Question

Two magnets of equal masses are joined at right angles to each other, as shown in the diagram. Magnet-$1$ has a magnetic moment $\sqrt{3}$ times that of magnet-$2$. This arrangement is pivoted so that it is free to rotate in the horizontal plane. Under equilibrium, what angle will the magent-1 subtend with the magnetic meridian?

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. $0^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is D ${30}^{\circ }$

Let $M_1$ and $M_2$ be the magnetic moments of the two magnets. Then,

$M_1=\sqrt{3}{M}_2$

Also, $B_{net}$ is in the direction of magnetic meridian.


For equilibrium of the system torques on $M_1$ and $M_2$ due to $B_{net}$ must counterbalance each other.

From the diagram,

${\overrightarrow{M}}_1\times {\overrightarrow{B}}_{net}=-{\overrightarrow{M}}_2\times {\overrightarrow{B}}_{net}$

$M_1{B}_{net}\sin \theta =M_2{B}_{net}\sin ({90}^{\circ }-\theta )$

$M_1\sin \theta =M_2\cos \theta$

$\tan \theta =\frac{M_2}{M_1}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^{\circ }$

Hence, option $\left(D\right)$ is the correct answer.