Oscillation Magnetometer

Oscillation Magnetometer

An oscillation magnetometer consists of a wooden box with glass windows at the front and at the sides as shown in the figure. A glass tube is fixed at the centre of its top surface and a silk thread, attached on a brass torsion head at the top of the tube passed through it. A brass hook is attached at the end of the silk thread on which a bar magnet is placed horizontally. At the bottom surface of the box, a plane mirror is set on which a straight line is scratched as a reference line. Two small glass windows are fixed to observe the oscillation of magnet at the upper wooden. The parallel base is made by two leveling screws at the bottom of the box. The front window can be slided up and down.

Before starting the experiment, the torsion in the silk thread is removed with the help of a brass strip and the scratch line in mirror is set in the magnetic meridian. A small bar magnet is now replaced on the hook symmetrically that initially aligns along the magnetic meridian. If the magnet is slightly deflected in horizontal surface and left it free, it oscillates in the earth’s magnetic field, H. For a small angular displacement q, the torque of the couple acting on the magnet is

$$\tau = MH\sin \: \theta $$

where M is magnetic moment of the bar magnet. The magnet tries to align itself along the field H but due to it’s moment of inertia I, it reaches to next extreme position, Y and returns back to original position, X. So the magnet in the box oscillates about H and its equation of motion is

$$ \tau = I\alpha $$

where a is the angular acceleration. Since the torque is restoring, so

$$\begin{align*} I \alpha &= -MH\sin \: \theta \\ \text {or,} \: \alpha &= \frac {MH\sin \: \theta }{I} \\ I\alpha &= - MH\sin \: \theta \\ \text {or,} \: \alpha &= \frac {MH\sin \: \theta }{I} \\ \text {For a small angle,} \: \sin\: \theta = \theta , \: \text {then} \\ \alpha &= - \frac {MH \: \theta}{I} \\ \end{align*}$$

As M, H and I are constant, the angular acceleration is directly proportional to the angular displacement q, and this shows that motion of the bar magnet in a uniform magnetic field is angular simple harmonic. Comparing this equation with \(\alpha = -\omega ^2\theta \), we have

$$\begin{align*} \omega ^2 &= \frac {MH}{I} \\ \text {and the period of oscillation is} \:\:\:T &= \frac {2\pi }{\omega } \\ &= 2\pi \sqrt {\frac {I}{MH}} \\end{align*}$$

Oscillation magnetometer is used to find the magnetic momentum M, of a bar magnet. For this, the period of oscillation of the magnet is noted and its moment of inertia is measured. From the above equation,

$$\begin{align*} T &= 2\pi \sqrt {\frac {I}{MH}} \\ \text {squaring both sides, we have} \\ T^2 &= 4\pi^2 \frac {I}{MH} \\ \text {or,} \: M &= 4\pi^2 \frac {I}{HT^2} \\\end{align*}$$

For a bar magnet of breadth b and thickness t , the moment of inertia about the axis passing perpendicularly through its length
$$ I = \frac {b^2 + t^2}{12} $$

Getting the values of H and I, and noting the period of oscillation of the bar magnet, the magnetic moment of the magnet and pole strength can be calculated.

To compare magnetic moments of two bar magnets, their time periods are separately measured using the oscillation magnetometer. Let T₁ and T₂ be the period oscillation of magnets having magnetic moment M₁ and M₂ respectively. We have

$$\begin{align*} T_1 &= 2\pi \sqrt {\frac {I_1}{M_1H}} \dots (i)\\ T_2 &= 2\pi \sqrt {\frac {I_2}{M_2H}} \dots (ii) \\ \text {Dividing equation (i) by equation (ii),} \\ \text {we have} \\ \frac {T_1}{T_1} &= \frac {\sqrt {I_1M_2}}{\sqrt {I_2M_1}} \\ \text {and on squaring both sides,} \\ \frac {T_1^2}{T_2^2} &= \frac {I_1M_2}{I_2M_1} \\ \text {or,} \: \frac {M_1}{M_2} &= \frac {I_1T_2^2}{I_2T_1^2} \\ \end{align*}$$

In this way, we can compare the magnetic moment of two bar magnets.

The period of oscillation T₁ of the bar magnet is noted at a place, and then the new period T₂ is at an other place with the same magnet and magnetometer. If the horizontal components are H₁ and H₂ at these places, we have

$$\frac {H_1}{H_2} = \frac {T_2^2}{T_1^2} $$

Comparison of magnetic moments of two magnets of different sizes

Suppose two bar magnets of magnetic moment M₁ and M₂ and moment I₁ and I₂ respectively. When these magnets are placed over the other magnet with like poles in the same direction, the moment of inerteia of the combination becomes I₁+ I₂ and magnetic moment M₁ + M₂. So , the time period is

$$T_1 =2\pi\sqrt {\frac {{I_1 +I_2)}}{M_1 + M_2)H}} \dots (i) $$

When the magnets are combined with unlike poles together, the moment of inertia remains same as before but the magnetic moment becomes M₁ – M₂ and the period of oscillation is

$$\begin{align*} T_2 &= 2\pi \sqrt {\frac {(I_1 + I_2)}{(m_1 – M_2)H}} \dots (ii) \\ \text {Dividing equation} : (i) \: \text {by} \: (ii),\text {we get} \\ \frac {T_1}{T_2} &= \sqrt {\frac {(M_1 –M_2)}{(M_1 + M_2)}} \\ \text {and on solving this equation,} \\ \text {we have}\\ \frac {M_1}{M_2} &= \frac {T_1^2 + T_2^2}{T_2^2-T_1^2} \\ \end{align*}$$

Putting the value of T₁ and T₂ the ratio of M₁ and M₂ is calculated.