Magnetic Field and potential difference

AI Thread Summary
Electrons are accelerated through a potential difference of 351 V and move in a circular path of radius 0.743 cm within a perpendicular magnetic field. The magnetic field strength was calculated to be 8.50 mT, and the current through the coils was found to be 6.15 A. For a proton to follow the same circular path, it would require a voltage of 0.190 V, which was initially deemed incorrect due to significant figure errors. The discussion highlighted the importance of considering the charge sign difference between protons and electrons when calculating voltage. Overall, careful attention to significant figures and charge polarity is crucial in these calculations.
strawberrysk8
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Homework Statement



Electrons are accelerated from rest through a potential difference of 351 V. Then these electrons enter a uniform magnetic field that is perpendicular to the their initial directon. Due to the interaction with the magnetic field, they move in a circular path of radius 0.743 cm.
a) What is the magnitude of the magnetic field?

The magnetic field was generated by a Helmhotz coil pair. You measure the coils and determine that the average diameter is 169 mm.

b) What is the magnitude of the current through the coils?

Assume a proton has a mass 1849 times greater than the electron. Its electric charge is the same magnitude as that of the electron, but it is positive instead of negative.

c) What voltage would the proton have to be accelerated through for it to move in the same circle as the electron in parts a and b?

Homework Equations



e/m = 2V/[(B^2)(r^2)] = [125(V)(R^2)] / [32(N^2)(u^2)(I^2)(r^2)]

N = 130
u = 4(3.14)(10^-7)
e = 1.6*10^-19C
m = 9.11*10^-31kg

The Attempt at a Solution



a) V = [e(B^2)(r^2)] / [2m] = 8.50mT

b) I = [125(m)(V)(R^2)] / [32(e)(N^2)(u^2)(r^2)]^(1/2) = 6.15A

c) V = [e(B^2)(r^2)] / [2(m)(1849)] = 0.190V

but the 0.190V part is wrong. why?
 
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If you are entering your answers online, you need to be careful with your significant figures. Using your answers from (a) and (b) I have a voltage of 0.189V to 3.s.f.
 
that still turned out to be wrong. hmm...what else could be the problem? can voltage be negative?
 
strawberrysk8 said:
that still turned out to be wrong. hmm...what else could be the problem? can voltage be negative?
That would make sense, since the proton and electrons have opposing signs.
 
thx so much!
 
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