Question

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 $\left(\theta \right)$ , the angular acceleration is (Magnetic induction of earths horizontal field $=$ $B_H$)

• A. $\frac{M{B}_H\theta }{I}$
• B. $\frac{I{B}_H\theta }{M}$
• C. $\frac{M\theta }{I{B}_H}$
• D. $\frac{I\theta }{M{B}_H}$

Answer 1

The correct option is A $\frac{M{B}_H\theta }{I}$

Torque acting on the magnet is $T=M{B}_{H}sin\theta$
Since $\theta$ is very small , $T=M{B}_H\theta$
$I\alpha =M{B}_H\theta$
$\alpha =\frac{M{B}_H\theta }{I}$