Question

Two magnets of equal mass are joined at right angles to each other, as shown in the figure. The magnet $1$ has a magnetic moment four times that of magnet $2$. This arrangement is pivoted so that it is free to rotate in the horizontal plane. In equilibrium, what will be the angle $\left(\theta \right)$ subtended by the magnet $1$ with the magnetic meridian.

• A. ${\tan }^{-1}\left(1\right)$
• B. ${\tan }^{-1}\left(\frac{1}{2}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{4}\right)$
• D. ${\tan }^{-1}\left(4\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{1}{4}\right)$


As per question,

$M_1=4M_2...\left(1\right)$

For the equilibrium of the system, torques on $M_1$ and $M_2$ should be of equal magnitude and in opposite direction.

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

$\Rightarrow \tan \theta =\frac{M_2}{M_1}$

$\Rightarrow \tan \theta =\frac{M_2}{4M_2}\left[\text{From (1)}\right]$

$\Rightarrow \tan \theta =\frac{1}{4}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{4}\right)$

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