PHYSICS 220/230 Lab 5: Magnetic Fields and Forces

Introduction:  In a series of experiments performed just prior to the turn of the century, J. J. Thompson was able to show that electrons behaved as particles of mass m that carried a fixed amount of (negative) electrical charge e. These particles moved in trajectories governed by the laws of electricity, magnetism and classical mechanics. Thompson received the Nobel Prize for this work in 1906. (J. J. Thompson's son, G. P. Thompson, received the Nobel Prize in 1937 for experiments that showed that electrons could exhibit wave-like properties as well.) Thompson was not able to measure the charge or mass separately but only the ratio of these two quantities, e/m.  In the following exercises you will investigate magnetic fields and forces by repeating some of Thompson's experiments.

Experimental Apparatus:  The apparatus you will use is similar to the apparatus used by Thompson.  It consists of an evacuated clear glass bulb containing an electron "gun" similar to that found in a modern TV picture tube. This gun is a heated filament that emits electrons that are then accelerated toward a metal plate kept at a positive potential relative to the filament. A narrow slit in the plate allows the electrons to pass through, and they eventually strike a flat mica sheet on which there is printed a centimeter scale. Metal electrodes above and below this mica sheet allow us also to apply electric fields to the region.

Apparatus for Measurement of e/m for electron

The glass bulb is supported between two large coils of wire. The arrangement of these two coils (suggested by Helmholtz) is such that the separation between them is equal to one-half of their average diameter. A current through these coils will create a magnetic field B which is substantially uniform over the central region between the coils. The strength of this magnetic field is given by

                             

where                                       N = number of turns of wire (320)
                                                 r = mean radius of coils (0.068m)
                                                 I = current in Amperes

There are two power supplies on the table in front of you, one for the Helmholtz coils and the other for the electron tube. For simplicity we will refer to these power supplies by brand name: GW and TEL-Atomic.

Procedures: Notice that each Helmholtz coil has a terminal labeled A and one labeled Z. Connect the Z terminals to one another and the A terminals to the GW power supply (one to the (-) and one to the (+)).  The knob labeled VOLTAGE/COARSE should be rotated to its extreme clockwise position and left there.  The AMPS HI/LO button also should be pushed in (to select the LO setting).  The current through the Helmholtz coils is controlled by rotating the FINE and COURSE CURRENT knobs and an approximate measure of the current may be read from the meter on the front panel of the GW power supply.  However, the current required is so small that you will want to include the multimeter in the circuit to give you a more precise measurement of the current.  Use the 10A setting on the multimeter throughout the lab.

The filament in the electron gun should be connected to the two yellow terminals labeled 0V and 6V4A on the TEL-Atomic power supply. Also, connect the 0V terminal, on the TEL-Atomic power supply, to the terminal that carries the symbol for electrical ground. The left-most black terminal on the TEL-Atomic supply (which is the negative terminal of the high voltage power supply) should be connected to this same ground. Finally, connect the electrode that sticks out of the neck of the glass tube to the terminal labeled +5kV on the TEL-Atomic power supply. The double female banana plug connector on the exposed end of this high voltage wire will protect against electrical shock.

Exercise 1: MEASURING e/m

Once your circuit has been approved, turn on the TEL-Atomic power supply and rotate the 5kV VOLTAGE CONTROL in a clockwise direction. When the voltage exceeds about 1500 Volts, you will begin to see the blue glow of the electron beam superimposed upon the light beam from the filament. (The electron beam is easier to see in a darkened room.) 

• Once you are able to see the beam, turn on the GW power supply.  Vary the accelerating voltage as well as the magnetic field.  Describe (in your lab book!) the effect of the accelerating voltage and the magnetic field on the moving electrons.
• Try reversing the direction of the current through the Helmholtz coils.  Describe the effect of reversing the direction of the magnetic field.

Check your knowledge of the direction of the magnetic field and the direction of the force it exerts on a moving negatively charged particle:

• Draw a picture showing the magnetic field between the Helmholtz coils, the trajectory of the electrons, and the direction of the observed force on the electrons.
• What is the direction of the current in the coils which produces this field?  Add to your drawing the direction of this current.

An electron moving at speed v through a uniform magnetic field B will trace out a circular path of radius R given by

                             

The speed v is obtained by recognizing that an electron which starts from rest and accelerates through a potential difference of V volts will acquire a kinetic energy given by

                          

We can eliminate v from these equations to obtain

                              

The voltage V in equation (4) is read directly from the TEL-Atomic power supply. For a given current, the magnetic field B may be computed from equation (1). All we need is a value for R in order to be able to compute the ratio of electronic charge to electronic mass. A little analysis will show that, for circles passing through the origin (which is at the exit aperture of the electron beam) and points (x,y) or (x,-y), the radius R is given by

(Note that R, the radius of the electrons' circular path, is not to be confused with r, the mean radius of the Helmholtz coils.)  For a given high voltage V, adjust the coil current I so that the electron beam passes through the point: x = 10, y = + 2.6.  Keeping this same voltage, reverse the magnetic field and vary its strength until the beam passes through the point: x = 10, y = - 2.6.

Analysis:  Use the average value of the magnitude of these two currents to calculate the magnetic field in equation (1). Repeat this process for four more values of the accelerating voltage and compute five different values for e/m. Do one complete sample calculation of e/m by hand to check your math; then use Excel to complete your calculations for the other accelerating voltages. (See below for a suggested spreadsheet to organize your data.)  Report your average e/m measurement and compare it to the accepted value of 1.7588 x 10⁺¹¹ Coulomb/kg.  Because of various misalignments during the manufacture of the tube, your value will probably differ from this.  As a consistency check (on the apparatus as well as on yourself), use the accepted values for the charge (1.60219 x10 ^-19 Coulomb) and mass (9.10953 x 10 ^-31 kg) to compute values for the velocity in equation (2) and equation (3).

Exercise 2: Crossed Electric and Magnetic Fields

Notice that the metal plates above and below the mica sheet are connected to electrodes on the top and bottom of the tube. If the lower electrode is connected to the electrode at the neck of the tube while the upper electrode is connected to that filament supply lead which is grounded, an electric field will be created which will deflect the electron beam downward along a parabolic path. Turn off the power to the Helmholtz coils to observe this effect.

A combination of electric and magnetic fields applied at right angles forms what is known as a velocity selector and is a common occurrence in mass spectrometers and particle accelerators. The physical principle is quite simple: an electric field E exerts a force qE on a charged particle while a magnetic field exerts a force qvB on the same particle. When these two forces are equal in magnitude but oppositely directed, the particle will move in a straight line. Its velocity will be given by v = E/B.

You can approximate this condition over a narrow portion of the electron beam.  To do this, adjust the potential difference applied to the electrodes above and below the beam and the magnetic field until you see the beam move in a straight line across the middle portion of the tube (where the magnetic field is most nearly uniform).  

Analysis:  See below for a table format that will help organize your results. Again, complete a sample calculation of v and then use Excel to complete your other calculations.  For parallel plates of infinite extent, the electric field is given by V/d, where d is the plate separation. The velocity selector selects particles whose velocity is given by v = E/B.  Compare this velocity with the velocity v' found in Exercise 1.

Discussion: 

• Draw a picture showing the magnetic and electric fields between the Helmholtz coils, the trajectory of the electrons, and the direction of the observed forces on the electrons.
• Does the equation E = V/d give a very accurate value for the electric field between the upper and lower electrodes?  Why or why not?
• Did the electric and magnetic forces cancel along the entire electron trajectory?  Explain.
• The magnetic field of the earth is roughly 0.5 gauss.  What effect might this field have on your results for both exercises?  Be as specific as you can.

Set up the following spreadsheets using Excel.  Remember to adjust the number format of your cells to ensure that a reasonable number of significant digits is reported.

MAGNETIC FIELD ONLY

Accepted value is: e/m = 1.7588 x 10⁺¹¹ C/kg

CROSSED ELECTRIC AND MAGNETIC FIELDS