Curvature of path made by electron in electromagnetism

[Mspike6]

I got this question on an Assigment, and ..man, am having really hard time with it

Here is how it start :

"A student used the apparatus shown below to measure the radius of the curvature of the path of electrons as they pass through a magnetic field that is perpendicular to their path. This experimental design has the voltage as the manipulated variable, the speed calculated from the voltage, and the radius as the responding variable.

Accelerating Potential Difference (V) |Speed (106 m/s)| Radius (10-2 m)
20.0 | 2.65 | 7.2
40.0 | 3.75 | 9.1
60.0 | 4.59 | 11.0
80.0 | 5.30 | 12.8
100.0 | 5.93 | 14.1
120.0 | 6.49 | 16.3A. Plot the graph of radius as a function of speed, and construct a best-fit line.

Solution:
I plotted the graph and it came out as a linear graph .

B. Using the slope or other appropriate averaging technique, determine the strength of the magnetic field.

solution

Slope = Rise/Run = R/v
Slop = (16.3-7.2)/(6.79-2.65)=2.20 <== i picked the first point and the last point

(Mv²)/r=qvB
(Mv)/r=qB
M(1/2.20)/q=B

B= (9.10*10^-31)(1/2.20)(1/-1.60*10^-19)Am pretty sure that i making something wrong here, but i can't figure it out

C. Derive the equation that would allow you to calculate the speed of the electrons from the accelerating potential.

Solution

qV=1/2 mv²

v²= (2qV)/m

v= [tex]\sqrt{\frac{2qV}{m}}[/tex]

EDIT:
i meant to ask if what i did is right or wrong !

Thank you, any help is really appreciated

 

[chaoseverlasting]

It looks right to me.

 

[kerimek]

.
It seems that you have the right idea, but there are a few errors in your calculations. Let's go through each part and correct them.

A. Plotting the graph:
Yes, the graph should come out as a linear graph. This is because the radius is directly proportional to the speed of the electrons.

B. Determining the strength of the magnetic field:
You are correct in using the slope of the best-fit line to determine the strength of the magnetic field. However, the slope should be calculated using the values for radius and speed, not the values for radius and voltage. Also, the unit for the slope should be in meters per second, since that is the unit for speed. So the correct calculation would be:
Slope = (16.3-7.2)/(5.93-2.65) = 2.73 m/s

Now, to calculate the strength of the magnetic field, we can use the equation you provided:
B = (m*v)/qr
But we need to rearrange it to solve for B:
B = (m*v)/(q*r)
Since we don't have the value for q, we can use the relationship between charge and voltage:
q = V/e
Substituting this value into the equation, we get:
B = (m*v)/(V/r*e)
Now we can plug in the values:
B = (9.10*10^-31 kg * 2.73 m/s)/((5.93 V/10^-2 m) * 1.60*10^-19 C)
B = 3.02*10^-5 T

C. Deriving the equation to calculate speed:
Yes, you are correct in using the equation v = √(2qV/m) to calculate the speed of the electrons. Just remember to use the correct values for q and m, and also to convert the units to match the ones in the equation (meters and kilograms).

Overall, your approach was correct, but there were some errors in your calculations. I hope this helps clarify things for you. If you have any further questions, please don't hesitate to ask. Good luck with your assignment!