Question

A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is ${40}^{\circ }$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of ${30}^{\circ }$ with the magnetic meridian. In this position the needle will dip an angle

• A. less than ${40}^{\circ }$
• B. more than ${40}^{\circ }$
• C. ${30}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is B more than ${40}^{\circ }$

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 ${30}^{\circ }$ with magnetic meridian $=B_H\cos {30}^{\circ }$

let $\theta and{\theta }^{\prime }$ are the angle of dip in magnetic meridian and another plane at angle ${30}^{\circ }$ with magnetic meridian respectively

In magnetic meridian $\tan \theta =\frac{B_V}{B_H}....\left(1\right)$
In other plane,
$\tan {\theta }^{\prime }=\frac{B_V}{B_H\cos {30}^{\circ }}$

$=\frac{B_V}{B_H\times \frac{\sqrt{3}}{2}}$

$\Rightarrow \tan {\theta }^{\prime }=\left(\frac{2}{\sqrt{3}}\right)\frac{B_V}{B_H}$

$=\frac{2}{\sqrt{3}\tan \theta }$ (from (1))

So, $\tan {\theta }^{\prime }>\tan \theta$

$\therefore {\theta }^{\prime }>\theta$
${\theta }^{\prime }>{40}^{\circ }$

Hence, the correct option is Option (c)