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)

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
$H^{\prime }=H\cos 30^{\circ }=\frac{H\sqrt{3}}{2}$
While vertical component is still V. Therefore, apparent dip will be given by
$\tan {\theta }^{\prime }=\frac{V}{H^{\prime }}=\frac{V}{H\sqrt{3}/2}=\frac{2V}{H\sqrt{3}}$ ...(ii)
Dividing Eq.(i) by Eq.(ii),we have
$\frac{\tan \theta }{\tan {\theta }^{\prime }}=\frac{\frac{V}{H}}{\frac{2V}{H\sqrt{3}}}$
$=\frac{\sqrt{3}}{2}$
or $\tan \theta =\frac{\sqrt{3}}{2}\tan {\theta }^{\prime }$
$=\frac{\sqrt{3}}{2}\tan 45^{\circ }\left(\because \theta =45^{\circ }\right)$
$=\frac{\sqrt{3}}{2}$
$\therefore \theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$