In his famous experiment with electrons, J.J. Thomson measured the \"charge-to-m
ID: 2144069 • Letter: I
Question
In his famous experiment with electrons, J.J. Thomson measured the "charge-to-mass ratio" r=e/m, where e is the electron's charge and m its mass. A modern classroom version of this experiment finds the ratio r by accelerating electrons through a voltage V and then bending them in a magnetic field. The ratio r=e/m is given by the formula
r=[(125)/(32*mew^2*N^2)]*[(D^2*V)/(d^2*I^2)]
In this equation mew is the permeabiliity constant of the vacuum (equal to 4*pi*10^-7 N/A^2 exactly) and N is the number of turns in the coil that produces the magnetic field; D is the diameter of the field coils, V is the voltage that accelerates the electrons, d is the diameter of the electrons curved path, and I is the current in the field coils. A student makes the following measurements.
N= 72
D=661 +/- 2 mm
V= 45.0 +/- .2 Volts
d=91.4 +/- .5 mm
I=2.48 +/- .04 amps
a.) FInd the students answer for the charge to mass ratio of the electron, with its uncertainty . [Assume all uncertainties are independent and random.
b.) How well does this answer agree with the accepted value r=1.759*10^11 c/kg?
note: it isnt necessary to understand the experiment or what is going on it is only about the uncertainty. Also the units will come out correctly when using SI units.
Explanation / Answer
first find the value
r = [(125)/(32*(4*pi*10^(-7))^2*72^2)]*[(661.0E-3^2*45)/(91.4E-3^2*2.48^2)]
=1.826E11
now with uncertainity if its square we put (2 dy/y)^2
so dr/r = sqrt( (2*dD/D)^2 + (dV/V)^2 + (dd/d)^2 + (dI/I)^2)
dr = 1.826E11*sqrt((2*2/661)^2 + (0.2/45)^2 + (0.5/91.4)^2 + (0.04/2.48)^2)=3E9
s answer is 183E9 +/- 3E9 c/kg
2) doesnt agree since outside of error.