Q. A bar magnet of magnetic moment M and moment of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle ($\theta$), the angular acceleration is (Magnetic induction of earth’s horizontal field $= B_{H}$)
A) $\frac{MB_{H\ }\theta}{I}$
B) $\frac{IB_{H}\theta}{M}$
C) $\frac{M\theta}{IB_{H}}$
D) $\frac{I\theta}{MB_{H}}$
Explanation
(a) When magnet is displaced by a very small angle $\theta,$ then restoring couple acting on the magnet is
$\tau = - MB_{H}\sin\theta$
Negative sing shows the restoring nature of torque. Now since $\tau = I\alpha$ and $\sin{\theta \approx \theta}$ for small angular displacement
$\therefore I\alpha = MB_{H}\theta$
Or $\alpha =$ angular acceleration$\ = \frac{MB_{H}\theta}{I}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.