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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
${\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
${\tan ^{ - 1}}\left( {\frac{1}{3}} \right)$
${\tan ^{ - 1}}(1)$
$0°$
Solution
(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 $(90 – \theta )$;
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)$
