Two Methods for Determining the Moment of a Magnet Inside a Cue Ball

Transcription

1 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 1 Two Methods for Determining the Moment of a Magnet Inside a Cue Ball Adam L. Fritsch and Thomas F. Pizarek Department of Physics, Wabash College, Crawfordsville, IN (Dated: April 26, 2008) By placing a magnet in the presence of an external magnetic field, one can calculate the magnet s moment through various experimental techniques. Balancing the gravitational and magnetic torques on the magnet yields a value of µ = ± A m 2 (95% CI), while a spherical harmonic oscillator approach gives the moment as µ = ± A m 2 (95% CI). These values agree in the range of µ = ± A m 2 (95% CI). While this magnet has a reported value of µ 0.4 A m 2 [1], we believe that our value is correct since the two methods are in agreement. There are many ways to describe the behavior of a magnet. One such way is to discuss its magnetic dipole moment µ. This property is useful because it can be used to find the magnetic field B of the magnet. It also describes how a magnet will react in an external magnetic field. This reaction is what we are able to use to find the moment. This experiment consists of two parts: static torque and spherical harmonic oscillation. For each method, we use a Magnetic Torque apparatus and power supply (Fig. 1). Each method gives a different way of calculating the magnetic moment µ of the magnet inside the cue ball. By using two separate methods, we hope to determine a more accurate value for µ, which we will compare to the accepted value of 0.4 A m 2 [1]. Now, we will explore each method in depth. The static torque method involves taking advantage of gravitational and magnetic torque equilibrium. We will have a light rod with an attached weight protruding from the cue ball, which contains the magnet. When a magnetic field B is applied to the cue ball, it will exert a torque on the weight because of the cue ball s magnet. Gravity will also exert a torque on the weight. If we can get the rod to stand still, we can equate the static torques due to the B field and gravity. We get τ = τmag + τ grav = 0 (1) for τ mag = µb sin θ (2) where θ is the angle between µ and B, and for τ grav = g sin θ(m w r r + m r r r ) (3) where m w is mass of the weight, g is 9.80 m/s 2, r w is the distance from the center of the ball to the weight, m r is the mass of the rod, r r is the distance from the center of the ball to the center of mass of the rod, and θ is the angle between the gravitational field and the rod and weight (Fig. 2). This gives, for θ = 90,

2 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 2 FIG. 1: This is the apparatus we will be using. On the right is the power supply. One can apply a current I of 0 to 4 amps to the coils. The analog readout of I is on the far left of the box. From left to right, we have the field direction switch (up or down), the field gradient switch (on or off), the strobe light frequency adjust knob, the strobe switch (on or off), and the air switch (on or off). The first two switches determine the direction of the magnetic field B from the coils and whether or not the B field is present. The strobe light is adjustable by frequency with the knob in the center of the board. It ranges from 0 to 10 Hz in 0.1 Hz steps. The air switch provides the cue ball-like object with an air cushion, making it float essentially frictionlessly. The object on the left is the experimental apparatus. The two coils have outer radii of 11.4 cm, inter radii of 10 cm, a separation of 12.4 cm, and 195 turns apiece. The cue ball contains a permanent magnet. It has an accepted value of µ of about 0.4 A m 2, a mass of 140 g, and a radius of 2.70 cm. The strobe light is place on top of the upper coil. It is seen on the left side of the upper coil in the photo. Also, the axis running through the center of the coils is perpendicular to the board they are mounted on. [1] µ = g m wr w m r r r. (4) B Also, for the magnetic field on axis equidistant from two Helmholtz coils [2], we get { [ B = µ 0NI R2 2 + ] R2 2 L 2 ln + L2 2 W T R1 2 + R1 2 + L2 2 [ R2 2 + ]} R2 2 L 1 ln + L2 1 R1 2 + R1 2 + L2 1 (5) where N is the number of turns in the coil, R 1 and R 2 are the inner and outer radii of the coil, I is the current passing through the wire, W is the width of the coil, T is its thickness, and L 1 and L 2 are the distances from the center of the coils to the inside and outside, respectively (Fig. A). All but I (and thus B) and r w are constants. So, for varying values of I and r w we can calculate µ. We took data for I from 2.1 A to 3.5 A in steps of 0.1 A. This resulted in enough measurements of r w to be able to plot Eq. (4) in the form

3 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 3 ~ B ~ ""C mag ~ mg FIG. 2: This figure shows how the torque due to the magnetic field and gravity are determined. Eqs. (2) and (3) give the functional form of the images. The torque s origin is located at the center of the cue ball in both cases. [2] m w gr w = µb m r gr r, (6) which allowed us to find µ as the slope (Fig. 3). For error, we used the least count of all measurements. For r w, we also considered the range over which we tried to determine if the rod and weight were actually still. The error propagation took the general form mgr mgr (J) y = x R= B (T) FIG. 3: This plot shows the line fit to the torque balancing data. The slope gives the value for µ as in Eq. (6). u(r) = for [ u(x) R ] 2 [ + u(y) R ] 2 [ u(z) R ] 2 x y z (7) R = R(x, y,..., z). (8)

4 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 4 For B, this becomes ( δb = δi ) 2 ( + δw ) 2 ( + δt ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 + δl 1 + δl 2 + δr 1 + δr 2. I W T L 1 L 2 R 1 R 2 (9) Then, just in terms of I, we have δb = T ( T A) 2 I 2. (10) For µ, this gives 1 δµ = I 4 ( ( r w)r w + I 2 ( r w ( r w ))). (11) After fitting the data to a line and determining the error on µ, this method provides a value of µ = ± A m 2 (95% CI). In the second method of the spherical harmonic oscillator, the rod is now without its weight. When a magnetic field is applied to the cue ball, it will oscillate if its magnet is not initially aligned with the B field. Alignment occurs in the +z direction, so by holding the rod off-axis in the presence of a B field and then letting go, the ball will oscillate back and forth (Fig. 4). By putting a light rod on the ball that is aligned with the ball s magnetic field, one can more easily see the oscillation. In this case, FIG. 4: This figure shows how the torque on ball s magnet is related to the angle the magnet is off of the apparatus s B field. Here, θ changes with time since the ball is oscillating due to an initial θ.[2] τ = d L dt (12) since the oscillation has a changing angular momentum. dl But, torque is also given by Eq 2. Also, dt = I dω dt = I d2 θ dt. Then, 2 µb sin θ = I d2 θ dt 2. (13)

5 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 5 Since θ is small, we have Also, ω = µb I d 2 θ dt 2 + µb I and T = 2π ω. So, µ = 1 B ( 4π 2 I T T 2 θ = 0. (14) + m r gr r ) (15) where I T is the total moment of inertia of the ball and rod, given by I T = 2 5 m br 2 b m rl 2 r + m r (d + l r /2) 2, (16) where m b is the ball s mass, r b is its radius, m r is the mass of the rod, l r is the length of the rod, and d is the distance from the end of the rod to the center of the ball when the rod is in place. In order to measure the period of the oscillation, we used a photogate attached to a laptop that was equipped with LoggerPro software. We once again took many data points, ranging from I = 2.5 A to 3.5 A. Another plot (Fig. 5) was made to find µ as the slope by fitting a line to pi^2I/T^2 4pi^2I/T^2 (T A m 2 ) y = x R= B (T) FIG. 5: This plot shows the line fit to the oscillator data. The slope gives the value for µ as in Eq. (17). 4π 2 I T T 2 = µb m r gr r. (17) Using the same error method as in the static torque technique, we find the error in µ to be δµ = B B 2 T B 2 T B 4 (18) ( ) T 2 ( I 2 ),

6 WJP X, XXXX.XX Wabash (20XX) Journal of Physics 6 where B is calculated for each I measurement. The error in T is the standard deviation for each value from LoggerPro. The rest of the error is from least count measurements as before. This error, along with the slope value from the plot, give the moment as µ = 0.450±0.011 A m 2 (95% CI). These two values of µ intersect at µ = ± A m 2 (95% CI). Since the two different methods are in agreement with each other, we are confident that our value is accurate. While the manual gives an approximate value of 0.4 A m 2, it does not give a confidence interval of any size. Thus, we hold that our calculated value is correct. [1] Teachspin, Inc., Magnetic Torque - A New Classic. [2] E. R. Andrew, I. Roberts, and R. C. Gupta, J. Sci. Instrum. 43, 936 (1966). [3] Brown and Caraher, Advanced Lab Manual, 2003.