User:Johnny Joe Gonzalez/Notebook/Physics 307L/2009/10/28

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E/M

SJK 18:49, 15 November 2009 (EST)
18:49, 15 November 2009 (EST)Excellent primary notebook, and looks like good data.  It looks like you ran out of steam on the analysis, though, in particular, missing any analysis of the slopes for constant V and I.
18:49, 15 November 2009 (EST)
Excellent primary notebook, and looks like good data. It looks like you ran out of steam on the analysis, though, in particular, missing any analysis of the slopes for constant V and I.

Safety

  • Check for exposed wires or loose connections as voltages as high as 5kV can be used.
  • Some of the corners on the power supplies are sharp.
  • The electron gun has several components made of glass.

Materials

  • Uchida e/m Experimental Apparatus Model TG-13
  • Two BK Precision Digital Multimeters, Model 2831B
  • Gelman Instrument Company Deluxe Regulated Power Supply
  • SOAR Corporation DC Power Supply Model 7403
  • HP DC Power Supply Model 6384A
This is a picture of the em device, the large circles surrounding the bulb are the Helmholtz coils, the bulb inside contains some helium.
This is a picture of the em device, the large circles surrounding the bulb are the Helmholtz coils, the bulb inside contains some helium.

Procedure

  • First connect all of the equipment, then turn it all on.
  • The heater for the e/m device will have a voltage reading of 6.3V (as recommended to us by Pranav).SJK 18:31, 15 November 2009 (EST)
    18:31, 15 November 2009 (EST)What does this mean, "the e/m device will have a voltage reading..."  I assume you mean the heater voltage, which actually should be at 6.3V?
    18:31, 15 November 2009 (EST)
    What does this mean, "the e/m device will have a voltage reading..." I assume you mean the heater voltage, which actually should be at 6.3V?
  • The two voltmeters should be set to measure amperes.
  • The E/M apparatus should be set to E/M, this switch is found left of the voltmeter and deflection plate connectors.
  • Once everything is connected and turned on then we wait for about five to ten minutes for the bulb to heat up.
  • Once the bulb is heated then we turn the current adjust knob in order to bend the beam of electrons around the bulb.
  • At this point if the beam forms a helix instead of a circle then try adjusting the bulb to align it until it forms a circle.
  • At this point make a reading of the current, accelerating voltage, left and right measurements off the mirror.
    • The size of the circle can be influenced by several factors: The focus, as this can bring the beam forward or backward on the bulb, the applied current on the Helmholtz coil, and the applied voltage on the electron beam being measured.

data

When taking the data on day on we varied the voltage but kept the current across the coils the same, on day two we varied the current across the coils, while keeping the accelerating voltage the same.

View/Edit Spreadsheet

Data Analysis

Using the equation: B=\frac{\mu R^{2}NI}{(R^{2}+x^{2})^{\frac{3}{2}}} we can find the B-field, several terms are already given to us for this experiment from the lab manual: R=.15, x^2=R/2, μ = 4π * 10 − 7(the permeability of free space, and N=130(the number of coils on the Helmholtz coils), as well as x=\frac{R}{2}.

The resulting B value then is shown to be: B=7.8*10^{-4}\frac{weber}{Amp*m^{2}}*I From the data above we can then measure the E/M ratio, since we only require the V, I, and radius of the electron beam. I have calculated the different B values below.

View/Edit Spreadsheet

However, we still need to relate are measured values with E/M, we can do this by looking at the Lorentz Force: F=e(\vec{v}X\vec{B})=m\frac{\vec{v^{2}}}{r},
we can then solve for the ratio e/m: \frac{e}{m}=\frac{\vec{v}}{r\left|\vec{B} \right|}
After which we can relate the velocity to eV: \frac{1}{2}mv^{2}=eV\Rightarrow v=\sqrt{\frac{2eV}{m}}.

We can then go back to the original equation and substitute v, this gives us the following: \frac{e}{m}=\sqrt{\frac{2eV}{m}}\frac{1}{r\left|\vec{B} \right|}\Rightarrow \frac{e^{2}}{m^{2}}=\frac{2eV}{m}\frac{1}{r^{2}B^{2}}\Rightarrow \frac{e}{m}=\frac{2V}{\left(rB \right)^{2}}

Now we can use the collected data to discover what the charge-to-mass ratio is.

\frac{e}{m}=\frac{2V}{\left(r7.8*10^{-4} \right)^{2}}

When V=200: \frac{e}{m}=\frac{2*200}{\left(r7.8*10^{-4} \right)^{2}} The data chart below shows the results for e/m with constant accelerating potentials, the current is negative, however, since the negative sign only dictates which direction the current is traveling I used its absolute value for my calculations.

SJK 18:36, 15 November 2009 (EST)
18:36, 15 November 2009 (EST)If you are going to take the mean of these individual e/m ratios (as opposed to fitting the slope), then you should use the standard error of the mean as opposed to the standard deviation as you did.  The difference is a factor of sqrt(N) in the denominator as we have seen in class.
18:36, 15 November 2009 (EST)
If you are going to take the mean of these individual e/m ratios (as opposed to fitting the slope), then you should use the standard error of the mean as opposed to the standard deviation as you did. The difference is a factor of sqrt(N) in the denominator as we have seen in class.
View/Edit Spreadsheet



View/Edit Spreadsheet

Below is a plot and data for the r^2 vs V plot, click on the chart tab to see the plot

View/Edit Spreadsheet
SJK 18:41, 15 November 2009 (EST)
18:41, 15 November 2009 (EST)It looks like you missed the point of the constant current and constant voltage measurements.  Read about it again in the lab manual.  The deal is that you can find a linear relationship between certain variables when the current or voltage is constant.  Then you can fit the data to a line and convert the slope into an e/m value.  I don't see you doing this anywhere on this page?
18:41, 15 November 2009 (EST)
It looks like you missed the point of the constant current and constant voltage measurements. Read about it again in the lab manual. The deal is that you can find a linear relationship between certain variables when the current or voltage is constant. Then you can fit the data to a line and convert the slope into an e/m value. I don't see you doing this anywhere on this page?


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