# Cyclotron physics problem

1. Apr 12, 2006

### cheez

The answer is 1.56 x 10^13

Fm = F centripetal Force
qvB = mv^2/r
v= rqB/m

Calculate energy

KE = 1/2 mv^2
I put in v = rqB/m

KE= 1/2 (rqB)^2/m

I plug in all the #, but still can't the answer

2. Apr 12, 2006

### Staff: Mentor

"The answer is 1.56 x 10^13

The answer is not 10^13V (not an energy) or 10^13eV (zowies!). What is the answer again?

3. Apr 12, 2006

### Integral

Staff Emeritus
What are the units of the given solution?

Are they the same?

What happens if you make them the same?

4. Apr 12, 2006

### cheez

The anwer doesn't have unit. (This is the sample test from my teacher, he doesn't type units for answer.) But in the question, it said volts. The unit I got is Newton x Seconds/ kg. Actually, I don't know how to make it into volts. I don't even know if it's the right method. :(

5. Apr 12, 2006

### cheez

Some misunstanding is here. Sorry about that. The answer my teacher gave is 1.56 x 10^13 V. I can't get the same answer as my teacher from the calculation above.

6. Apr 14, 2006

### arunbg

You probably also have to take into account the role of gravitational force in providing centripetal acceleration apart from magnetic force field.

7. Apr 15, 2006

### Beam me down

Cheez's question doesn't include the mass of the earth, so it seems it should be disregarded. From my perspective it seem the question is asking:

Through what potential difference should the protons be accelerated through for the protons to just orbit the earth (at the same height)?