PCS Lab 5

pdf

School

York University *

*We aren’t endorsed by this school

Course

125

Subject

Electrical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by DeanIbis2539

Charge to Mass Ratio Veerinder Angroya, Kyle Viernes PCS125-302 Instructor: Amin Jafari Sojahrood TA: K Abdullah

Introduction The purpose of this laboratory experiment is to determine the ratio between the charge of an electron and its mass. A charged particle traveling in a uniform magnetic field which is perpendicular to its velocity will travel in a circular path. The circular motion of a charged particle moving in a magnetic field depends on both the charge of the particle and the mass of the particle. It’s important to note that it's not dependent on the properties separately, but rather the ratio of the two. This ratio is measured by examining the motion of the particle. Theory From existing formulas and we can equivalate them and get the 1/2 𝑚𝑣 2 𝑒∆𝑉 relationship = , which nets us, 1/2 𝑚𝑣 2 𝑒∆𝑉 (1) 𝑣 = 2𝑒∆𝑉 𝑚 This formula will help us determine the speed of the particle while it's traveling in the magnetic field. In order to find the radius of its path we need to equivalate and to get the ?𝑣𝐵 𝑚𝑣 2 ? relationship, = , which nets us, ?𝑣𝐵 𝑚𝑣 2 ? (2) ? = 𝑚𝑣 ?𝐵 Combining (1) and (2) gets us, ? 2 = 2( 𝑚 𝑒 )∆𝑉 𝐵 2 Where, B H = 8µ??𝐼 𝑎 125

The acceleration of a particle traveling in a circle is given by . If r was small, it 𝑎 𝑐 = 𝑣 2 ? would increase the acceleration since r is inversely proportional to acceleration, and if r was large the acceleration would be lower for the same reason. Similarly the force of a particle traveling in a circle is given by . If r was small the force would increase since r 𝐹 𝑐 = 𝑚𝑎 𝑐 = 𝑚𝑣 2 ? is inversely proportional to force, and if r was large the force would decrease for the same reason. This was proven in our experiment as the electron path was more defined at a smaller radius at a higher current (which is represented by Coulombs/Second) or in other words a higher speed. The force was also proven this way as the spiral that went towards/away from the magnet was getting more defined as the radius was getting smaller. If the accelerating voltage was at ΔV 1 and the current was set so that the diameter was exactly 6 cm, the resulting diameter when increasing the voltage to ΔV 2 (400 V) would be around √2 x 6 or more generally r√2. This is because r is proportional to √2 x ΔV from the equation which when increasing ΔV should increase the radius by a factor of √2. ? 2 = 2( 𝑚 𝑒 )∆𝑉 𝐵 2 Materials ● e/m Tube (may or may not include in- puts for heater, and accelerating voltage). ● Set of Helmholtz coils ● High voltage power supply (for accelerating voltage) - may or may not be integrated into e/m apparatus. ● Low voltage variable power supply (for helmholtz coils) - may or may not be integrated into e/m apparatus. ● Banana cables ● Bar Magnet ● Wooden cover (to block out light)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Procedure First, the power supply was turned on and 30 seconds were awaited before the accelerating voltage was slowly increased until it reached ΔV 1 (250 Volts). Next, a bar magnet was brought close to the electron beam. The resulting beam had broken and started to spiral away from the magnet due to it being the negative side. When the positive side brought closer the electron beam was broken and started spiraling towards the magnet. The higher the amperage was, the more defined the spiral was. After, the current was slowly increased and the electrons were bending into a circular path inside the tube. The current was then adjusted so that the circular path of the electrons had the smallest possible radius, this radius and its associated current was recorded. This was repeated from 10 more times to get unique radius and current pairs. Which were then also recorded. Once done, the knobs were turned down to the lowest setting and then powered off. Results and Calculations After proceeding through the procedure, with the radius of the Helmholtz coils being 0.15 ± 0.0075 m, the following values were obtained for radius and current: Radius (m) Current (A) 0.05 ± 0.005 1.18 ± 0.04 0.0475 ± 0.005 1.24 ± 0.03 0.045 ± 0.005 1.37 ± 0.03 0.0425 ± 0.005 1.41 ± 0.04

0.04 ± 0.005 1.57 ± 0.03 0.0375 ± 0.005 1.77 ± 0.02 0.035 ± 0.005 1.89 ± 0.02 0.0325 ± 0.005 2.03 ± 0.01 0.03 ± 0.005 2.2 ± 0.02 0.0275 ± 0.005 2.36 ± 0.02 0.025 ± 0.005 2.61 ± 0.03 The magnetic field for each current could be found using the formula , given that the number of turns in each coil (N) is 130, and the radius of 𝐵 = (8µ?𝐼)/(𝑎 125 ) the Daedalon Apparatus coils is 0.15 m. For example, the magnetic field strength for a current of 1.18 A can be found: 𝐵 = (8µ?𝐼)/(𝑎 125 ) 𝐵 = (8 * (4π * 10 −7 ) * 130 * 1. 18)/(0. 15 * 125 ) 𝐵 = 0. 0009195576 Doing this for the rest of the currents provides the following table of values:

Current (A) Magnetic Field (T) 1.18 0.0009195576 1.24 0.0009663148 1.37 0.001067622 1.41 0.0010987934 1.57 0.0012234792 1.77 0.0013793364 1.89 0.0014728508 2.03 0.0015819508 2.2 0.0017144295 2.36 0.0018391152 2.61 0.0020339368 Using the equation derived in the pre-lab, r 2 = 2 * (m/e) * ΔV/B 2 , where r was the radius of the observed beam, m/e was the mass to charge ratio, ΔV was the voltage used, and B was the magnetic field strength, the data can be plotted in a linear fashion, where 2*(m/e)*ΔV serves as the slope of the graph.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Reciprocal of Magnetic Field Strength Squared (1/T 2 ) Radius of Observed Ring Squared (m 2 ) 1182611.571 0.0025 1070934.076 0.00225625 877334.0217 0.002025 828262.3414 0.00181625 668046.694 0.0016 525605.1425 0.00140625 460980.4315 0.001225 399589.4884 0.00105625 340220.7034 0.0009 295652.8926 0.00075625 241726.9609 0.000625

Using the prior equation, and the slope (M) given (M = 2 * 10 -9 ), the charge to mass ratio can be found by isolation: M = 2*(m/e)*ΔV (e/m) = 2*ΔV/M (e/m) = 2 * 250 / (2 * 10 -9 ) (e/m) = 2.50 * 10 11 C/kg Comparing this with the accepted value for the charge mass ratio, e/m = 1.76 * 10 11 C/kg, some error was observed during the experiment. The percent error between the two values was calculated to be 42%.

With the results found, it was also able to determined whether this particle was a proton or not, as determined by J.J. Thomson the first time this experiment was performed. At his time, the only particle discovered was a proton, with mass 1.67 * 10 -27 kg, and charge of 1.602 * 10 -19 C. The charge to mass ratio of the proton was 9.59 *10 7 C/kg, which is vastly different from the observed charge to mass ratio during the experiment, 2.50 * 10 11 C/kg. Using the uncertainties for a and I, the uncertainties for magnetic field strength B can also be determined, and are displayed as δB. B (T) δB (T) 0.0009195576 0.00006 0.0009663148 0.00005 0.001067622 0.00006 0.0010987934 0.00006 0.0012234792 0.00007 0.0013793364 0.00007 0.0014728508 0.00008 0.0015819508 0.00008 0.0017144295 0.00009

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

0.0018391152 0.00009 0.0020339368 0.0001 The graph was then replotted with error bars for both 1/B 2 and r 2 : 1/B^2 (1/T^2) r^2 (m^2) δ(1/B^2) δ(r^2) 1182612 0.0025 154327.9 0.0005 1070934 0.002256 110826.6 0.000475 877334 0.002025 98611.76 0.00045 828262.3 0.001816 90455.11 0.000425 668046.7 0.0016 76443.1 0.0004 525605.1 0.001406 53347.91 0.000375 460980.4 0.001225 50077.62 0.00035 399589.5 0.001056 40414.86 0.000325 340220.7 0.0009 35720.18 0.0003 295652.9 0.000756 28936.48 0.000275 241727 0.000625 23769.37 0.00025

Discussion and Conclusion The direction of the magnetic field would be perpendicular to the path of the electrons. If the electrons were going in a counterclockwise direction the magnetic field would be pointing towards us. This is proven due to the right hand rule as the curl of the fingers represents the path of the electrons and the thumb represents the direction of the magnetic field. When putting the hypothesis to test we can see that the prediction was somewhat correct as the diameter went from 6 cm at 250 V to 8 cm at 400 V. This was an increase of about a factor 1.33 which is close to the predicted factor of √2 or 1.41. The overall experiment went well, there were no large human errors with the recording of the values and there weren't any systematic errors to throw off the predictions we had made before the experiment. The main things to take away from this experiment is that magnetic fields and electric fields go hand in hand, voltage has a multiplying factor of 1.33 when comparing to radius, and the relationship between current and radius of the electron path.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version