Question

Two magnets of equal mass are joined at right angles to each other as shown. The 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. $0^{\circ}$

Answer 1

Answer: (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 $B_{H}$ will be $\left(90^{\circ}-\theta\right) ;$ so $M_{1} B_{H} \sin \theta=M_{2} B_{H} \sin$
$\left(90^{\circ}-\theta\right)$
$\Rightarrow \tan \theta=\frac{M_{2}}{M_{1}}=\frac{M}{3 M}=\frac{1}{3}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{3}\right)$