# Radius of Curvature of Ion in Magnetic Field

1. Feb 14, 2009

### Nexest

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

A doubly ionized atom (charge = +2e) whose mass is 5.15E-26 kg is accelerated by a voltage of 3450 V and enters a region where a uniform magnetic field B = 0.100 T acts perpendicular to its motion. What is the radius of curvature of the path of the ion in the B-field?

2. Relevant equations

r=$$\frac{mv}{qB}$$
e = 1.602 x 10-19

3. The attempt at a solution

Given the equation above this is a very simple problem except I don't know how to calculate the velocity of the ion from voltage. The only idea I had was conservation of energy: Kf = Ui or $$\frac{1}{2}$$mv2 = q$$\Delta$$V. But this ended up wrong and in retrospect I'm not sure if it entirely makes sense to begin with...

2. Feb 14, 2009

### LowlyPion

Looks like you are on the right track. The Δ Voltage will result in the ΔPE which becomes KE.

Looks to me like ½mv² gets related to the Lorentz force component and the centripetal acceleration as you've already written.

I end up with an equation that looks like:

R = 1/B*(2 * m * ΔV/ q)1/2