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Note: Items in red are measurements that must be made as you do the experiment.

Introduction

In 1897 J. J. Thomson was able to measure the charge to mass ratio of the electron using a cathode-ray tube. In this lab, we use a modern version of the cathode-ray tube found in an oscilloscope to do the same thing. The oscilloscope has been modified so it can deflect the electron beam with an electric field, a magnetic, or both. When both fields are used together so that the electron beam travels through the tube undeflected, the device functions as a velocity selector. The currently accepted value for e/m is 1.758820 × 1011 C/kg.

Figure 1 -J. J. Thomson with a cathode-ray tube.. http://www.udayton.edu/~cps/cps460/notes/displays/crt/Disc-of-Electron-Images.html

Our experimental device is illustrated (somewhat schematically) in Figure 2. Electrons are accelerated in an electric potential. Then the electrons pass through a region of space where we can produce an electric field with  metal plates - essentially a capacitor - or a magnetic field with an electromagnet. The deflection region, of length L, is small compared to the distance the electrons travel to the screen, S. Because of this, we can conveniently relate the deflection distance, h, to the x and y components of the electron's velocity. Eq. 5-1 We should define a few other variables and write the relations between them that you will need to do this lab.

Figure 2 - Diagram of the e/m apparatus.

Accelerating the electrons: The electrons gain energy by being accelerated through an electric potential. This potential is called the "accelerating potential" and is written as V₀. This acceleration produces the x component of the electron velocity which remains nearly unchanged through the experiment. (The magnetic field will change it slightly, but we ignore this.) Conservation of energy gives:

1) First, record the values of b, d, L, and S  which are written on the oscilloscope case, and V₀ from the voltmeter. Note that the voltmeter reads in kV.

Enter b:  T/A    (Use values such as 1.23 or 3.57E-3)

Enter d:  m

Enter L:  m

Enter S:  m

Enter V₀: V   [about 2000 V]

2) Get a feeling for how the apparatus works. Adjust V_v and I  to see how the beam deflects as you vary each field in turn.

3) We will now use the oscilloscope as a velocity selector in order to determine v_x.
  Carefully note the position of the beam when V_v = I = 0. 
  Turn up V_v until you deflect the beam 4cm. Enter your value of V_v: V  [about 50 V]
  Then turn up I until the beam is back in its original location. Enter your value of I: A  [about 0.15 A]
   Please read the yellow sheet taped to the desk to see how to read the ammeter!
  You now know that the forces from the electric and magnetic fields are equal and opposite. Equate the two forces in 5.3 and 5.4 to obtain an expression for v_x.

Enter your algebraic equation for v_x: 

Determine a numerical value for v_x: m/s

Determine e/m

4) Now you know vx, you can calculate e/m from the energy equation, Eq. 5-2. Numerical value of e/m from Eq. 5.2:  C/kg 5) Again turn Vv and I  to zero. Adjust Vv until the beam is deflected 4cm. Enter your value of Vv:  V Use Eq. 5-3 and 5-5 to find an algebraic expression for e/m. Algebraic expression for e/m = Calculated value of e/m from this expression: C/kg 6) Again turn Vv and I  to zero. Adjust I until the beam is deflected 4cm. Enter your value of I: A Use Eq. 5-4 and 5-5 to find an algebraic expression for e/m. Algebraic expression for e/m = Calculated value of e/m from this expression: C/kg Results Which method gave you the best result? Select one Energy (Eq.5-2) E field (Eq. 5-4) B field (Eq. 5-5) Enter your best value of e/m:  C/kg Using the accepted value of e/m given above, calculate the percentage error: % Comments for the grader: