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 Thanks in advance!