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Mspike6

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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.

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

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

Thank you, any help is really appreciated

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.

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

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

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

(Mv^{2})/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.

EDIT:Solution

qV=1/2 mv^{2}

v^{2}= (2qV)/m

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

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

Thank you, any help is really appreciated

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