Download Ratio of Charge to Mass for the Electron

Ratio of Charge to Mass for the 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 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.
Figure 1
The magnitude of this force is:
(1)
F = qvBsin(! )
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 =
(2)
mv 2
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.
Therefore,
!KE = "!PE
or
1 2
mv = eV
2
(3)
From Equations (2) and (3),
e
2V
= 2 2
m Br
(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 are 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=
8 µ0 I
125R
(5)
where N is the number of turns per coil, I is the current in the coils, R is the radius of
the coils, and µo is the magnetic permeability of free space. (µo = 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 100 - 150 V and vary the grid potential to bring the beam into
sharper focus.
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
1.758 x 1011 C/kg.
Data
Trial
Beam Path
Plate Voltage
Field Current
Magnetic
e/m of
Radius (m)
(V)
(A)
Field (Tesla)
electron
1
2
3
e/m accepted value
= ___________________
Average e/m experimental value = ___________________
Percent error
= ___________________
Questions
1.
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?
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?