Question

A small compass needle of magnetic moment 'm' is free to turn about an axis perpendicular to the direction of uniform magnetic field 'B'. The moment of inertia of the needle about the axis is 'I'. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.

Answer 1

Conditions for $S.H.M$ :-

$\bullet$ An Elastic restoring force most be there

$\bullet$ Acceleration $\left(a\right)$ of a system should be directly proportional to its displacement $\left(x\right)$

i.e, $a\alpha -x$

$(-)$ve sign indicate opposite to direction of motion.

External Torque on compass needle $m\times B$

$M=m\theta \sin \theta$

For $\theta \to$ small, $\sin \theta \simeq \theta$

$\Rightarrow T=mB\theta ⟶\left(i\right)$

Also, it rotates about centre $O$ due to inertia $T=-Id⟶\left(ii\right)$ $\{d=\frac{d^2\theta }{{dt}^2}=\theta \}$

$(-)$ve indicator restoring torque.

From $\left(i\right)$ & $\left(ii\right)$

$Il=mB\theta$

$\Rightarrow l\alpha -\theta$ (condition of $SHM$)

$\Rightarrow Il+mB\theta =0$

Normal frequency, $f=\frac{W_n}{2\pi }$

$W_n=\sqrt{\frac{mB}{I}}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{mB}{I}}$

& Time period, $T=\frac{1}{f}=2\pi \sqrt{\frac{I}{mB}}$