< User:Chad A McCoy‎ | Notebook‎ | Jr. Lab‎ | 2008‎ | 11
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## e/m ratio Lab: 10/26-11/9/2008

### Data Analysis

• For this lab we were calculating the ratio e/m for an electron using a deflection tube in helmholtz coils.

${\displaystyle B={\frac {\mu \,{R}^{2}NI}{({R}^{2}+{x}^{2})^{3/2}}}}$

• Therefore, for our helmholtz coils, such that N=130, R=.15m, and x=R/2:

${\displaystyle B={{7.8}*10}^{-4}{I}{\frac {Wb}{m^{2}}}}$

• Using this, we can calculate the magenetic field and with that, the ratio e/m.
• We know that

${\displaystyle {e}{V}={\frac {1}{2}}{m}{v}^{2}}$, and ${\displaystyle {F}_{B}={q}{v}{B}}$

• Therefore:

${\displaystyle {\frac {e}{m}}={\frac {{2}{V}}{{r}^{2}{B}^{2}}}={\frac {{2}{V}}{{r}^{2}{({{7.8}*10}^{-4}{I})}^{2}}}}$

• Using Excel, I calculated e/m and the error in the ratio, finding the value

${\displaystyle {\frac {e}{m}}=3.64(11)*10^{11}{\frac {C}{kg}}}$

SJK 15:36, 16 November 2008 (EST)
15:36, 16 November 2008 (EST)
Good work w/ analysis...you are a master of Excel array functions
• The second calculation that was requested was to take data with constant voltage and fit a line between that and r^2, using the line to calculate e/m.

• I fixed the regression line to go through zero, because with no magnetic field created by the current, there would be no deflection in the electron beam.
• For the regression, the slope was m=.0667, which led to the calculation:

${\displaystyle {\frac {e}{m}}={\frac {{2}{V}}{{7.8*10^{-4}*{0.0667}}^{2}}}=2.952*10^{11}}$

• After calculating the error I got a final value of:

${\displaystyle {\frac {e}{m}}=2.95(8)*10^{11}{\frac {C}{kg}}}$

• The third calculation that was requested was to take data with constant voltage and fit a line between that and r^2, using the line to calculate e/m.

• I fixed the regression line to go through zero, because with no accelerating voltage, the electrons would have no velocity into the magnetic field and the magnetic field is velocity dependent.
• For the regression, the slope was m=3.68*10^-6, which led to the calculation:

${\displaystyle {\frac {e}{m}}={\frac {2}{{{7.8*10^{-4}}{I}}^{2}*{3.68*10^{-6}}}}=3.971*10^{11}}$

• After calculating the error I got a final value of:

${\displaystyle {\frac {e}{m}}=3.97(10)*10^{11}{\frac {C}{kg}}}$

• With these three values and errors for the ratio e/m I averaged the ratios and summed the errors in quadrature to come up with a final value of the ratio e/m, being:

${\displaystyle {\frac {e}{m}}=3.52(17)*10^{11}{\frac {C}{kg}}}$

• Comparing this value to the known value of ${\displaystyle {\frac {e}{m}}=1.76*10^{11}{\frac {C}{kg}}}$, I was able to see that my with my calculated value, taking a 99.7% confidence interval doesn't come close to the true value.
• A 99.7% confidence interval would be:

${\displaystyle {\frac {e}{m}}=3.52(51)*10^{11}{\frac {C}{kg}}}$

• The minimum value in the interval is still more than 5 standard deviations (.85) above the true value.
• This means that the most likely reason for the measured value is systematic error that either results in a smaller radius or current, or a higher voltage than what is measured.

### Qualitative Analysis

• For qualitative analysis, we were asked to rotate the bulb and observe the spiral that forms, which is due to the aspect of the magnetic force that results in it acting only in a direction perpendicular to the velocity.
• By rotating the bulb, the velocity is no longer perpendicular to the field, and so the parallel component stays and the perpendicular component is acted upon by the field.
• Putting a voltage on the plates with the coils disconnected deflected the beam upwards, meaning that assuming it was connected in the same polarity as notated on the leads, the beam is negatively charged as it wanted to move towards the positively charged plate.
• Reversing the charge on the plates bent the electron beam down, because the negatively charged beam was attracted to the positively charged plate which was now at the bottom.

### Questions

• We see the electron beam because the beam is comprised of highly energized electrons that give off photons as they travel, going down to a lower amount of kinetic energy in doing so.
• Ignoring the Earth's magnetic field would lead to an error of approximately 4.6*10^-5 Tesla in the magnetic field we use for the measurements, which is a value approximately 1.5% that of the magnetic field generated with the coils.
• Using protons instead of electrons would result in smaller radii, because the protons have a higher mass, making their velocity be lower, also they would be bent in the opposite direction, because they are positively charged.
• A relativistic correction would not be appreciable, because the velocities of the electrons in such an apparatus have an average velocity in the range of.01c and peak velocity around .04c.