# Charge to Mass Ratio for An Electron Formal Report Lab Report

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Abstract: a summary of the experiment Theory: (must contain the following) 2 pages • • • • About energy Explain each equation clearly Explain the relation between the first 4 equations About the experiment 1 1 --- (potential energy for voltage) 𝑞∆𝑉 = 2 𝑚𝑣 2 (kinetic energy) 2 ---(Magnetic force) 𝐹𝐵 = 𝑞𝑣𝐵 3 --- 𝐹 = 4 --- 𝐵 = 𝑚𝑟 2 𝑟 𝜇0 𝑁𝐼 𝐿 = 𝜇0 𝑛𝐼 𝑞 𝑧∆𝑉 5 --- 𝑚 = 𝑟2 𝐵2 Make sure the theory is clear and simple to any one can understand Procedure: describe briefly what was done in the lab (short paragraph) Data: in a table plug r, current (I) and voltage (V). Analysis: in a table plug magnetic field (B) and charge to mass (q/m) then the average of charge to mass and find the % error by using theorical mass ratio 𝐵= 𝜇0 𝑁𝐼 𝐿 Give just one example for the calculation Dissection: answer the discussion questions in the report. There are four Conclusion: what you learned from the experiment 1 2 3 4 5 r (m) I (A) 0,00875 0,0075 0,0075 0,0055 0,0055 ∆V 3,79 4,2 4,64 5,08 4,67 B 172 218 150 150 200 0,01452 0,016091 0,017777 0,019462 0,017892 e/m 2,13E+10 2,99E+10 1,69E+10 2,62E+10 4,13E+10 L = 15.9 cm N= 97 x 5 L = 15.9 cm 1 2 3 4 5 r (m) I (A) 0,00875 0,0075 0,0075 0,0055 0,0055 ∆V 3,79 4,2 4,64 5,08 4,67 B 172 218 150 150 200 0,01452 0,016091 0,017777 0,019462 0,017892 e/m 2,13E+10 2,99E+10 1,69E+10 2,62E+10 4,13E+10 L = 15.9 cm N= 97 x 5 L = 15.9 cm Abstract: a summary of the experiment Theory: (must contain the following) 2 pages • • • • About energy Explain each equation clearly Explain the relation between the first 4 equations About the experiment 1 1 --- (potential energy for voltage) 𝑞∆𝑉 = 2 𝑚𝑣 2 (kinetic energy) 2 ---(Magnetic force) 𝐹𝐵 = 𝑞𝑣𝐵 3 --- 𝐹 = 4 --- 𝐵 = 𝑚𝑟 2 𝑟 𝜇0 𝑁𝐼 𝐿 = 𝜇0 𝑛𝐼 𝑞 𝑧∆𝑉 5 --- 𝑚 = 𝑟2 𝐵2 Make sure the theory is clear and simple to any one can understand Procedure: describe briefly what was done in the lab (short paragraph) Data: in a table plug r, current (I) and voltage (V). Analysis: in a table plug magnetic field (B) and charge to mass (q/m) then the average of charge to mass and find the % error by using theorical mass ratio 𝐵= 𝜇0 𝑁𝐼 𝐿 Give just one example for the calculation Dissection: answer the discussion questions in the report. There are four Conclusion: what you learned from the experiment
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1 Lab Report
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I just read this information from you,, could you add another 2 hours? hence I just received these useful information
Here is the report :) aIso attached plagiarism result for this ;)

Charge-to-Mass Ratio of the Electron
Name:
Partners:
Date

Abstract
Measurements were taken of the current decreasing from 11 cm to 5 cm and
increasing. Then the apparatus had to be rotated by 180 degrees and measured again for both
decreasing and increasing current. Four measurements of the current had to be measured seven
𝑇

times each. The electron charge mass ratio was identified as 5.352 ∙ 10−5 ± 4.395 ∙ 10−7 𝑀. For
the second part of the lab, the current of the coils had to be set at 1.5 A. This time the
measurements had to be measured in voltage following the same procedures as in part 1. The
electron charge to mass ratio in part 2 was 1.325 ∙ 105 ± 1880

𝑉
𝑚

.

Procedure
The red lamp was lightened and was used to read the measurements. We had to fix the green
aura was that was coming out in order it to be as straight and sharpest as possible. Then, all four
knobs on the supply unit were turned fully counterclockwise to ensure that the initial output
voltages to zero. Then, a current to the Helmholtz coils was supplied. The current was increased
until the diameter of the electrons orbit is 10cm (and other diameters as specified). Then, the
beam voltage, diameter of the electron beam, and coil current were recorded before using
Equation 5 to calculate the charge to mass ratio, e/m.
Experiment Data

Outer x-axis

Inner x-axis

Outer y-axis

Inner y-axis

Diameter

Diameter (cm)

Diameter (cm)

Diameter (cm)

(cm)

Coil

30.7cm

28.7cm

30.7cm

28.3cm

Part 1
Set voltage at 300 V

𝐼1 ↑ (A)

𝐼2 ↓ (A)

𝐼3 ↑ 180°(𝐴) 𝐼4 ↓ 180°(𝐴) 𝐼𝑎𝑣𝑒𝑟𝑎𝑔𝑒 (A)

0.11

1.21

1.20

1.22

1.21

1.21

0.10

1.30

1.33

1.35

1.35

1.33

0.09

1.48

1.51

1.49

1.50

1.50

0.08

1.67

1.70

1.67

1.69

1.67

0.07

1.89

1.94

1.91

1.93

1.92

0.06

2.22

2.22

2.23

2.24

2.23

0.05

2.65

2.65

2.66

2.66

2.66

Part 2
Set current at 1.5 A
𝑉1 ↑

𝑉2 ↓

𝑉3 ↑ 180°

𝑉4 ↓ 180°

𝑉𝑎𝑣𝑒𝑟𝑎𝑔𝑒

0.11

426

448

448

450

443

0.10

356

362

373

379

367.5

0.09

299

303

304

305

302.75

0.08

262

264

244

244

253.5

0.07

221

225

185

192

205.75

0.06

175

176

145

146

160.5

0.05

121

121

116

116

118.5

Results
Equation used to find the radius a of the coils

outer diameter + innerdiameter 30.7 + 28.7
=
= 14.85𝑐𝑚 = 0.1485 ± 5 ∙ 10−4
4
4
Equation to find slope with respect to the magnetic field strength average
𝑒
2𝑣
= 2 2
𝑚 𝑟 𝐵
𝑒
2(300𝑉)
=
= 2.095 ∙ 1011
−5
2
𝑚 (5.352 ∙ 10 )

Equation to find the slope for the voltage average

𝐵2𝑒 2
𝑉=
∙𝑟
2𝑚

𝑒 𝐵2𝑒
=
𝑚 2𝑚

𝑒 2 ∙ (1.325 ∙ 105 )
=
= 2.19 ∙ 108
𝑚
0.001209

Equation to find the slope of the magnetic field strength average

𝑒
2𝑉
= 2 2
𝑚 𝜋 𝐵

2𝑉
𝐵2 = 𝑒
⁄𝑚 𝑟 2

2𝑉 1
𝐵 = √𝑒 ∙
⁄𝑚 𝑟

𝐵=√

2(300)
1

= 3.604 ∙ 10−4
2.095 ∙ 1011 0.1485

Equation to calculate current

𝜇𝑁𝐼 4
𝐵=(
)( )
𝑎
5
𝐵𝑎 5
𝐼 = ( )( )
𝜇𝑁 4

3⁄
2

2⁄
3

2⁄
3

(3.604 ∙ 10−4 )(0.1485) 5
𝐼=(
)( )
(4𝜋 ∙ 10−7 )(132)
4

= 0.374 𝐴

Error Propagation from part 1

2

𝑒

𝜕
𝑒
𝑚

∆( ) = (
∗ ∆𝑠𝑙𝑜𝑝𝑒) =
𝑚
𝜕𝑠𝑙𝑜𝑝𝑒

𝑒

−4𝑉

𝑚

𝑠𝑙𝑜𝑝𝑒

∆ ( ) = √(

2

2

4(300V)

∗ ∆𝑠𝑙𝑜𝑝𝑒) = √((5.352∗10−5)3 ∗ 4.395 ∗ 10−7 ) = 3.44 ∙ 109
3

% Error b/t Theoretical and Experimental

Theoretical − Experimental
|
| ∗ 100
Theoretical

|

1.759∗1011 −2.095∙1011
1.759∗1011

| ∗ 100 = 19.1%

Error propagation from part 2

𝑒

2

𝑒

𝜕
𝜕
𝑒
𝑚
∆ ( ) = √(
∗ ∆𝑠𝑙𝑜𝑝𝑒) + ( 𝑚 ∗ ∆𝑎)
𝑚
𝜕𝑠𝑙𝑜𝑝𝑒
𝜕𝑎

2

2

2
𝑒
125𝑎2
𝑎 ∗ 𝑠𝑙𝑜𝑝𝑒 ∗ 125
∆ ( ) = √(
∗ ∆𝑠𝑙𝑜𝑝𝑒) + (
∗ ∆𝑎)
𝑚
32(𝜇0 𝑁 ∗ 𝐼)2
16(𝜇0 𝑁 ∗ 𝐼)2

2

2

𝑒
125(0.1485)2
0.1485 ∗ (1.325 ∗ 105 ) ∗ 125
∆ ( ) = √(

1880)
+
∗ (5 ∙ 10−4 )
(
𝑚
32((4𝜋 ∗ 10−7 )(132) ∗ 1.5)2
16((4𝜋 ∗ 107 )(132) ∗ 1.5)2

𝑒
9
∆ ( ) = 2.896 ∙ 10
𝑚

% Error b/t Theoretical and Experimental

|

𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 − 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙
| ∗ 100
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙

1.759∗1011 −2.19∙108

|

1.759∗1011

| ∗ 100 = 99.8%

Discussion
The lab began with getting the apparatus started, it needed to be left on for about 5 to 10
minutes to warm up before starting the experiment. The external and internal diameter of the
Helmholtz coils needed to be measured on numerous axes. The mean of both the external and
internal diameter needed to be identified.

The first part of the lab, began with setting the apparatus with the accelerating voltage at
300 V. A greenish light will then appear inside the tube. The beam needed to be measured
varying the diameter. First the current was measured from decreasing the diameter from 11 cm to
5 cm in 1 cm increments. Then the current also had to measure from increasing the diameter
from 5 cm to 11 cm. Once the data was gathered, the apparatus had to be rotated 180 degrees and
follow the same procedures for measuring the current. The second part of the lab, consisted of
mostly the same step. The voltage was then measured with setting the apparatus to a current of
1.5 A. The voltage would be varied, measuring the circles of the beam between 11 to 5 cm; two
voltage readings of the decreasing and increasing the diameter. The apparatus then had to be
rotated 180 degrees and measure two different reading of the diameters. In part 1 of the lab, the
3

value of 𝐵 had to be calculated using the formula𝐵 =

𝜇0 𝑁∗𝐼 4 2
(5) .
𝑎

The value of B was then plotted

1

into Graphical Analysis along with the radius inverse 𝑟 . The results of the linear slope the
𝑇

calculated charge to mass ratio part 1, which was5.352 ∙ 10−5 ± 4.395 ∙ 10−7 𝑀. In part 2 of the
lab, it was similar while plotting the graph of 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑣𝑠. 𝑟𝑎𝑑𝑖𝑢𝑠 2. The calculated charge to
mass ratio was 1.325 ∙ 105 ± 1880

𝑉
𝑚

.

Conclusion
The main objective was to calculate the charge to mass ratio. The first part of the lab
consisted of measured current with a set voltage of 300 V. Then the second part of the lab was to
measure with the voltage change at a set current of 1.5 A. The electron charge mass ratio for part

𝑇

1 was identified as 5.352 ∙ 10−5 ± 4.395 ∙ 10−7 𝑀 and for part 2 as 1.325 ∙ 105 ± 1880
experiment was an overall success.

𝑉
𝑚

. This

here is the second report with plagiarism result

1

Charge to Mass Ratio of Electron

Name
Affiliation
Date

2

Abstract
Through this examination, the electron to mass ratio was dictated by understanding the
connection between the electric potential used to quicken the electron at a specific speed, the
quality of the size field that influences the electron's movement, and the range of the roundabout
way that the electron pursues. It was additionally discovered that the charge to the mass ratio of a
molecule is regularly called the particular charge-which was resolved by means of the
investigation.
Procedure:
The current was expanded until the measurement of the electrons circle is 10cm (and different
breadths as determined). At that point, the pillar voltage, breadth of the electron beam, and curl
currently were recorded before utilizing Equation 5 to ascertain the charge to mass proportion,
e/m. The pillar voltage to were shifted and the information were gathered for every breadth: 300
V, 250 V, 200V, and 150V, and after that the e/m esteems were acquired. The red light was
helped and was utilized to peruse the estimations. We needed to fix the green atmosphere was
that was turning out a...

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Anonymous
This is great! Exactly what I wanted.

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