# Homework Help: Magnetic Field and potential difference

1. Oct 4, 2008

### strawberrysk8

1. The problem statement, all variables and given/known data

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?

2. Relevant 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

3. 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?

2. Oct 4, 2008

### Hootenanny

Staff Emeritus
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.

3. Oct 5, 2008

### strawberrysk8

that still turned out to be wrong. hmm...what else could be the problem? can voltage be negative?

4. Oct 5, 2008

### Hootenanny

Staff Emeritus
That would make sense, since the proton and electrons have opposing signs.

5. Oct 6, 2008

thx so much!