Can you do the highlighter question 10,11,12?

Beatriz Fernandez Lab 1? \"Charge to Mass Ratio of the electron." The electron is the rst elementary particle discovered. It was discovered by J. J. Thompson in the late l9Ih century. In this experiment we will use the charge to mass ratio apparatus based on a cathode ray tube inside two Helmholtz's coils. The accepted value of the charge to mass ration is :7 = 1.759 x 1011 C/kg A current is passed through the lament of the cathode ray tube, heating up the lament causing it to emit thermal electrons, which are then accelerated across a potential difference V, before being injected into a uniform magnetic eld B, created from the superposition of the elds from each coil at the center of the separation between the coils. Materials: Properties of the em apparatus Coil Radius, a = 147.5 mm Coil Separation, d = 147.5 mm Turns per Coil, N =132 Preliminary Questions: 1. Do magnetic elds accelerate charge particles? Explain :2 Yes, the magnetic eld does accelerate the particles because it applies a force on the charged particles, and Newhm's Laws suggest that that forces cause accelerations. 2. If a particle of mass m and charge q moves with velocity v0 inside a uniform magnetic eld 30, with the direction of motion perpendicular to the eld, deduct the expression for the radius of the trajectory in terms of the given quantities. 2. The particle executes circular motion due to the magnetic force as the direction of the particle is perpendicular to the magnetic field B\". The particle of mass m and charge q has a velocity Va The centripetal force experienced by the particle is F _ HIV-F: C _ r r is the radius of the circular trajectory This force is a consequence of the magnetic force experienced by the particle. Fm = cI'r'o X B0 = quBoSing = CIVoBo 1('9 = 90} At equilibrium, the centripetal force must be balanced by the magnetic force Fr 2 F"I Invm r unBu therefore, r ulvu til-Bo The radius of trajectory is r = 1(qu an 3. Suppose that the particle in the previous question was accelerated across a potential difference Vo, use conservation of energy, and determine the expression for the velocity at the moment it exists the potential difference, in terms of the given quantities. 3. The particle has a kinetic energy in the presence of the potential Vo, the kinetic energy balances the potential energy as mv = qVo 2qVo VO =1 m Therefore the velocity at the moment particle exists in potential difference is Vo 2qVo m Procedures: 1. Turn on the e/m ratio apparatus. Please wait the 30 seconds before doing anything else! 2. Inspect the tube and find the graduated in centimeter bar (ruler), notice the labeling. Part A: Constant Voltage 3. Set the accelerating potential to 300 V. 4. Adjust the current on the coils until you see a perfect circle for the particle trajectory. 5. Now adjust the current until the diameter of the circle matches a labeled marking in the ruler. 6. Record the current and the corresponding radius in table A 7. Repeat for five more currents, record your results in table A. Part B: Constant Current 8. Set the current to a constant value, such that when the voltage changes is able to go through the range of available radii. Record that current. 9. Adjust the voltage on the coils until you see a perfect circle for the particle trajectory. 10. Now adjust the voltage until the diameter of the circle matches a labeled marking in the ruler. 1 1. Record the voltage and the corresponding radius in table B. 12. Repeat for five more currents, record your results in table B. 2Data: Table A: V = 300 V Current I Radius r (cm) Radius r (m) Magnetic field B (T) e/m (C/kg) 1.35 0.05 1.09E-03 2.04E+11 1.55 4.5 0.045 1.25E-03 1.918+11 1.75 0.04 1.41E-03 1.89E+11 1.95 3.5 0.035 1.57E-03 1.99E+11 2.15 3.25 0.0325 .73E-03 1.90E+11 2.35 0.03 .89E-03 1.87E+11 Average 1.93E+11 Standard deviation 6.618+09 The average of the e/m is calculated as (2.04 + 1.91 + 1.89 + 1.99 + 1.90 + 1.87) x 1011 Average e m Average- e m -= 1.93 x 1011 C kg The standard deviation is calculated as (2.04 - 1.93)? + (1.91 - 1.93)" + (1.89 - 1.93)? + (1.99 -1.93)? + (1.90 -1.93)? + (1.87 -1.93)2 10" Stan dard deviation = Standard deviation = 6.61 x 10" Current I Radius r (cm) Radius r (m) Magnetic field B (T) e/m (C/kg) 1.35 5 0.05 .09E-03 2.04E+11 1.55 4.5 0.045 1.25E-03 1.91E+11 1.75 4 0.04 1.41E-03 1.89E+11 1.95 3.5 0.035 1.57E-03 1.99E+11 2.15 3.25 0.0325 1.73E-03 1.90E+11 2.35 3 0.03 1.89E-03 1.87E+11 Average 1.93E+11 Standard deviation 6.61E+09 3Table B: I = 1.30 A For table B similarly, use the formula for magnetic field and e/m to complete the data and find the average and standard deviation Voltage radius (cm) radius r (m) Magnetic field (T) e/m 300 5.5 0.055 0 . 001045562222 1.818+11 280 5.25 0.0525 0 . 001045562222 1.86E+11 260 5 0.05 3. 00104556222: 1.90E+11 240 4.25 0.0425 8. 001045562222 2.43E+11 220 4.5 0.045 3. 001045562222 1.99E+11 200 4.25 0.0425 8. 001045562222 2.03E+11 Average 2.00E+11 Standard deviation 2.24E+10 Analysis: 1. Does the radius of the trajectory depend on the Current? Explain = The current increases the magnetic field strength; the radius of the trajectory will depend inversel y on themagnetic field. If the current increases, the radius will be less. 2. Does the radius depend on the trajectory on the magnetic Field? Explain = Yes, as shown by the collected data, an increase in the magnetic field resulted in a decrease of the radius, reflecting an inversely proportional relationship. 3. Does the radius depend on the accelerating voltage? Explain The radius of the electron depends on the accelerating voltage. The accelerating voltage will increase the velocity of electron and the radius is proportional to the velocity. 4. Does the radius depend on the speed of the charges? Explain = Yes, radius is dependent on the speed, they are proportional and an increase in the speed will result in an increase in the radius. Part A:5. Calculate the corresponding magnetic field using the equation for the magnetic field of the Helmholtz Coils: B = ()2 HON equation 1 6. Calculate the charge to mass ratio of the electron using the following equation: 2V m 72BZ equation 2 7. Determine the average and standard deviation for e/m: From table A, i) average e/m ratio is 1.93 x 1011 C ii) the e/m ratio obtained from the graph is 1.83 x 1011 C From table B, i) average e/m ratio is 2.00 x 1011 c kg ii) the e/m ratio obtained from the graph is 1.23 x 1011 C 8. Equation 2 could be manipulated and written as: " 2V e B2 , create a new table with the proper columns, m such that you are able to determine the e/m ratio of the electron from the slope of a line. Two new columns for _ and B2 are added to the table Current I Radius r (cm) Radius r (m) Magnetic field B (T) |aim (C/kg) 2Vir#2 (V/m*2) B*2 (T*2) 1.35 5 0.05 1.09E-03 2.04E+11 240000 1,18E-06 1.55 0.045 1.25E-03 1.916+11 296296.2963 1.75 0.04 41E-03 1.89E+11 375000 1.98E-06 1.95 3.5 0.035 1.57E-03 99E+11 489795.9184 2.46E-06 2.15 3.25 0.0325 1.73E-03 1.90E+11 568047.3373 2.99E-06 2 35 0.03 1.89E-03 1.87E+11 666666 6667 3.57E-06 UTAdd two more columns 2V and Bar and complete the table B Voltage radius (cm) radius rim) Magnetic field (7) elm 2VIV) B*2 2(T*2 m*2) 300 5.5 0.055 0. 081845562222 1.818+1 600 0.00000000330693 109 280 5.25 0.0525 0. 081845562222 1 86E+11 560 0.000000003013133493 260 0.05 8, 081845562222 1.90E+11 520 0.000000002733000901 4.25 0.0425 8. 881845562222 243E+11 190 0.000000001974593 151 220 4.5 0.045 8- 081845562222 1.89E+11 140 0.00000000221373073 200 4.25 0.0425 8. 081845562222 203E+11 403 0.000000001874593 151 The graph of _ and B' is plotted as follows to obtain the slope Plot 2V and Ber to obtain its slope e/m B42 vs. 2V/r*2 - 2I - LIE-11% . 21238 2V and B*2 r^2 - 2010 - 1.235+11 - 183 406080 1.SE-6 2 0E-6 2.SE-6 3.0E-6 3.SE-6 I DOES 2.256-9 1 90E-9 3 0CE-9 The slope obtained is 183 x 10ll the slope obtained from this graph is 1.23 x 1011 C the e/m ratio for table A is 183 x10 Therefore from the data of table B the e/m ratio is 1.23 x 1011 0 6