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 three 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. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• C. ${\tan }^{-1}\left(1\right)$
• D. $0$

Answer 1

The correct option is B ${\tan }^{-1}\left(\frac{1}{3}\right)$

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

$M_1=3M_2$

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



Components of $M_1$ and $M_2$, in the direction perpendicular to magnetic meridian, must be cancelled.

From the diagram,

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

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

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

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