NAME: Phys 3B - E&M Magnetic Fields Experiment 7: Charge to Mass Ratio Magnetic Forces In 1897, J. J. Thomson conducted an experiment that discovered the electron. Up to that point, the smallest particle was believed to be the atom. By using a cathode ray and magnetic fields, Thomson was able to find the e/ m ratio of this mysterious particle we now call the electron. You will be doing a similar experiment today. Apparatus: e/m vacuum tube, Helmholtz coils, high voltage power supply, low voltage power supply, and ammeter. The apparatus will already be set-up for this lab. It consists of a filament that will release electrons when heated. Once the high voltage is applied (~200 V), the electrons will accelerate from the filament towards the positively charged. The electrons will pass through a small opening (see Fig. 1), exiting the anode with a downward velocity. With the appropriate current through the Helmholtz coils, the magnetic field will cause the electron to undergo circular motion. Figure 1 - Front View Figure 2 - Side View Electron trajectory Filame R R Prelab Analysis 1. Use conservation of energy to find an expression of an electron's velocity once it has passed through the small opening of the anode. 4 , + K , = Us + K F LV, = 1mv 2 V =12qV m 2. In Figure 1, draw the velocity acting on the electrons on various positions of the trajectory. 3. In Figure 1, draw the magnetic forces acting on the electrons on the same positions from the previous step. 4. Using the cross-product, determine the direction of the magnetic field that will create these forces. Draw the magnetic field vectors in figure 1 and figure 2. Professor Orozco Last Revised: Spring 2019Phys 3B - E&M Experiment 7: Charge to Mass Ratio Magnetic Fro Magnetic For 5. Once the electron is ejected from the filament, it undergoes circular motion. Use Newton's laws to show that the e/m ratio is =, B272, Where V is the accelerating voltage, B is the magnetic field due to the coil, and r is the radius of the electron's trajectory. Imv 2 = qv F = Qv B sina Femy q VB sin @ = mu 2 = qV 2y V = q Br S m Bar2 6. Using the equation of the magnetic field due to a single ring of charge, derive the formula for the magnetic field at the midpoint between the two coils (see Fig. 2). You expression should be in terms of N (the number of turns in one coil), I, and R. MoIR 2 = 6 4 R Bloop -2(R2 + 22) 3/2 HoIRA B= do 2CRE 3/2 2 ( R + (1 R) B = 2N X MOIR 2 2 [ Q 2 + ( R ) 3 3/2 Measuring the e/m Ratio B = (4)3/2 MONI R Set the accelerating voltage to 180 V. Adjust the current to the Helmholtz coils until the electron beam strikes the 7-cm mark on the scale. Record the current in the table below. Continue to record the currents for different diameters. V=180 V Diameter (cm) Current (A) Trial 1 7 160 Trial 2 8 140 Trial 3 9 123 Trial 4 10 1 1 1 3.5 0.035 2 Professor Orozco Last Revised: Spring 2019. Charge to Mass Ratio Magnetic Fields Magnetic Forces Keeping the current the same value as your last trial, you will now change the accelerating voltage to change the trajectory. Observe the behavior of the electrons and enter the values of the voltage in the table below. I = ( 1) magnetic Diameter (cm) Trial 1 Voltage (V) 9 Trial 2 150 8 Analysis 125 1. Explain why you needed to increase the current to obtain trajectories with lower diameters. electrons circular trajectory is inversely proportional to the magnetic field 2. What did you observe as you kept the current constant but changed the voltage? Explain why this occurs. Calculations 1. Calculate B for every trial using your derived equation. Show one calculation and present your results in a neat table. Values of Helmholtz Coils: N = 130, R = 0. 150 B40op - 4xx 10-1X 160 1 40 1 10 -1x 130*721 0.0865 4 .4 x TO- 4 SV'S 0. 150 Bul Bwhen udtage is same loop -47210-7 8160*0- 1503 -4 79 5xx BHoop = 0. 125 T 2 (read 1502 (0150) 27 3/ 2 B, = 8 4x10 x 130 x 160 0. 125 B when current the same O . ISO = 0. 0865 T 2. Calculate e/m for each trial. Show one calculation and present your results in a neat table. Take the average of these values. e = 2cleo ) = 1. 80 x 107 e = 2 (180 ) (0 . 125 ) (0. 02532 m ( 4.468 * 10- 4 ) 2 (0. 035 ) 2 = 1.98 x 10? = 1 . 27 8 x 10 0 12 (0.0865 ) ( 4.5x 10-2, 1. 98 * 10 + 1.85 x 107 3. The accepted value for e/m is 1.76 x 1013C/kg. Calculate the % error between the = 1.93x10% accepted value and your average value. Professor Orozco Last Revised: Spring 2019