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.
(b) A compass needle , free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. find out the values of (i) horizontal component of earth's magnetic field and (ii) angle of dip at the place.

Answer 1

(a)This is done by placing a small compass needle of known magnetic moment m and moment of inertia I and allowing it to oscillate in the magnetic field.
The torque on the needle is,

$\tau =m\times B$

In magnitude, $\tau =mBsin\theta$

Here, $\tau =$ Restoring torque, $\theta =$ angle between $m$ and $B$

Therefore, in equilibrium,

$I\frac{d^2\theta }{d{t}_2}=-mBsin\theta$

Negative sign indicates that the restoring torque is in opposition to deflecting torque.

This represents a simple harmonic motion.

The square of angular frequency is ${\omega }^2=mB/I$ and the time period is

$\tau =2\pi (I/mB{)}^{-1/2}$

(b) Since the compass needle is oriented vertically,

(i) Horizontal component of earth's magnetic field will be zero.

(ii)Angle of dip will be ${90}^0$.