Calculating Proton Energy for a Large Cyclotron Using Earth's Magnetic Fields

[cheez]

You plan to build a large, cheap cyclotron using the Earth's magnetic fields(1x10^-4 T) and orbiting just above the Earth's atmosphere (radius 5.7 x10 ^6 m) What energy, in volts, should protons be given to just circle the earth? q= 1.6 x10^-19 C, m= 1.67x10^-27

The answer is 1.56 x 10^13
I can't get the answer.

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

please help, thanks so much!

 

[berkeman]

"The answer is 1.56 x 10^13
I can't get the answer."

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

 

[Integral]

What are the units of the given solution?

What are the units of your answer?

Are they the same?

What happens if you make them the same?

 

[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. :(

 

[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.

 

[arunbg]

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

 

[Beam me down]

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

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)?

Which would lead to an answer in Volts.