Question

A magnetic needle is free to rotate in vertical plane. Initially plane of rotation of needle is perpendicular to magnetic meridian and needle is in vertical position (i.e. angle of dip is ${90}^{\circ }$). Now if plane of rotation of needle is rotated by ${60}^{\circ }$ about a vertical axis and angle of dip measured in this position is found to be ${30}^{\circ }$, then true angle of dip at this place is

• A. $\varphi ={\tan }^{-1}\left(0.5\right)$
• B. $\varphi ={\cot }^{-1}\left(0.5\right)$
• C. $\varphi ={\sin }^{-1}\left(0.5\right)$
• D. $\varphi ={37}^{\circ }$

Answer 1

The correct option is A $\varphi ={\tan }^{-1}\left(0.5\right)$

Let true angle of dip be $\varphi$, which is obtained when plane of rotation of needle is along magnetic meridian

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



Given is $\Rightarrow tan{30}^{\circ }=\frac{B_v}{B_{H}cos{30}^{\circ }}$

$tan{30}^{\circ }cos{30}^{\circ }=\frac{B_v}{B_H}=tan\varphi \Rightarrow tan\varphi =\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{2}=0.5$