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 by an angle

• A. ${40}^{\circ }$
• B. ${30}^{\circ }$
• C. More than ${40}^{\circ }$
• D. Less than ${40}^{\circ }$

Answer 1

The correct option is C More than ${40}^{\circ }$

We know that,

$\tan \delta =\frac{B_V}{B_H}$

When the angle of dip is ${40}^{\circ }$, $\delta ={40}^{\circ }$

$\Rightarrow B_V=B_H\tan {40}^{\circ }.......\left(1\right)$

Let the new dip angle be ${\delta }^{\prime }$

$\tan {\delta }^{\prime }=\frac{B_V^{\prime }}{B_H^{\prime }}=\frac{B_V}{B_H\cos {30}^{\circ }}$

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

Using equation $\left(1\right)$ and $\left(2\right)$ we have,

$\therefore \tan {\delta }^{\prime }=\frac{2}{\sqrt{3}}.\left(\tan \delta \right)$

$\Rightarrow \tan {\delta }^{\prime }>\tan \delta$

$\therefore {\delta }^{\prime }>\delta$

$\therefore {\delta }^{\prime }$ will be more than ${40}^{\circ }$

Hence, option $\left(C\right)$ is the correct answer.

Why this Question? Key point: The angle of dip, horizontal component of earth's magnetic field and vertical component of earth's magnetic field are related by, $\tan \delta =\frac{B_V}{B_H}$