In this experiment we will study how a magnetic dipole moment responds to an external magnetic field. In particular, we will look at how the addition of angular momentum changes this response and therefore understand more about the concept of quantum mechanical spin. The first part of the experiment will work with a classical “particle” whose full properties can be measured and observed. This macroscopic study will help us understand the physics which will be involved in the second (quantum) part where we study electrons; in this latter case, the particles are not visible, and we can only deal indirectly by observing responses of a large ensemble of particles.
Complete this question before coming to lab on Day 1.Late work will not be accepted.
DAY 1 QUESTION: Watch the video below:
Write a brief (~1 paragraph) explanation of why a strobe light is useful for measuring rotation speeds.
Complete this question before coming to lab on Day 2.Late work will not be accepted.
DAY 2 QUESTION: From your precession data, fit a curve for $\omega$ vs B to get the gyromagnetic ratio $\gamma$ for the cue ball. If you used SI units, $\gamma$ should have units of (rad/s)/Tesla.>
In theory, how quickly could the cue ball precess in an extremely strong man-made field? (Google can help you get an idea of what we’re capable of nowadays)
For the classical particle, you will use a spinning billiard ball, containing a magnet embedded at its center. The objectives are to:
Consider a loop of positive current $I$ whose path encloses area $A$ , as in Fig. 1.
The area enclosed by the loop may be considered a vector: the magnitude is the area and the direction is given by the right-hand rule, with the fingers pointing in the direction of positive current flow and the thumb points in the direction of the resulting magnetic dipole moment. Then, the magnetic dipole moment is given by
| $\overrightarrow{\mu} = \overrightarrow A I$ | (1) |
Consider a magnetic dipole moment placed in an external magnetic field $\overrightarrow B$ , as shown in Fig. 2.
The dipole will experience a torque given by
| $\overrightarrow\tau = \overrightarrow \mu \times \overrightarrow B$, | (2) |
and the energy of interaction between the dipole and the field is given by
| $E = - \overrightarrow\mu \cdot \overrightarrow B$ . | (3) |
Note that the minus sign indicates that the energy is minimum when $\overrightarrow\mu$ and $\overrightarrow B$ are parallel.
CONSIDER: If an object having a magnetic dipole moment (but no angular momentum) is free to move, how will it move in the presence of the magnetic field?
CONSIDER: If the object is given angular momentum, parallel or anti-parallel to its magnetic dipole moment, and is placed it in a magnetic field, how will it move?
To help us answer these questions, consider Fig. 3.
The magnetic field acts on the magnetic dipole moment to produce a torque, given by Eq. (2). This torque gives rise to a change in the angular momentum $d \overrightarrow{L}$ during the time $dt$ such that
| $\overrightarrow \tau = \dfrac{d \overrightarrow{L}}{dt}$ . | (4) |
The change $d \overrightarrow{L}$ causes $\overrightarrow L$ to precess through an angle $\omega dt$ , where $\omega$ is the precession angular velocity. Note from Fig. 3 that $dL = L \sin \theta$ . Therefore,
$\tau = \mu B \sin \theta = \dfrac{dL}{dt} = \omega L \sin \theta$ and
| $\omega = \dfrac{\mu}{L} B$ . | (5) |
Eq. (5) is often re-stated as
| $\omega = \gamma B$ | (6) |
where $\gamma$ is called the gyromagnetic ratio.
We will use the TeachSpin Magnetic Torque apparatus for this “classical” part of the experiment, shown in Fig. 4. The apparatus consists of the following:
| $B = \left[(1.36 \pm 0.03) \times 10^{-3} \frac{\textrm T}{\textrm A}\right]I$ | (7) |
For this part of the experiment, keep the angular momentum of the ball equal to zero.
Level the air table using the small bubble level mounted at one corner of the base. Use gravitational torque balancing the magnetic torque to determine $\mu$ . To do so:
NOTEBOOK: Draw a sketch of the ball showing the magnetic and gravitational torques. Derive the equation relating these torques at equilibrium.
From the torque diagram, you should see that
| $\overrightarrow \mu \times \overrightarrow B = \overrightarrow r \times m \overrightarrow g$, | (8) |
where $r$ is the distance from the center of the ball to the center of the sliding mass $m$ , and $g$ is the acceleration due to gravity. Eq. (8) reduces to
| $\mu B = rmg$ | (9) |
As the mass position $r$ is linearly related to the magnetic field $B$ , we can determine the magnitude of the magnetic dipole moment $\mu$ from the slope of a vs. $r$ plot.
In practice, precisely measuring $r$ would be difficult to do directly since you don't have access to the center of the cue ball. You could calculate $r$ by combining measurements of the diameter $d$ , the black handle $h$ , and the position of the mass $r^\prime$ : $r = d/2 + h + r^\prime$ You'll notice that if this is used in place of $r$ in eq. (9), you will be able to separate out an intercept (due to $d$ and $h$ ) and a slope (due to $r^\prime$ ). Since the information about $\mu$ can be found from the slope, a plot of $r^\prime$ versus $B$ will work equally well.
NOTEBOOK: Move the sliding mass to multiple positions along the aluminum rod and, at each position, determine the magnetic field needed to balance the ball. Record the values of current and mass position (with uncertainties) in a table.
NOTEBOOK and ANALYSIS: Carefully plot magnetic field as a function of position and determine $\mu$ from the slope (roughly for the notebook, and by a fit (with uncertainties) for the report).
You may provide angular momentum to the ball by spinning it using its black handle. Recall that, for a uniform solid sphere, the angular momentum $L$ is given by
| $\overrightarrow L = \dfrac{2}{5}MR^2 \overrightarrow\Omega$ , | (10) |
where $M$ is the mass, $R$ is the radius, and $\Omega$ is the spin angular velocity of the sphere. Note that for your sphere, the angular momentum and magnetic dipole moment are either parallel or anti-parallel.
NOTEBOOK: Measure the mass and radius of the ball.
If our ball has angular momentum and we apply an external field, we predict that the ball will precess about the field vector. The precession frequency depends (according to Eq. (5)) on the magnitude of the magnetic dipole moment (measured above), the angular momentum and the magnitude of the magnetic field. We will keep the angular momentum fixed, and measure the precession frequency as a function of field strength.
Angular momentum depends on the spin angular velocity of the ball. In order to keep angular momentum fixed, we must always make our measurements with the ball is spinning at the same angular velocity. In order to achieve this, we will use the following procedure:
NOTEBOOK: Repeat steps 1 through 7 at for a range of currents, recording the current and precession period with uncertainties.
NOTEBOOK and ANALYSIS: Calculate the angular momentum, Eq. (10), from your measurements of the mass and radius as well as the known spin frequency. Propagate uncertainties.
NOTEBOOK and ANALYSIS: Plot $\omega$ vs.$B$ and find the gyromagnetic ratio $\gamma$ from the slope (roughly for the notebook, and by a fit (with uncertainties) for the report).
This part of the experiment provides a qualitative demonstration of how the ball, having both angular momentum and magnetic dipole moment, behaves in a rotating magnetic field, perpendicular to the constant, vertical B field.
Remove the position indicator and install the magnetic field saddle. This saddle provides a field of constant magnitude, which may be rotated in the horizontal plane.
CONSIDER: What are the conditions that lead to a “spin flip?” How is this connected to the physics behind the resonance?
For the quantum system, we will apply a magnetic field and observe the response of free electrons in a sample of DPPH (diphenyl-picri-hydrazyl). The goals are to:
The electrons in atoms are bound in discrete energy states. Magnetic fields are generated within the atom by the
If atoms are placed in an externally applied magnetic field, the interactions of the applied field with the internal fields listed above cause the energy levels of the atoms to shift. Similarly, if the atoms are placed in a solid, the magnetic fields produced by neighboring atoms will also contribute to energy level shifts.
Electron spin resonance (ESR) is a technique for inducing and detecting transitions among energy levels. Energy level shifts are induced by application of a known magnetic field, while transitions among energy levels is induced by application of electromagnetic radiation of a known frequency. It is found that only for particular combinations of magnetic field and frequency are transitions induced.
Detection is accomplished by measuring the slight decrease in energy in the electromagnetic field which occurs when the energy is absorbed during the transition. A large ensemble of atoms is needed to absorb sufficient energy to be detectable.
It should be noted that the net energy shifts are due to the total field: applied and nearest neighbor. Since we know the value of the applied field, it follows that measuring the frequencies at which resonances occur is a probe into the details of the environment of the solid sample at the atomic scale.
The detailed study of solids using ESR is complex and beyond the scope of this course. Therefore, for simplicity we will study a much simpler system: “free” electrons, not bound to an atom. The sample we will use is the molecule DPPH (diphenyl-picri-hydrazyl), which has one, nearly free, electron per molecule.
The electron is a spin 1/2 particle. Thus, if an electron is placed in a steady magnetic field, the electron can be in only two possible energy states. In the semi-classical model we may think of the magnetic dipole moment (or the angular momentum) of the electron precessing about the applied magnetic field with two possible orientations, one as shown in Fig. 3 and the other with the $\mu$ and $\vec L$ vectors reversed relative to $\overrightarrow B$ .
NOTE: This model is useful, but has some limitations: The presence of a well-defined vector implies knowledge of the x and y-components of the vector, but quantum mechanics tells us that these components are really indeterminate.
In these two states, the magnitudes of the components of spin angular momentum parallel to the field (in the z-direction) are $\pm \hbar / 2$ . The magnetic dipole moments associated with these states are
| $\vec\mu_Z = \pm g \overrightarrow \mu_B /\,2$ , | (11) |
where $\mu_B = e\hbar/2m$ is called the “Bohr magneton.” For the free electron – for which all the angular momentum is spin (rather than orbital) angular momentum – we have $g = 2.0023$ . The energy of a magnetic dipole moment in a magnetic field is given by
| $E = - \overrightarrow\mu \cdot \overrightarrow B$ . | (12) |
Thus, the energy difference between the two states is
| $\Delta E = 2 \mu_ZB = g\mu_BB$ . | (13) |
Energy level splitting in an atom with one optically active electron outside a closed subshell-pure spin angular momentum state: $S = 1/2$ , $\ell = 0$ , $j = 1/2$ placed in a magnetic field.
If we apply electromagnetic radiation with frequency $f$ , such that
| $hf = \Delta E$ . | (14) |
where $\Delta E$ is given by Eq. (13), we should induce transitions between the two energy states. Eq. (14), however, can also be written
| $\hbar \omega = \Delta E$ . | (15) |
At resonance, it turns out that $\omega$ of Eq. (15) is the same as that of Fig. 3, i.e., the precession angular velocity of the electron around the magnetic field. Thus, photons of angular frequency $\omega$ carrying angular momentum $\hbar$ , cause the electron's spin to change state. Combining Eqs. (13) and (15) gives
| $\omega = \dfrac{g\mu_B}{\hbar}B$ . | (16) |
It should be noted that, for a free electron in a magnetic field, the magnitude of the spin magnetic dipole moment is
| $\left\lvert{ \overrightarrow{\mu}}\right\rvert = g \mu_B\sqrt{s(s+1)}$ . | (17) |
and the magnitude of the spin angular momentum is
| $\left\lvert \overrightarrow{S}\right\rvert = \sqrt{s(s+1)}\hbar$ . | (18) |
Thus, the ratio of magnetic dipole moment to angular momentum is
| $\dfrac{\left\lvert \overrightarrow\mu\right\rvert}{\left\lvert \overrightarrow S\right\rvert} = \dfrac{g\mu_B}{\hbar},$ | (19) |
which appears also in Eq. (16).
Eq. (16) is often re-written as
| $\omega = \gamma B$ | (20) |
where $\gamma$ is called the gyromagnetic ratio. Eq. (20) is analogous to Eq. (6) for the classical case. In both cases $\gamma$ is the ratio of magnetic dipole moment to angular momentum. The factor $g$ is required by quantum mechanics.
Above, we make the argument that a spin flip occurs when the system absorbs a photon of energy $E = \hbar \omega$ which carries with it one unit of angular momentum $\hbar$ (in the correct direction). In practice, however, we do not use photons, but instead a radio frequency (RF) oscillating magnetic field
| $\vec{B}(t) = 2B_0\cos(\omega t)\hat{x}$ |
which possess intrinsic energy and angular momentum. This oscillating field can be decomposed into two counter-rotating fields in the x-y plane as
| $\overrightarrow{B}(t) = B_0\left[\cos(\omega t)\hat x + \sin(\omega t)\hat y \right] + B_0 \left[ \cos (\omega t)\hat x - \sin (\omega t)\hat y \right]$ . |
In the classical case, we saw that a rotating magnetic field in the direction of precession at the correct frequency could induce a spin flip, whereas a rotating magnetic field in the wrong direction (even with the correct frequency) had no effect; thus, we see that our oscillating field has a resonance-inducing part and an ignorable part.
The red and blue arrows are both unit-length vectors, rotating in opposite directions at the same rate. The green arrow is their sum, which is a vector in the $\hat y$ direction with a sinusoidally varying magnitude. To experiment a bit with the code that generates this animation, open this notebook.
We use the Daedalon ESR apparatus (shown in Fig. 5), consisting of the following:
| $B = \left[ 0.48\, \mathrm{ mT/A}\right] I$ | (21) |
*The sample we will use is the molecule DPPH (diphenyl-picri-hydrazyl), which has one, nearly free, electron per molecule.
The user's manual for the apparatus can be found here
Make the electrical connections as shown in Fig. 6:
We will now try to find an electron spin resonance signal:
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| Figure 7: An example of the scope trace as viewed in “time versus voltage” mode with resonance “spikes” visible. |
Once you've found a clear signal, we can make a measurement of the magnetic field at the time of the resonance so that we can pair this value of $B$ with the supplied value of $f$ :
NOTEBOOK: For one or two frequencies, record the voltages (four per cycle) that correspond to resonance. Calculate the corresponding $B$-field values.( You will measure these values (and more) again using a different procedure below.)
There is an alternate way to measure the voltages at which resonance occur. If we place the oscilloscope into “XY mode” (selectable in the “Display” menu on the scope under the “Format” option), then the trace which appears will be a “loop” such that channel 1 will serve as the (time-dependent) x-coordinate and channel 2 will serve as the (time-dependent) y-coordinate. Since both signals are periodic with the same frequency (60 Hz), the loop traces the same pattern each cycle and the image is stable.
Let us follow the trace for one of these cycles. (See Fig. 8). Beginning at the leftmost point, our Helmholtz voltage (x-coordinate) is at the minimum value. As this voltage increases (following the trace to the right), we pass through a resonance point and the voltage on the oscillator signal (y-axis) rises. Continuing, we go off resonance (y falls), we pass through zero field, and then we come to the corresponding resonance point for positive Helmholtz voltage (y rises again). Once we reach the maximum voltage in the Helmholtz coils (maximum x-value), the voltage decreases and the the trace reverses direction. Moving now from right to left, we see the resonance on the positive half, pass through zero, then see the resonance on the negative half before we finally return to where we started.
Likely, you will initially see a trace which looks like the image on the left in Fig. 8 with four distinct peaks. This means that the four resonances are appearing on the scope at four different voltages within the cycle. We know that these resonances in fact should occur only at two different voltages ( $+V_0$ and $-V_0$ ). Therefore, the oscillator voltage signal must be out of sync with the Helmholtz voltage signal. To adjust the phase relationship between the two signals, adjust the “Phase Null” knob on the front of the power supply until the pairs of peaks overlap (as seen in the right image of Fig. 8). In XY-mode, the voltage measured along the x-axis from the center of the loop to each peak corresponds to the voltage in the Helmholtz coil at the time of resonance. However, since there is some ambiguity in the “center” of the loop, a more accurate measurement would be to measure the voltage along the x-axis from peak to peak (which corresponds to twice the resonant voltage) and divide by two.
NOTEBOOK: Using either the “time vs. voltage” or “XY” mode, record values of the voltage in the coils at each resonance for fixed frequency. Collect data over the full range of frequencies at as many values as you can.
NOTEBOOK and ANALYSIS: Convert from $V$ to$I$ to $B$ , and from $f$ to $\omega$. Plot $\omega$ vs. $B$ and find the gyromagnetic ratio $\gamma$ from the slope (roughly for the notebook, and by a fit (with uncertainties) for the report). Is this value consistent with Eq. (16)? Note that if your dominant uncertainties are in the magnetic field, you may with to invert the axes and fit to find $1/\gamma$ .
We now want to test the relation between the direction of the RF field and the applied Helmholtz field:
CONSIDER: What do you observe? Does this make sense with your understanding of the physics of the spin flip resonance? In particular, what effect does rotating the field have on angular momentum in our system?
As the TA looks over your notebook, they will be looking for general good practices and for specific things. Be prepared to point to (and explain) the appropriate places in your lab notebook where the following information is contained.
| Setup | Data | Commenting |
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| Item | Good (4) | Missing (0) |
| Day 1 question | Question answered completely and carefully, even if not correct. | No attempt made, attempt is incomplete, or effort is clearly last-minute and insufficient. |
| Day 2 question | Question answered completely and carefully, even if not correct. | No attempt made, attempt is incomplete, or effort is clearly last-minute and insufficient. |
| Notebook (setup) | Notebook includes sketches, values of apparatus settings, comments on experiment conditions, and any other necessary information about the setup and procedure. | Most or all key features are missing, illegible, or incorrectly recorded. |
| Notebook (data) | Notebook includes data in a legible format (or information about how data are stored digitally), descriptions of the conditions under which such data were collected, and values and details of the uncertainty estimates. | Most or all key features are missing, illegible, or incorrectly recorded. |
| Notebook (commenting) | Notebook includes comments, calculations or plots where appropriate, and provides summary conclusions or observations. | Most or all key features are missing, illegible, or incorrectly recorded. |
Electron Spin Resonance Analysis Rubric
| Item | Good (4) | Adequate (3) | Needs Improvement (2) | Inadequate (1) |
| Plot for determining classical magnetic dipole moment | Plot of $B$ vs $d$ is made and a fit is performed. $\mu$ is calculated from fit parameters and includes uncertainties. | As Good but plot is missing one elements or has numerous unneeded elements. | Plot is missing uncertainties, fits, or is difficult to interpret. | Plot is missing uncertainties, fits, and is difficult to interpret. |
| Plot for finding cue ball (classic) gyromagnetic ratio | Plot of $\omega$ vs $B$ is made and a fit is performed. Gyromagnetic ration $\gamma$ is found with uncertainties. | As Good but plot is missing one elements or has numerous unneeded elements. | Plot is missing uncertainties, fits, or is difficult to interpret. | Plot is missing uncertainties, fits, and is difficult to interpret. |
| Discussion of electron measurement choices | Method for finding resonant frequencies is described. Choice between 'V vs t' or 'XY' mode is justified. | As good but description is missing information Justification is unclear. | Method description is missing a key step or multiple minor steps. Justification is superficial. | Method description is missing significant information. Justification is completely arbitrary. |
| Plot for finding electron (quantum) gyromagnetic ratio | Plot of $\omega$ vs $B$ is made and a fit is performed. Gyromagnetic ration $\gamma$ is found with uncertainties. $\gamma$ is compared to expected theoretical value (eq 16) | As Good but plot is missing one elements or has numerous unneeded elements. Comparison is done with overly simple methods (e.g. % difference) | Plot is missing multiple element of adequate plot or no comparison is made. | Plot is missing multiple element of adequate plot and no comparison is made. |