# Magnetic Torque Physics 2150 Experiment No. 11 University of Colorado

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1 Experiment 11 1 Introduction Magnetic Torque Physics 2150 Experiment No. 11 University of Colorado In this experiment, you will study how a magnet moment interacts with a magnetic field. You will measure the magnetic moment of a magnet in three different ways. Let us begin by reviewing some basic theory. Classically, a magnetic moment is a loop of electrical current. If the loop has area, and the direction normal to the plane of the loop is given by the unit vector, then the magnetic moments is defined as the vector =, (1) where the direction of is given by right-hand-rule (with your right hand, curl fingers along the current; your thumb points along ). Examples of objects with magnetic moments include: Electrons and protons in which the current is a result of the quantum mechanical spin of the charged particle Permanent magnets in which an effective current is formed by the alignment of atomic currents associated with spin or orbital angular momentum Loops of current-carrying wire Note that a few of these examples are not physical loops of current, but still produce magnetic moments. The current loop is simply an analogy for a magnetic moment. A magnetic moment in a uniform magnetic field experiences no net force but it does experience a torque = (2) that tends to align the moment with the field. A magnetic moment in a non-uniform magnetic field experiences a net force given by =. (3a) In the case that the gradient is aligned with the z-axis, this last equation becomes

2 Experiment 11 2 =. (3b) Finally, a magnetic moment in a -field has a potential energy given by =, (4) so the energy is minimum when the moment is aligned with the field. A spinning object that has charge possesses a magnetic moment. Similarly, a spinning object that has mass possesses angular momentum. For instance, protons and electrons have both mass and charge and they both have a quantum mechanical spin, which can be thought of very roughly and somewhat inaccurately as a rotation of the particle. So both protons and electrons have a magnetic moment as well as angular momentum. Classically, the angular momentum of an object spinning about a fixed axis can be defined as =. (5) Here I is the moment of inertia, defined as = =! ", where is the distance from the axis of rotation. The moment of inertia of a uniform sphere about an axis through its center is Isphere = # \$%. The vector is the angular velocity, whose magnitude is =2'( = ), where ( is the frequency * and + is the period of rotation. The direction of and is along the axis of rotation, given by the right hand rule. Part I: Harmonic Oscillation In this part, you will measure the magnetic moment of a small, powerful magnet embedded in the center of a white billiard ball. The magnetic moment of the magnet is aligned with a small black plastic cap on the ball. The ball can be placed in an air bearing (the brass cup in the center of the apparatus) which allows it to rotate freely with almost no friction. The air bearing is located in the center of a pair of Helmholtz coils which create a nearly uniform -field at the location of the ball. The -field is proportional to the current through the coils according to =,, where, = (1.36±0.03) x 10-3 T/A. (6)

3 Experiment 11 3 [See additional remarks about the Helmholtz coils in the Appendix.] When the magnetic field is on, the torque on the magnet due to the field will tend to align the magnetic moment with the field. If the ball is released from rest with its axis not aligned with the field, the ball will oscillate back and forth about the field direction. From the period of oscillation, it is possible to compute the magnetic moment as we show below. A torque causes a change in the angular momentum according to = /. (7) In this case, the magnetic torque has magnitude = sin3 and / leading to the equation = 4 =5 6 5, sin3 = (8) [Question: Why is there a minus sign in this equation? Answer this in the introduction/theory section of your lab report] In the special case of small angles, 3 << 1 radian, sin3 3, and Eq. (8) can be rewritten as = (9) Though you may not recognize it, this is the equation of motion of a simple harmonic oscillator. To see this, recall that for a mass on a spring with spring constant :, we can write =; :< = 5 = 5 5 = 5 = >? This last equation has the solution < CD, where = <. (10) ) * = E>? are arbitrary constants determined by initial conditions. (You should verify this!) and and C Comparing Eqs. (9) and (10), we notice they are exactly the same equation, except with different symbols. The symbol 3 corresponds to x and the symbol B/I corresponds to k/m. Since (9) and (10) are the same equation, they have the same solution. Hence the solution of Eq. (9) is 3 =3?F= where = ) * =E8 9. (11) We emphasize that this solution is only valid in the limit of small amplitude oscillations.

4 Experiment 11 4 The last equation can be rewritten as G =H 8 J. (12) * 5 I) 5 9 Thus a plot of (1/+ ) vs. should be a straight line with a slope DQ. The moment of inertia of the ball is readily determined by making careful measurements of its mass and its diameter. [Recall that for a sphere, ]. Thus, one can solve for the magnetic moment. Procedure (for Part I) Begin this experiment as you would begin any experiment by thoroughly inspecting the apparatus. Find the on/off switch; figure out what all the switches and knobs do; find the air supply and inspect the controls; play with the billiard ball on the air bearing; give it a spin; see what happens with the field on and off. Get the big picture before taking any data! The air bearing is turned on with the building air supply, not with the switch on the control panel. Carefully measure the mass and diameter of the billiard ball. Use the electronic scales and the calipers. From these measurements, compute the moment of inertia of the ball. As usual, estimate the uncertainty of your result. Make careful measurements of the period of oscillation of the ball for several values of the S-field. Be careful to keep the amplitude of the oscillation small. At a given value of S, determine the period T by measuring the time for several periods (at least 10) and dividing by the number of periods. (Don t measure the period 10 times; measure the time for 10 periods once and then divide by 10.) Note: you do not need the rod inserted for part I. Make a plot of (U/T V ) vs. S. The points should fall on a straight line. Determine the slope of this line with the Mathematica program Wlinfit.nb (which is on the hard disks of the PC s in the 2150 lab). From your data, determine the magnitude of the magnetic moment W and make a clear determination of the uncertainty in W. Part II: Mechanical Torque vs. Magnetic Torque In this part, we will again measure the magnetic moment of the magnet in the billiard ball, but this time with a completely different method. Here we will balance the magnetic torque on the ball with a mechanical torque.

5 Experiment 11 5 There is a small weight on a thin aluminum rod which can be placed in the hole in the cap of the billiard ball. One end of the aluminum rod has a tiny steel pin in it, so the magnet will grab it when the rod is inserted into the ball. The small weight on the rod is used to exert a mechanical torque about the center of the ball, as in the diagram. When the ball is in the air bearing and is given a spin about its axis (by twirling the rod), the ball has an angular momentum parallel to the axis of rotation. When the -field is zero, the only torque of the ball is the mechanical torque. Under the action of this torque, the angular momentum will precess like a top. This precessional motion is described by Eq. 9: = / X/ X [Can you see in the diagram at the right that the vectors and Δ are indeed parallel, as demanded by Eq. (9)?] Now if a magnetic field is applied, and the field has the correct magnitude and direction, then the magnetic torque = can cancel the mechanical torque = Z leading to no net torque and no precession. The condition for no net torque is [ [= Z sin3 = Zsin3 = 8?^. (13) Thus a plot of vs. should be a straight line with slope /Z. We have been slightly glib in discussing the mechanical torque. To be precise, we should take account of the mass _` of the aluminum rod as well as the mass of the small weight on the rod. The torque due the weight of the rod alone has magnitude (% a? _` Zsin3) where % a? is the distance from the center of the ball to the center-ofmass of the rod. This torque due to the rod should be added to the torque due to the small weight ( Zsin3) to get the total torque. Eq. (13) should be rewritten as sin3 = Z sin3+ % a? _` g sin3 =?^ 8 H? cdef gh J. (14)? From this equation, we see that the mass of the rod has no effect on the slope of i vs. S curve. So we can completely ignore the mass of the rod when taking measurements. It is essential to make a plot of vs. and measure the slope; we cannot determine by simply taking one measurement and assuming = 8?^.

6 Experiment 11 6 Procedure (for Part II) Begin by carefully determining the mass j of the small weight by using the high precision digital scale. When inserting the aluminum rod into the ball, make sure the end with the steel pin is inserted: you can feel the magnet grab the rod when it is all the way in. To determine the length i, you will need to measure the diameter of the billiard ball the length of the black plastic cap on the ball, the width of the small weight, and the distance from the top of the cap to the bottom of the weight. With the -field set to zero, and at some particular value, place the ball with its rod and small weight in the air bearing and give it a good, strong twirl. It will precess and will continue to precess as long as the ball is spinning about its axis. Adjust the magnitude and direction of the -field until the precession stops completely. (There is a field polarity switch for changing the direction of the field.) This gives you one (i,s) data pair. Repeat this procedure for several values of i and plot i vs. S. Determine the slope of the line and, from your measurements, determine the magnetic moment W and its uncertainty. Be sure to compare your results from Parts I and II. Part III: Field Gradient Method Remember this part is optional, but by doing this part you will have completed one of the two required starred labs. In this part, you will determine the magnetic moment of a small magnet by measuring the force on the magnet in a field gradient as given by equation (3b). = (3b) This is called the field gradient method and is a technique frequently used by chemists and physicists to determine the magnetic moment of an unknown sample. In our case, the unknown is a small magnet (in a white plastic case), which, according to the manufacturer, is identical to the magnet in the billiard ball. (You will determine how identical it actually is!)

7 Experiment 11 7 There is a switch on the apparatus which turns on the field gradient by switching the current direction in one of the Helmholtz coils. The formula for the gradient produced by two single loops of wire of radius %, separated by a distance ", and each carrying a current is = k `% " (15) G lf 5 m e5 p 5 n o Each of our coils has 195 turns. The average radius of the coils is q = m and the average separation of the coils is r = m. (The uncertainty on each number is one place in the last digit.) For our magnet, the resulting field gradient is =s, where D = 1.69 x 10-2 T / (m A) (16) The magnet is attached to a spring of unknown spring constant : all in a plastic holder which fits neatly over the air bearing, so you can suspend the magnet by the spring in the field gradient. (The air bearing is OFF throughout this part.) To determine the spring constant :, you can use the 5 small steel balls as weights (they will stick to the magnet). By measuring the stretch of the spring as a function of the number of balls hanging from the magnet at the end of the spring, determine the spring constant t. (HINT: You do not need the mass of the magnet in your calculations if you make the correct graph and take the slope.) With the spring constant known, measure the force on the magnet when it is suspended in the field gradient. (Be sure the magnet is always centered in the Helmholtz coils.) Take many measurements with different values of the field gradient. Make the necessary plot to get the desired slope. Now, boldly proceed to determine W! Appendix I: Helmholtz Coils Helmholtz coils are a pair of current-carrying loops of wire designed to produce a magnetic field of very high homogeneity. When the separation " of the coils is equal to the

9 Experiment 11 9 In magnetic resonance imaging (MRI), the sample is usually a human being and the NMR signal comes from the protons of the water molecules in the patient. The patient is placed in a slightly non-uniform magnetic field. Since the NMR signal frequency is proportional to the -field, which varies with position in the patient, different parts of the patient s body will be producing different-frequency signals. The frequency of the signal gives information on the location in the body, and a picture of the water density in the patient can be reconstructed.

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