User:Nathan Giannini/Notebook/Physics 307L/101028

Jump to: navigation, search

Equipment

SJK 22:30, 13 October 2010 (EDT)
22:30, 13 October 2010 (EDT)
This a a good start on a primary notebook. The inclusion of spreadsheet for data is good, as is the dicussion of the analysis. A little more detail on analysis methods would be good to help me know what you did, but this was complicated a little by the incorrect notation of uncertainty. I'd like more information about procedure while taking data. Linking to Gold's manual is good, but for sure it doesn't tell us everything you did. Not to mention when you varied from the stated procedure. Remember that the purpose of the primary lab notebook is to allow for you or someone else to reproduce the experiment several months later, once memories have faded.
SJK 22:13, 13 October 2010 (EDT)
22:13, 13 October 2010 (EDT)
Pictures would help!
• 6-9V Power supply rated at 2A (Gelman Deluxe Regulated Power Supply)
• Power supply of 6.0V max rated at 1.5A (Soar DC Power Supply model PS 3630)
• 150-300V power supply rated at 40mA (Hewlett Packard 6236B)
• 3 Voltmeters
• e/m Experimental Apparatus

Safety

Beware of electric shock!

Setup and Procedure

The setup and procedure can be found here under the e/m section.

My Experience

In the lab we found this a very easy setup to work with. To measure the radii, Richard overlapped the bending electron beam with its reflection and then looked at the ruler (which is built into our Helmholtz coils) to determine the radius of one side and then the other. Next, we adjusted the voltage and current to get multiple radius measurements.

Data

 View/Edit Spreadsheet

I've accidentally deleted my right radius in my google spreadsheet. The data can be most likely found in my lab partner's data section.

Equations

To calculate e/m we first determined B, the magnetic field, by the equation:

$B=\frac{\mu R^2NI}{(R^2+x^2)^{3/2}}\,\!$
$B=(7.793*10^-4weber/A*m)*I\,\!$

We used the fact that the potential energy of the electrons in the electron beam equals the kinetic energy

$eV=\frac{mv^2}{2}\,\!$

We then calculated v, the velocity of the electrons, by equating two equations:

$F_B=evB\,\!$
• $F_B\!$ is the force of the magnetic field.
$F_c=mv^2/r\,\!$
• $F_c\!$ is the centripetal force.
$F_B=evB\,\!$

Solving for v:

$v=\frac{eBr}{m}\,\!$

Substitute v into energy equation and simplifying:

$e/m=\frac{2V}{r^2*B^2}\,\!$

From the plot of r^2 verse V with constant I:

$r^2=\frac{2Vm}{(7.793*10^-4I)^2*e}\,\!$

This is a line with slope:

$slope=\frac{2m}{(7.793*10^-4I)^2*e}\,\!$

From the graph $slope=2.3(93)*10^{-5}+/-8.1(96)*10^{-6}m^2/V\,\!$

• Plugging this result into the slope equation:
$e/m=6.73(90)*10^{10}+/-3.36(80)*10^{10}\frac{C}{kg}\,\!$

From the plot of r verse 1/I with constant V

$1/r=\sqrt{\frac{(7.793*10^-4)^2*e*I}{2Vm}}\,\!$

This is a line with slope:

$1/slope=\sqrt{\frac{(7.793*10^-4)^2*e}{2Vm}}\,\!$

From the graph $slope=0.048(98)+/-0.0032(15)/m*A\,\!$

• Plugging this result into the slope equation:
$e/m=2.98(45)*10^{11}+/-1.53(94)*10^{10}\frac{C}{kg}\,\!$

The currently accepted value is:

$\frac{e}{m}=1.76\times10^{11}\frac{C}{kg}\,\!$

Qualitative Experiments

• Rotating the glass envelope causes our beam to spiral to some extent. This is easily explained in that we are changing the direction vector of our magnetic field while the position, in space of our electron beam is the same, and is, therefore, being accelerated in a different direction then before by our magnetic field.
• Reversing the charges on our coils causes our beam to point downwards.
• Turning on the deflection plates causes our beam to bend away from horizontal with an increasing angle as the voltage increases.
• Reversing our deflection plates causes our beam to point downwards and then slowly turn upwards as the voltage increases.

Answers to Questions

These are the questions found at the end of Gold's E/M experiment. Those that are answered I thought were more interesting.

1. We can see the electron beam b/c it is not a perfect vacuum inside of the bulb and, therefore, experiences diffraction SJK 22:15, 13 October 2010 (EDT)
22:15, 13 October 2010 (EDT)
Actually, it's due to collisions with low pressure helium gas that's in the tube!
, also it is possible that photons are being emitted by the electrons in all directions. These photons would be carrying part of the electrons kinetic energy it came from.
2. If we had photons instead of electrons, we would find that our magnetic field would either effect our photons to a lesser extent or to an invisible extent. Also, the beam would probably appear dimmer as we would not have any particles radiating photons in some direction.

Citations

• R. Lafler for the equipment list and equations (we used the same equipment they did)