=\frac{\tan \delta }{\cos \theta }$

$\tan {\delta }_1=\tan \delta .\sec \theta$

Hence, option (b) is correct