Pre Lab for Ratio of Mass to. Charge of an Electron

Transcription

1 Pre Lab for Ratio of Mass to Charge of an Electron The direction of the magnetic force on a charged particle moving in the magnetic field is given by the right hand rule. Students need practice using the right hand rule. Find the direction of the magnetic forces for the following situations. 1. Charges 1, 2, 3 and 4 are positive or negative charges as indicated in the figure below. The arrow on each charge is the direction of the velocity of the charges and magnetic field B is uniform and parallel to the xaxis. Using the right hand rule, find the direction of the magnetic force acting on the charges.z y 3 1 B z 2 4 x 2. Repeat problem 1 with the magnetic field parallel to the zaxis. y 3 1 B z 2 4 x 48

2 Ratio of the Charge to the Mass for an Electron For a positive charge moving in a uniform magnetic field B with velocity v, the force F on the charge is always perpendicular to both the magnetic field and the velocity. The direction of the force is given by a right hand rule. Place your fingers in direction of the velocity and rotate them in the direction of the magnetic field. Then your thumb points in the direction of the magnetic force. The magnitude of this force is: Figure 1 F = qvbsin(θ) (1) Since the direction of the force is always perpendicular to the velocity, the force is a centripetal force and the charge will travel in circular motion as given by: qvb = mv2 r For the electron with a negative charge of e, the direction of the force is in the opposite direction. A positive charge will go in clockwise circles and a negative charge will go in counterclockwise circles. The velocity of the electron can be found by knowing that the change in kinetic energy of the electron is equal to the negative change in electric potential energy. (2) 49

3 Therefore, ΔKE = ΔPE or 1 2 mv2 = ev From Equations (2) and (3), e m = 2V B 2 r 2 (3) (4) The magnetic field which bends the beam of electrons is produced by a current in a Helmholtz coils. These coils are mounted vertically, so they produce a magnetic field horizontally. The geometry of the coils is such that at the center between the coils a uniform magnetic is produced. The magnitude of the magnetic field is given by Equation (5). B = 8µ oni 125R (5) where N ( N = 130) is the number of turns per coil, I is the current in the coils, R (R = 15 cm) is the radius of the coils, and µ o is the magnetic permeability of free space. (µ o = 4 π x 10 7 T m/a ) For our apparatus, B = (7.80 x 10 4 T/A) x (Field Coil Current). Apparatus 1. e/m Apparatus 2. High Voltage (500V/50V) power supply 3. Low Voltage, High Current power supply Procedure 1. Study the circuit shown in Figure 2. Normally the electrical connections are made by the instructor. However, you must study the circuit to know how to control the grid voltage, plate voltage, and the field current. Allow the tube to warm up for about a minute. Apply the plate voltage and observe a blue stream of electrons streaming from the right side of the electron gun. Adjust the plate potential to V and vary the grid potential to bring the beam into sharper focus. 50

4 Figure 2 2. Turn on the field coil current and adjust the variable control to the Helmholtz coil until the beam bends into a complete semicircle. The field current should be less than 3.0 A. Adjust the plate potential to vary the accelerating potential and the field current to vary the magnetic field until the beam makes a complete circle. The grid potential must be adjusted when the plate potential is changed in order to keep the beam in focus. 3. Record the plate voltage, the field current, and the radius of electron beam. 4. Repeat the observations two more times with different values of plate potential and/or field current to obtain more sets of data. 5. Calculate B from the formula B = (7.80 x 10 4 T/A) x (Field Coil Current). Use this to calculate the e/m ratio from equation (4). The accepted value for e/m is x C/kg. 51

5 Data Trial Beam Path Plate Voltage Field Current Magnetic e/m of Radius (m) (V) (A) Field (Tesla) electron e/m accepted value = Average e/m experimental value = Percent error = Questions 1. Show that B = 8µ o N 125R I = (7.8x10 4 T / A)I 2. In this experiment, the effect of the earth s magnetic field was neglected. The magnetic field on the earth surface is typically close to 5 x 10 5 T and the direction is 70 below the horizontal at Winona. How does neglecting the earth s magnetic field affect your results? What significant digit in your result does it affect? How can the experimental setup be changed so that the earth s magnetic field has no effect on the results? 52

6 2. Find the work done on an electron by the magnetic field for a semicircular path. 3. Suppose the experimental apparatus was changed to replace the beam of electrons by a beam of singly charged isotopes of the same element. How would this change the experimental results? 53