Q 12-05-050JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
A magnet hung at $45^\circ$ with magnetic meridian makes an angle of $60^\circ$ with the horizontal. The actual value of the angle of dip is
Answer: (A) $\tan^{-1}\left(\sqrt{\dfrac32}\right)$
In a vertical plane at angle $\theta$ to the meridian, the horizontal field seen is $B_H\cos\theta$ while the vertical field is unchanged, so the apparent dip $\delta'$ satisfies
$$\tan\delta' = \frac{\tan\delta}{\cos\theta}$$
$$\tan\delta = \tan60^\circ\cos45^\circ = \frac{\sqrt3}{\sqrt2} \Rightarrow \delta = \tan^{-1}\sqrt{\frac32}$$
Solution by Sreeraj P, M.Sc Physics