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# 6.1: Background Material

## Magnetic Dipole Oscillation

If you have ever used a compass, you may have noticed that when you look at it right after taking it out of your pocket, the needle does not immediately settle to pointing north – it oscillates a bit first. The Earth's magnetic field applies a torque to the magnetic dipole (the needle), pulling the needle into alignment with the external field, but angular momentum acquired by the needle then carries it past the north direction. The torque due to the Earth's magnetic field then slows the needle down by pulling it back toward north again. This continues until friction dissipates the needle's energy, at which point the oscillations damp down to zero, with the needle finally pointing north.

The torque applied by the constant magnetic field of the Earth has a restoring influence, dragging the system back toward an equilibrium state, in the same manner that a spring provides a restoring force for an attached mass, always pulling or pushing it toward the equilibrium length of the spring. It is therefore not surprising that this torque is responsible for simple harmonic motion. Unlike the case of a spring, this harmonic motion is rotational, but all the same mathematics holds.

Following the same process to get a differential equation for rotational motion as in the case of the pendulum, but replacing the torque introduced by gravity with the torque that acts on a magnetic dipole in a field gives:

$\left. \begin{array}{l} \tau = -\mu B\sin\theta \\ \tau = I\alpha = I\dfrac{d^2\theta}{dt^2} \end{array} \right\} \;\;\;\Rightarrow\;\;\; \dfrac{d^2\theta}{dt^2}+\dfrac{\mu B}{I}\sin\theta=0$

[The direction of the torque vector given by $$\overrightarrow\tau = \overrightarrow \mu \times \overrightarrow B$$ is expressed above in the negative sign, which indicates that the direction of the torque $$\tau$$ is always the opposite of the direction of the angular displacement $$\theta$$.]

The $$I$$ in the equation above is the moment of inertia of the magnetic dipole about its axis of rotation, not the electric current that defines the dipole moment.

Now we take the exactly the same step as in the case of the pendulum – note that for small angles (measured in radians), the sine of an angle is approximately equal to the angle itself. The resulting differential equation is exactly that of a harmonic oscillator:

$\dfrac{d^2\theta}{dt^2}+\dfrac{\mu B}{I}\theta=0$

From this we can extract the frequency of the oscillator (the square of which is the coefficient of the second term):

$\omega^2 = \left(2\pi f\right)^2 = \dfrac{\mu B}{I} = \alpha B$

[We have renamed the ratio of the dipole moment and the moment of inertia to "$$\alpha$$" to simplify the bookkeeping, and to remove the variable $$I$$ from our equations, which could later be confused for electric current. We will not need to compute the constant $$\alpha$$ for this experiment.]

This result shows that there is a linear relationship between the square of the oscillation frequency and the magnetic field strength. If we can apply a known magnetic field to a dipole, we can measure its corresponding oscillation frequency. If we do this for several magnetic fields, we can plot the linear relationship on a graph. So we need a way to apply measurable magnetic fields...

## Applying a Known Magnetic Field

Magnetic fields come from moving charges as currents, so if we choose a geometry of conductors for which we know the field, we can measure the current with an ammeter, and we'll know the magnetic field. Here is the set up we will use for this experiment:

Figure 6.1.1 – Experimental Apparatus This apparatus has several components (a few of which are not depicted above):

• magnetic field source – two coaxial coils of wire that carry equal currents in the same direction
• magnetic dipole – a small, doughnut-shaped magnet that hangs from a thread to a position along the axis of symmetry that is halfway between the two coils, which oscillates back-and-forth as it spins around the axis of the thread
• power source – a variable power supply that allows us to change the amount of current passing through the coils, thereby varying the magnetic field affecting the magnet
• ammeter – a means of measuring the current, giving us a way to indirectly measure the magnetic field
• timing device – used to measure the frequency of oscillation of the magnet

All that remains is to make the link between the current measured by the ammeter and the field of the coils that we can infer exists at the position of the magnet. Fortunately, the field on the axis of a single circular loop of wire has been worked out for us in the third text reference, the result being Equation 4.4.10. In our particular case, we have two coils, but currents in these are in the same direction, and the magnet is placed in a position where each coil provides an equal field, thus doubling the field. Furthermore, each coil actually has several turns (we'll call it $$N$$), which further multiplies the field strength. In the end we have the field at the position of the magnet in terms of the current through the coils:

$B_{loops}\left(I\right) = 2N\cdot\dfrac{\mu_oIr^2}{2\left(r^2+z^2\right)^{\frac{3}{2}}} = \left(\dfrac{\mu_oNr^2}{\left(r^2+z^2\right)^{\frac{3}{2}}}\right)I$

While this is the field we generate with the loops, it is not the total field affecting the oscillation frequency of the magnet. For that, we have to superpose the magnetic field from the coils with the (uniform) field of the Earth. We can therfore also write the magnetic field of the loops in terms of the total field and the Earth's field. When we include Equation 6.1.3, we get a linear relationship between the field of the loops and the square of the frequency of oscillation:

$B_{loops} = \left(\dfrac{4\pi^2}{\alpha}\right)f^2 - B_{earth}$

We can measure the current through the coils ($$I$$) and the corresponding frequency of oscillation ($$f$$), which means we make a plot of $$B_{loops}$$ vs. $$f^2$$, and according to our result above, this should form a line. We seek the magnetic field of the Earth, and we can determine this directly from the best fit line to this plot.