Oscillations of a compass needle

A compass needle is, essentialy, a (ferro)magnetic bar. One of its main properties is to keep its magnetization, which gives a constant magnetic moment $\vec{\mu}$ to the needle. Its direction (the same of the needle) aligns with the Earth's magnetic induction field, $\vec{B}$ .

When $\vec{\mu}$ and $\vec{B}$ are not aligned, there is a torque $\vec{\tau}$ brought by this interaction, given by:

$\vec{\tau} = \vec{\mu}\times\vec{B}$

If the needle's angle is displaced by an angle $\theta$ out of the alignment, this torque will make the needle come back to the initial position, and its value is:

$\tau = -\mu B sin\theta$

Consider that $\theta$ is very small, so $sin\theta \approx \theta$ . If the moment of inertia of this needle is $I$ , then the movement of the needle is given by:

$\tau \approx -\mu B \theta = I\ddot{\theta}$

This is a simple harmonic motion equation, whose period $T$ is given by:

$T = 2 \pi \sqrt{\frac{I}{\mu B}}$

If we are analyzing two different periods of different locations, $T_{location1}$ and $T_{location2}$ , using the previous equation, we may see that:

$\frac{T_{location1}}{T_{location2}} = \sqrt{\frac{B_{location2}}{B_{location1}}} = \eta$

Therefore:

$\frac{B_{location2}}{B_{location1}} = x = \eta^2 = 2.25$ .

Let $I$ be the moment of inertia of the needle on an axis passing through it's center. Let $\mu$ be the magnetic moment of the needle. The torque acting on a magnetic moment $\mu$ under the effect of a magnetic field $B$ is given by the equation: $\overrightarrow\tau$ = $\overrightarrow\mu\times\overrightarrow B$ = $I\overrightarrow\alpha$ . When you swing the needle by a small angle $\theta$ , the torque acting on the magnet will be (we'll treat these physical quantities as scalar): $-\mu B \sin \theta$ = $I\alpha$ . As $\alpha$ is the second time derivative of $\theta$ , for small $\theta$ we can write $\sin \theta \approx \theta$ . So, $\alpha = -\frac {\mu B \theta}{I}$ , which represents the differential equation of a Simple Harmonic Motion with period $T = 2 \pi \sqrt{\frac {I}{\mu B}}$ . Dividing the period on the first location by the period on the second location (note that the magnetic moment and the moment of inertia are the same on both the locations), we can say that: $\sqrt {\frac {B_2}{B_1}} = 1.5$ $\Leftrightarrow$ $\frac {B_2}{B_1} = 2.25$ .

The compass needle in this problem is a magnetic dipole. Let the dipole moment of the needle be $m$ . Torque generated by external magnetic fields on a dipole is $\tau=\vec{m} \times \vec{B}$ . Here, the magnetic field and dipole moment point in directions which the cross product of these terms produces a torque that points opposite the direction of movement. Hence, there will be oscillation about the NS line. Mathematically,

$\tau=I\alpha=\vec{m} \times \vec{B}$

$I\alpha=-|\vec{m}||\vec{B}|\sin\alpha$
For small oscillation $\sin\alpha\simeq\alpha$ , hence
$\ddot{\alpha} + \frac{mB}{I}\alpha=0$
This is a simple harmonic equation with $\omega=\sqrt{\frac{mB}{I}}$ , thus $T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{mB}}$
Hence, we get
$x=\frac{B_2}{B_1}=(\frac{T_1}{T_2})^2=2.25$

$ 1.Oscillation of a bar magnet suspended by a small angle is given by T=2\pi \sqrt {\frac {L}{MB} } $ $ $ 2. T is the period of oscillation; L is the length of the bar magnet; M is the moment of inertia; and B is the magnetic flux density. $ $ $ 3. T \alpha \sqrt {\frac {1}{B} } \Rightarrow T^2B=const \Rightarrow T 1^2B 1=T 2^2B 2. $ $ $ (Only B and T are variable). $ $ $4. ({\frac{B 2}{B 1}}) = ({\frac{T 1}{T 2}})^2$ $5. Therefore ({\frac{B 2}{B 1}}) = ({\frac{3}{2}})^2 = {\frac{9}{4}}$

It is simple to deduce that such a oscillation is simple harming motion, or SHM. This is in essence an oscillation describes purely by trigonometric functions.

We know that in SHM the equation describing the oscillating variable is given in the form

d2x/dx2=-kx

which when solved gives

x=A*Sin(wt+p), where A is the amplitude, w is the angular frequency, and p is the phase.

It is also known that w is given by sqrt(k/m), where k is the force constant and m is the effective mass of the system.

Since w1/w2=1.5, we can easily deduce that k1/k2=1.5^2=2.25

T^2 \alpha \frac {1}{B} (B->magnetic field) \Rightarrow \frac {T 1}{T 2} = \frac {\sqrt{B 2}{\sqrt{B 1} \Rightarrow \frac {\sqrt{B 2}{\sqrt{B 1}= 1.5^2 = 2.25

In order to determine the period of oscillations we need to know the needle's moment of inertia and its magnetic properties (magnetic moment). Magnets have permanent currents which interact with external magnetic fields. In any case, it is clear that the restoring force must be proportional to the magnetic field B. Recall that for a physical pendulum the restoring force (gravity) is proportional to g and we have $T \propto \sqrt{ 1/g}.$ Thus, by analogy we can write $T \propto \sqrt{1/B}$ and therefore $x=\frac{B_{location 2}}{B_{location 1}}= (\frac{T_{location 1}}{T_{location 2}})^{2}= 2.25.$