# Electric potential difference to accelerate electron

jegues

## Homework Statement

What electric potential difference is required to accelerate and electron, initially at rest, to the speed at which it will move in a circular path with radius 1cm when the electron enters a region of uniform magnetic field of 1.0T, if the field is perpendicular to the electron's velocity? (Ignore gravity)

## The Attempt at a Solution

We know that,

$$F_{B} = qvB = \frac{mv^{2}}{r}$$

So,

$$v = \frac{qBr}{m}$$

The part I'm confused about is how to relate this knowledge the the potential difference?

Any suggestions?

EDIT: I figured it out.

$$\Delta V = \frac{qB^{2}r^{2}}{2m}$$

cupid.callin
change in the potential energy is equal to gain in kinetic energy

qV = 0.5mv2