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, angle with which the magnet 1 subtend with the magnetic meridian is ${\tan }^{-1}\left(\frac{1}{x}\right)$, where $x$ is (Write upto two digits after the decimal point)

Answer 1

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 $M_1{B}_H\sin \theta =M_2{B}_H\sin (90-\theta )$
$\Rightarrow \tan \theta =\frac{M_2}{M_1}=\frac{M}{3M}=\frac{1}{3}\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{3}\right)$