What Determines the Smallest Orbit Radius for a Proton in a Magnetic Field?

AI Thread Summary
The discussion focuses on determining the smallest orbit radius for a 2 MeV proton in a 2 T magnetic field using the formula r = mv/qB. To find the radius, the velocity of the proton must be calculated from its kinetic energy, which requires converting MeV to Joules. Participants note that while the formula suggests a relationship between radius and velocity, there is no definitive "smallest" radius since velocity can approach zero. The necessary parameters include the mass of the proton, its charge, and the magnetic field strength. Overall, the key to solving the problem lies in accurately calculating the proton's velocity from its kinetic energy.
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Question:

What is the radius of the smallest possible circular orbit that a 2MeV proton can have in a 2 T magnetic field?

Answer (what I've got so far, at least):

I assume that r = mv/qB is the formula I use for this problem, due to being given mass, B, q and all that, but I can't figure out how to apply it. I assume that v sould be replaced with an equation, but I cannot find that equation.

Plus, it doesn't appear that there can be a "smallest" radius because as long as v is above zero, there will be a radius, but v can get infinitely small.

Any suggestions?
 
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The 2 MeV is the kinetic energy of the proton. So, you can work out the velocity v of the proton from the kinetic energy. You also know the charge q on a proton, and you're given B. You'll need the proton mass m, too.

In calculating v, you'll need to convert the energy in MeV into an energy in Joules.
 
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