Question

A magnetic needle suspended in a vertical plane at $30^{\circ}$ from the magnetic meridian makes an angle $45^{\circ}$ with the horizontal. What will be the true angle of dip ?

• A. $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. $\tan ^{-1}(\sqrt{3})$
• C. $45^{\circ}$
• D. $30^{\circ}$

Answer 1

Answer: (A)

$\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 2

Let $B_{e}$ be the magnetic field at some point. H and V be the horizontal and vertical components of $B_{e}$ and $\theta$ is the actual angle of dip at the same place.
$H =B_{e} \cos \theta$
and $V =B_{e} \sin \theta$
$\therefore \frac{V}{H} =\tan \theta$
or $\tan \theta =\frac{V}{H}$ ...(i)
In a vertical plane at $30^{\circ}$ from the magnetic meridian, the horizontal component is
$H'=H \cos 30^{\circ}=\frac{H \sqrt{3}}{2}$
While vertical component is still V. Therefore, apparent dip will be given by
$\tan \theta'=\frac{V}{H'}=\frac{V}{H \sqrt{3} / 2}=\frac{2 V}{H \sqrt{3}}$ ...(ii)
Dividing E(i) by E(ii),we have
$\frac{\tan \theta}{\tan \theta'} =\frac{\frac{V}{H}}{\frac{2 V}{H \sqrt{3}}}$
$=\frac{\sqrt{3}}{2}$
or $\tan \theta =\frac{\sqrt{3}}{2} \tan \theta'$
$=\frac{\sqrt{3}}{2} \tan 45^{\circ} \left(\because \theta=45^{\circ}\right)$
$=\frac{\sqrt{3}}{2}$
$\therefore \theta =\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$