Question

If a magnet is suspended at an angle ${30}^o$ to the magnetic meridian, the dip needle makes an angle of ${45}^o$ with the horizontal. The real dip is:

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

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

As in a vertical plane at an angle $\alpha$ to the magnetic meridian,
$\tan \theta =\frac{B_V^{\prime }}{B_H^{\prime }}=\frac{B_V}{B_H\cos \alpha }=\frac{\tan {45}^{\circ }}{\cos {30}^{\circ }}=\frac{1}{\frac{\sqrt{3}}{2}}$
$\therefore \theta ={\tan }^{-1}(\frac{2}{\sqrt{3}})$