Physics307L F07:People/Dougherty/Notebook/071008
Cary M. Dougherty 15:34, 8 October 2007 (CDT)
Setup/Prodedure
SJK 00:52, 13 November 2007 (CST)
Good description, though you can add a little more specifics about what you are actually doing (e.g., below, "same current" etc. will be meaningless in the future as you will forget what you were actually doing). You should link to the lab manual, and state that you are following it, "except with these changes..." Figure is nice, but caption could explain the parts. Also, you need to record the specific model numbers and manufacturers of the parts you are using.
The heating element connected to the anode gun is heated to the point of boiling off electrons which are then accelerated by a voltage across a circular plate. the electrons in motion then move in a "curled" path by a magnetic field caused by a current in the helmholtz coils. this magnetic field caused by the current constantly puts a force on the electrons making them turn in a circular "orbit." then the helium in the bulb causes collisions with the electrons causing a bluish radiation of visible light. we can measure the radius of the beam by a ruler behind the bulb. by measuring the voltage applied, the heat applied, and the current, along with the radius of the beam, we can come up with the e/m ratio.
same voltage, different current 1
| Accelerating Voltage | Coil Current | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg Radius (cm) +/- |
|---|---|---|---|---|
| 246 | 1.02 | 4.5 | 4.9 | 4.7 |
| 246 | 1.04 | 4.4 | 4.8 | 4.6 |
| 246 | 1.06 | 4.4 | 4.6 | 4.5 |
| 246 | 1.08 | 4.4 | 4.6 | 4.5 |
| 246 | 1.10 | 4.4 | 4.6 | 4.5 |
| 246 | 1.12 | 4.4 | 4.5 | 4.45 |
| 246 | 1.14 | 4.4 | 4.5 | 4.45 |
| 246 | 1.16 | 4.4 | 4.5 | 4.45 |
| 246 | 1.18 | 4.4 | 4.4 | 4.4 |
| 246 | 1.20 | 4.4 | 4.3 | 4.35 |
| 246 | 1.22 | 4.3 | 4.2 | 4.25 |
| 246 | 1.24 | 4.2 | 4.2 | 4.2 |
| 246 | 1.26 | 4.2 | 4.0 | 4.1 |
same voltage, different current 2
| Accelerating Voltage (V) | Coil Current (A) | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg. Radius (cm) +/- 0. |
|---|---|---|---|---|
| 246 | 1.02 | 4.5 | 4.6 | 4.55 |
| 246 | 1.04 | 4.5 | 4.6 | 4.55 |
| 246 | 1.06 | 4.5 | 4.5 | 4.5 |
| 246 | 1.08 | 4.5 | 4.5 | 4.5 |
| 246 | 1.10 | 4.5 | 4.5 | 4.5 |
| 246 | 1.12 | 4.5 | 4.5 | 4.5 |
| 246 | 1.14 | 4.5 | 4.4 | 4.45 |
| 246 | 1.16 | 4.4 | 4.4 | 4.4 |
| 246 | 1.18 | 4.4 | 4.3 | 4.35 |
| 246 | 1.20 | 4.4 | 4.3 | 4.35 |
| 246 | 1.22 | 4.3 | 4.3 | 4.3 |
| 246 | 1.24 | 4.3 | 4.2 | 4.25 |
| 246 | 1.26 | 4.2 | 4.1 | 4.15 |
change heater
Changed heater voltage to 6.0 from 6.2
| Accelerating Voltage (V) | Coil Current (A) | Left Radius (cm) +/-0.1 | Right Radius (cm) +/-0.1 | Avg. Radius (cm) +/- |
|---|---|---|---|---|
| 246 | 1.02 | 3.9 | 4.0 | 3.95 |
| 246 | 1.04 | 3.9 | 4.0 | 3.95 |
| 246 | 1.06 | 3.9 | 3.9 | 3.9 |
| 246 | 1.08 | 3.9 | 3.9 | 3.9 |
| 246 | 1.10 | 3.9 | 3.9 | 3.9 |
| 246 | 1.12 | 3.9 | 3.9 | 3.9 |
| 246 | 1.14 | 3.9 | 3.9 | 3.9 |
same current, different voltage
| Accelerating Voltage (V) | Wire Current (A) | Left Radius (cm) +/- .1 | Right Radius (cm) +/- .1 | Avg Radius | Radius SqauredAvg |
|---|---|---|---|---|---|
| 250 | 1.31 | 4.3 | 4.0 | 4.15 | 17.2225 |
| 255 | 1.31 | 4.3 | 4.1 | 4.2 | 17.64 |
| 260 | 1.31 | 4.4 | 4.1 | 4.26 | 18.0625 |
| 265 | 1.31 | 4.5 | 4.2 | 4.35 | 19.9225 |
| 270 | 1.31 | 4.5 | 4.2 | 4.35 | 19.9225 |
| 275 | 1.31 | 4.5 | 4.3 | 4.4 | 19.36 |
| 280 | 1.31 | 4.5 | 4.3 | 4.4 | 19.36 |
| 285 | 1.31 | 4.5 | 4.4 | 4.45 | 19.8025 |
| 290 | 1.31 | 4.6 | 4.4 | 4.5 | 20.25 |
| 295 | 1.31 | 4.6 | 4.5 | 4.55 | 20.7025 |
| 300 | 1.31 | 4.6 | 4.5 | 4.55 | 20.7025 |
Calculating e/m
To calculate e/m ratio we can use a sum of forces (ignoring gravity) and with Newton's 2nd law and centripetal motion we have
[math]\displaystyle{ F=ma }[/math]
[math]\displaystyle{ evB=\frac{mv^2}{r} }[/math]
solve for q/m using:
[math]\displaystyle{ qV=\frac{1}{2}mv^2 }[/math]
and solve for v and plugging in:
[math]\displaystyle{ e/m=\frac{2V}{(B*I)^2R^2} }[/math]
where B=7.8 x 10^-4 (weber/amp-meter^2)
e/m calculations
SJK 00:54, 13 November 2007 (CST)
you should write that you're using Matlab, and any other notes that would allow you to repeat this analysis in the future.
Set 1
SJK 00:56, 13 November 2007 (CST)
Especially in this case, where you have lots of data, and measurements taken in different manners, plots of the data would be very nice. Also, you can plot the data in such a way as to obtain an expected linear relation, and then do a linear fit to extract your e/m ratio.
V=[246,246,246,246,246,246,246,246,246,246,246,246,246];
I=[1.02,1.04,1.06,1.08,1.10,1.12,1.14,1.16,1.18,1.20,1.22,1.24,1.26];
r=[4.7,4.6,4.5,4.5,4.5,4.45,4.45,4.45,4.4,4.35,4.25,4.2,4.1];
B=7.8e-4.*I;
e_m=(2.*V)./((B.^2).*(r.^2))
avg=mean(e_m)
e_m =
1.0e+011 *
Columns 1 through 5
3.5187 3.5334 3.5542 3.4238 3.3004
Columns 6 through 10
3.2555 3.1423 3.0349 2.9999 2.9678
Columns 11 through 13
3.0080 2.9815 3.0302
avg =
3.2116e+011
Set 2
close all;clear all;
V=[246,246,246,246,246,246,246,246,246,246,246,246,246];
I=[1.02,1.04,1.06,1.08,1.10,1.12,1.14,1.16,1.18,1.20,1.22,1.24,1.26];
r=[4.55,4.55,4.5,4.5,4.5,4.5,4.45,4.4,4.35,4.35,4.3,4.25,4.15];
B=7.8e-4.*I;
e_m=(2.*V)./((B.^2).*(r.^2))
avg=mean(e_m)
e_m =
1.0e+011 *
Columns 1 through 5
3.7545 3.6115 3.5542 3.4238 3.3004
Columns 6 through 10
3.1836 3.1423 3.1042 3.0693 2.9678
Columns 11 through 13
2.9385 2.9118 2.9576
avg =
3.2246e+011
Set 3
close all; clear all;
V=[246,246,246,246,246,246,246];
I=[1.02,1.04,1.06,1.08,1.10,1.12,1.14];
r=[3.95,3.95,3.9,3.9,3.9,3.9,3.9];
B=7.8e-4.*I;
e_m=(2.*V)./((B.^2).*(r.^2))
avg=mean(e_m)
e_m =
1.0e+011 *
Columns 1 through 5
4.9817 4.7920 4.7319 4.5583 4.3940
Columns 6 through 7
4.2385 4.0911
avg =
4.5411e+011
Set 4
close all;clear all;
V=[250,255,260,265,270,275,280,285,290,295,300];
I=[1.31,1.31,1.31,1.31,1.31,1.31,1.31,1.31,1.31,1.31,1.31];
r=[4.15,4.2,4.26,4.35,4.35,4.4,4.4,4.45,4.5,4.55,4.55];
B=7.8e-4.*I;
e_m=(2.*V)./((B.^2).*(r.^2))
e_m =
1.0e+011 *
Columns 1 through 5
2.7806 2.7691 2.7444 2.6827 2.7333
Columns 6 through 10
2.7210 2.7705 2.7569 2.7433 2.7296
Column 11
2.7759
avg =
2.7461e+011
Standard error of the mean
Standard Error :[math]\displaystyle{ s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \overline{x})^2} }[/math]
SET 1: s=2.3052e+010
SET 2: s=2.8116e+010
SET 3: s=3.1873e+010
SET 4: s=2.9017e+009
Standard error of the mean: [math]\displaystyle{ SE = \frac{s}{\sqrt{N}} }[/math]
SET 1: SE= 6.3934e+009
SET 2: SE= 7.7978e+009
SET 3: SE= 1.2047e+010
SET 4: SE= 8.7489e+008
[math]\displaystyle{ %error= \frac{|Actual-Experimental|}{|Actual|}x100 }[/math]
SET 1: e= 82%
SET 2: e= 83%
SET 3: e= 158%
SET 4: e= 56%
Possibilities of errors
The lab manual discusses why the e/m ratio we came up with is higher than normal. it says that the voltage across the anode creates an un-uniform field which causes the [math]\displaystyle{ V_e }[/math] to slow down and also the collisions with the helium and electrons also causes [math]\displaystyle{ V_e }[/math] to slow down. and since that is directly related to [math]\displaystyle{ 1/(R^2) }[/math], then it greatly effects the radius measured.
another few problems is the width of the beam, although the manual says the best way to measure is the outside of the beam. also it is hard to line up the beam with the ruler behind the bulb because of light and the length measured (mm). another couple problems is the bulb itself. the circular bulb creates a parallax of the light viewed and can distort your measurements. and also the bulb itself is very unstable. lining it up so its center is directly over the origin of the ruler is tough and also if the table is bumped or the apparatus is bumped it can move the bulb and skew measurements.
as of the voltmeter and ampmeter. as discussed in class. they are directly hooked up and directly measured so the probability of error is surely less then 1%.
