Magnetic Field, potential difference and speed

[physgrl]

Homework Statement

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.

Homework Equations

r=mv/(qB)
F=qvB

The Attempt at a Solution

I though of relating the potential difference (3000V) with the speed using U=E=mv^2 but I don't think its right. I don't 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
Thanks in advance!

 

[I like Serena]

Homework Statement

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.

Homework Equations

r=mv/(qB)
F=qvB

The Attempt at a Solution

I though of relating the potential difference (3000V) with the speed using U=E=mv^2 but I don't think its right. I don't 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
Thanks in advance!

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.

 

[physgrl]

Thanks! :)