# Homework Help: Magnetic Field, potential difference and speed

1. Feb 26, 2012

### physgrl

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

An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm. Find the speed of the electron as it enters the magnetic field.

2. Relevant equations

r=mv/(qB)
F=qvB

3. The attempt at a solution

I though of relating the potential difference (3000V) with the speed using U=E=mv^2 but I dont think its right. I dont know how else to approach the problem because in r=mv/qB I have two unknowns and in F=qvB I have three.
The answer key says: 3.25 x 107 m/s

2. Feb 26, 2012

### I like Serena

Hi physgrl!

You're on the right track.
You only need that the energy of the electron when accelerated by a potential difference V is charge of the electron times the potential difference.

3. Feb 26, 2012

Thanks!! :)