Q. A dip circle is so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is $40{^\circ}.$ Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $30{^\circ}$ with the magnetic meridian. In this position, the needle will dip by an angle
A) $40{^\circ}$
B) $30{^\circ}$
C) More than $40{^\circ}$
D) Less than $40{^\circ}$
Explanation
(c) Angle of dip
$\therefore\ \ \tan{\delta = \frac{B_{V}}{B_{H}}}$
In new position.
$\Rightarrow \ \delta^{'} = \tan^{- 1}{\left( \frac{{B^{'}}_{V}}{{B^{'}}_{H}} \right) = \tan^{- 1}\left( \frac{B_{V}}{B_{H}\cos{30{^\circ}}} \right)}$
$\Rightarrow \ \delta^{'} = \tan^{- 1}\left( \frac{2}{\sqrt{3}}\ \delta \right)\ $
Hence, $\theta = \ $true dip $= 40{^\circ}$
$\therefore\ \ \theta^{'} > \theta$
Hence , in this position, the needle will dip by an angle more than $40{^\circ}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.