PCS Lab 5

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Electrical Engineering

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Apr 3, 2024

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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)

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:

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