Question

$M$ and $M\sqrt{3}$ are the magnetic dipole moments of the two magnets, which are joined to form a cross figure. The inclination of the system with the field, if their combination, is suspended freely in a uniform external magnetic field $B$ is

• A. $\theta$ = ${30}^{\circ }$
• B. $\theta$ = ${45}^{\circ }$
• C. $\theta$ = ${60}^{\circ }$
• D. $\theta$ = ${15}^{\circ }$

Answer 1

The correct option is C $\theta$ = ${60}^{\circ }$

Given, the magnet having magnetic moment $M$ makes an angle $\theta$ with the magnetic field.

Then, the magnet having magnetic moment $M\sqrt{3}$ makes an angle $({90}^{\circ }-\theta )$ with the magnetic field.

Torque ${\tau }_1$ acting on the magnet having magnetic moment $M$ is,

${\tau }_1=MB\sin \theta$

Similarly, torque ${\tau }_2$ acting on the magnet having magnetic moment $\sqrt{3}M$ is,

${\tau }_2=\sqrt{3}MB\sin ({90}^{\circ }-\theta )=\sqrt{3}MB\cos \theta$

In equilibrium, both the torques are balanced.

$\therefore MB\sin \theta =\sqrt{3}MB\cos \theta$

$\Rightarrow \tan \theta =\sqrt{3}$

$\therefore \theta$ = ${60}^{\circ }$

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