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 $(\theta)$, the angular acceleration is (Magnetic induction of earth's 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

Answer: (A)

When magnet is displaced by a very small angle $\theta$, then $r \in$ storing couple acting on the magnet is
$\tau=-M B_{H} \sin \theta$
Negative sign shows the restoring nature of torque.
Now since $\tau=I \alpha$ and $\sin \theta \approx \theta$ for small angular displacement
$\therefore I \alpha=M B_{H} \theta$
or $\alpha=$ angular acceleration
$=\frac{M B_{H} \theta}{I}$