Question

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

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

Answer 1

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

The plane of suspension of needle is making an angle $\theta ={30}^{\circ }$ with magnetic meridian and in this plane, the angle between needle and horizontal plane is ${45}^{\circ }$. That means the apparant dip $={45}^{\circ }$

${\delta }_{app}$ = 45${}^{\circ }$

The angle between the plane of needle and magnetic meridian is, $\theta ={30}^{\circ }$

$\therefore \tan \left({\delta }_{app}\right)=\frac{\tan \left({\delta }_{Real}\right)}{\cos \theta }$

$\tan \left({\delta }_{Real}\right)=\tan {45}^{\circ }\times \cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

$\Rightarrow {\delta }_{Real}={\tan }^{-1}\left(\sqrt{3}/2\right)$

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

Why this Question? Note: The relation between apparent dip and true dip is given by, $\tan \left({\delta }_{app}\right)=\frac{\tan \left({\delta }_{Real}\right)}{\cos \theta }$