Question

(a) 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

Hint: For the first part of the question, find out deflecting torque and restoring torque. At equilibrium equate them and find the relation between angular acceleration and angular displacement. In the second part of the question, find the horizontal component of Earth’s magnetic field analyse the plane in which the needle freely moves and the plane of the horizontal component of the magnetic field. Finally using the formula calculate the angle of dip at that place.

Complete answer step by step:
(a) Given,
Magnetic moment = m
Strength of magnetic field = B
Moment of inertia = I
Let the needle be displaced by an angle \[\theta \] with respect to the axis of magnetic field.
The deflecting torque is given by the expression,
\[{{\tau }_{d}}=I\dfrac{{{d}^{2}}\theta }{d{{t}^{2}}}\]
Due to this displacement, a restoring torque will act on the needle it is given by the expression,
${{\vec{\tau }}_{R}}=\vec{m}\times \vec{B}$
In magnitude, it is given as
${{\tau }_{R}}=mB\sin \theta $
If $\theta $ is very small then we can write, $\sin \theta \approx \theta $.
Therefore,
${{\tau }_{R}}=mB\theta $
This restoring torque acts in the opposite direction to that of deflecting torque.
Therefore,
Deflecting torque = - Restoring torque
\[{{\tau }_{d}}=-{{\tau }_{R}}\]
$\begin{align}
  & I\dfrac{{{d}^{2}}\theta }{d{{t}^{2}}}=-mB\theta \\
 & \dfrac{{{d}^{2}}\theta }{d{{t}^{2}}}=-\dfrac{mB}{I}\theta \\
\end{align}$
We have, ${{\omega }^{2}}=\dfrac{mB}{I}$
Thus, $\dfrac{{{d}^{2}}\theta }{d{{t}^{2}}}=-{{\omega }^{2}}\theta $
Since ${{\omega }^{2}}$is a constant, we have,
$\dfrac{{{d}^{2}}\theta }{d{{t}^{2}}}\propto -\theta $
i.e. the angular acceleration of the needle is directly proportional to its angular displacement and it is directed in the opposite direction of angular displacement.
Thus, the needle performs Simple harmonic motion.
(b) Since the compass needle rotates in a vertical plane orients itself with its axis vertical at a certain place on the earth, the horizontal component of earth's magnetic field = 0 and the vertical component of earth's magnetic field = |B|.
\[\begin{align}
  & Angle\text{ }of\text{ }dip={{\tan }^{-1}}(\dfrac{{{B}_{v}}}{{{B}_{H}}}) \\
 & Angle\text{ }of\text{ }dip={{\tan }^{-1}}(\dfrac{B}{0}) \\
 & \therefore Angle\text{ }of\text{ }dip={{90}^{{}^\circ }} \\
\end{align}\]

Note: Remember the condition for simple harmonic motion which are (i) there must be an elastic restoring force (ii) the acceleration must be proportional to the displacement and direction of acceleration must be opposite to the direction of displacement. Correctly find out the plane of the horizontal and the vertical component of Earth’s magnetic field. Do not forget that deflecting and restoring torque acts in the opposite direction.