Q. Two magnets of equal mass are joined at right angles to each other as shown. Magnet 1 has a magnetic moment 3 times that of magnet 2. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium what angle will the magnet 1 subtend with the magnetic meridian
A) $\tan^{- 1}\left( \frac{1}{2} \right)$
B) $\tan^{- 1}\left( \frac{1}{3} \right)$
C) $\tan^{- 1}(1)$
D) $0{^\circ}$
Explanation
(b) For equilibrium of the system torques on $M_{1}$ and $M_{2}$ due to $B_{H}$ must counter balance each other, $i.e.,\ M_{1} \times B_{H} = M_{2} \times B_{H}$. If $\theta$ is the angle between $M_{1}$ and $B_{H}$, then angle between $M_{2}$ and $B_{H}$ will be $(90 - \theta)$; so $M_{1}B_{H}\sin\theta = M_{2}B_{H}\sin{(90 - \theta)}$
$\Rightarrow \tan\theta = \frac{M_{2}}{M_{1}} = \frac{M}{3M} = \frac{1}{3} \Rightarrow \theta = \tan^{- 1}\left( \frac{1}{3} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.