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ELECTRON BEAM IN A MAGNETIC FIELD Introduction: A charged body moving relative to a magnetic field experiences a force which is perpendicular to both the velocity of the particle and to the magnetic field. This phenomenon is exploited in modern technology in electric motors, generators, meters, sensors, cathode ray tubes (CRTs - used in visual displays such as television and computer monitors), advanced rocket motors, and a variety of both low and high power devices. Such devices move objects, ‘paint’ pictures on a screen, and weigh atoms. In this exercise, you will observe an electron beam moving with constant speed through a region of magnetic field. As a result of this laboratory, you should: 1. Know what is meant by ‘Helmholtz coils’, and be able to calculate the magnetic field (magnitude and direction) produced by such a system. 2. Experimentally verify the magnetic force (direction and magnitude) on a moving charge. 3. Be able to calculate the kinetic energy of a charged body accelerated through a known potential. 4. Measure the charge of the electron (e), an elementary particle. Theory: An arrangement of wire coils, called ‘Helmholtz coils’ will be used to provide a magnetic field, B, perpendicular to the initial velocity of electrons moving in an electric field free region. For such an arrangement (two identical coils of wire with N turns each placed parallel to each other, separated by the radius of a coil) the field produced within the volume between the coils is fairly uniform. At the geometric center of the system it has the value: B = 80NI/[r (5)3/2], Where: 0 = 4 x 10-7 T m/A, the permeability of free space, N = 320, the number of turns in one of the coils, I is the current passing through the coil in amperes, r = 0.069 m, the radius of the coil. Electrons are fired into this field perpendicular to B with a kinetic energy: mv 2/2 = eV. The theoretical force resulting is perpendicular to both v and B, and is given by: F = e v x B (the direction of v x B is given by the right hand rule, but remember that e is negative, so that F is in the opposite direction). Since F is always perpendicular to v, and since vis constant, the electron’s path is a circle. The radius of the circle must be the one for which the magnetic force provides exactly the right centripetal (pointing toward the center of the circular orbit) force to keep the orbit stable with radius R: eBv = mev2/R. Adjusting magnet current I and accelerating voltage V to values that keep R, the radius of the orbit constant, should then yield the relationship: 64 R 2 ( o ) 2 N 2 e 2 V ( ) I 250r 2 m That is, V is a linear function of I2, with slope proportional to the charge-to-mass ratio (e/m), and with zero y-intercept. In this exercise, you will try to verify these theoretical predictions: 1. That you will observe the orbit of an electron moving perpendicular to a magnetic field to be circular. 2. That the magnet current is a linear function of the square root of the electron’s accelerating voltage V, provided R is constant. 3. That the y-intercept of the V vs I2 line is zero within error. 4. That the slope of the line gives the electric charge, e. During the first session, you will record 5 different values of V and I for which R = 40 mm. During the second session, you will carry out the graphical analysis. Setup: The CRT used in this experiment (Figure 11) was made for the specific purpose of making visible the path taken by electrons. It does this with a lowpressure helium filling. When energetic electrons strike helium atoms they excite them, causing them to emit light. Free electrons are boiled off an electrically heated filament (powered by 6.3 V.A.C.). They are then accelerated in an electron gun by a voltage difference V (adjustable 0-400 volts) between the filament and grid. Those electrons which pass through the hole in the grid have acquired kinetic energy eV, where e (= 1.602x10-19 coulomb) is the charge on an electron. They form a collimated beam of electrons, all moving with approximately the same velocity: v = (2eV/me)1/2, where me is the electron mass (approximately 9.109x10-31 Kg) . Since producing light robs an electron of its energy and removes it from the beam, the beam becomes dimmer with distance. When the initial accelerating voltage is low enough, it doesn’t even reach the wall of the tube. Figure 1: Components and operation of the CRT CAUTION: BEFORE TURNING ANYTHING ON, MAKE SURE ALL ADJUSTING KNOBS ARE SET TO MINIMUM (FULL COUNTERCLOCKWISE) POSITION. a) b) Figure 2: Power supplies used for a) CRT and b) Helmholtz coils The circuit will be set up when you arrive: Sketch the apparatus in your notebook, indicating the positions of the connectors on the coils. Turn the system on and observe the electron beam: With all knobs set to minimum, turn on the ‘power’ switch of the CRT power supply and turn on the digital voltmeter. What are the uncertainty and zero of the voltmeter? You should see a glow from the filament of the tangential electron gun. NOTE: AS YOU INCREASE THE B+ VOLTAGE, ALWAYS WATCH THE AMMETER AT THE LEFT OF THE SUPPLY. NEVER ALLOW THIS TO EXCEED 25 mA.(Danger Sticker) When you see the electron beam, maximize the potential (volts) of the beam by increasing the CRT voltage until the clamp on the knob stops against the bolt, but be sure this does not exceed 25mA. Turn on the power supply to the coils, set the voltage to the coils to a maximum, and then adjust the current on the coils to get the beam to form a circle with an 80 mm diameter. (Align the beam orbit with the marks on the tube.) This is your maximum value of V for the electron gun. Once you have this, turn the B+ knob back down to zero while you complete the next few bullets. Sketch the direction of motion of the electrons, and the direction of the centripetal force. From these directions and the sign of the electronic charge, INFER the direction of the applied magnetic field, and indicate it with a labeled arrow (and/or short description next to your sketch). Estimate the accuracy (in mm) with which you think you can set the circle on the marks. This will be R, the uncertainty in the radius of the orbit R. You will need it to find the uncertainty in the predicted slope of the I vs. V1/2 curve. Now increase B+ to the maximum again. Slowly decrease it and the current across the coils to find the minimum value of V (the setting of B+ for which you cannot adjust the coil current to form an 80mm diameter electron orbit. Divide the range of accessible potential (Vmax-Vmin) into about 10 evenly spaced values. All setups are different and some will give a larger range than others – do your best. Ten intervals is not a golden rule, but the more data you have, the less a single outlier can throw off your results. For each of these values of V, find the coil current I which produces a circle with diameter 80 mm. Record V and I for each. When you have finished, return all adjustment knobs to minimum, and turn off both power supplies. Analysis Calculate the predicted value of e from the slope of your plot. Calculate the uncertainty in the predicted value of e using your estimated value of R. Assume that everything in the expression for the slope (except R) is known to a very high degree of precision. (If you’d like a ‘reminder’ of how to do this, look at the Appendix on ‘Propagation of Errors’.) Curve fitting: In this section, you will fit your data, V vs. I2, to a straight line to see if the predictions are correct. If necessary, review the appendix article on Curve Fitting. Turn on the computer and log on. Single click on the Logger Pro icon (looks like a Vernier caliper). Enter your raw data in the x and y columns, I in x, V in y. Double click on each column heading to bring up its ‘Column Options. . .’ dialog box. In this, name the data and units. Create a new calculated column containing I2. (In the ‘Data’ menu, choose ‘new calculated column’ to bring up the appropriate dialog box. Once the column exists, if you need to change its function, just double click on the column heading, and go on from there.) Be sure to set the column title and units correctly. Select columns for graphing by double clicking on the graph itself. Set the "I2" column as ‘x’ and the "V" column y as ‘y’. Now fit a straight line to the data, determine if the line is a good fit within its error, and find its slope and y-intercept and their errors to compare with the predicted values. Review the last experiment if necessary to review how to use the fitting algorithms to find the slope, y-intercept, and their errors, and the mean square error of the fit. After performing the necessary fits, state whether the fit is a good one or not. If so, this supports the original picture of the force exerted on a charge as it moves through a magnetic field. (This was the basic assumption used in the above derivation of e/m.) Using a linear comparison graph, compare your experimental vs. the accepted value of e. In your conclusion, restate and answer the questions: Is V a linear function of I2? (Is the fit a ‘good one’? How do you know?) Is the intercept of V vs. I2 equal to zero within error? Is your value of e equal to the accepted value within error?